2015-02-21 04:30:47 +08:00
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//===-- LowerBitSets.cpp - Bitset lowering pass ---------------------------===//
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//
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// The LLVM Compiler Infrastructure
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//
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// This file is distributed under the University of Illinois Open Source
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// License. See LICENSE.TXT for details.
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//
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//===----------------------------------------------------------------------===//
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//
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// This pass lowers bitset metadata and calls to the llvm.bitset.test intrinsic.
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// See http://llvm.org/docs/LangRef.html#bitsets for more information.
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//
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//===----------------------------------------------------------------------===//
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#include "llvm/Transforms/IPO/LowerBitSets.h"
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#include "llvm/Transforms/IPO.h"
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#include "llvm/ADT/EquivalenceClasses.h"
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#include "llvm/ADT/Statistic.h"
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2015-03-14 08:00:49 +08:00
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#include "llvm/ADT/Triple.h"
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2015-02-21 04:30:47 +08:00
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#include "llvm/IR/Constant.h"
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#include "llvm/IR/Constants.h"
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#include "llvm/IR/GlobalVariable.h"
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#include "llvm/IR/IRBuilder.h"
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#include "llvm/IR/Instructions.h"
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#include "llvm/IR/Intrinsics.h"
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#include "llvm/IR/Module.h"
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#include "llvm/IR/Operator.h"
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#include "llvm/Pass.h"
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#include "llvm/Transforms/Utils/BasicBlockUtils.h"
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using namespace llvm;
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#define DEBUG_TYPE "lowerbitsets"
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LowerBitSets: Use byte arrays instead of bit sets to represent in-memory bit sets.
By loading from indexed offsets into a byte array and applying a mask, a
program can test bits from the bit set with a relatively short instruction
sequence. For example, suppose we have 15 bit sets to lay out:
A (16 bits), B (15 bits), C (14 bits), D (13 bits), E (12 bits),
F (11 bits), G (10 bits), H (9 bits), I (7 bits), J (6 bits), K (5 bits),
L (4 bits), M (3 bits), N (2 bits), O (1 bit)
These bits can be laid out in a 16-byte array like this:
Byte Offset
0123456789ABCDEF
Bit
7 HHHHHHHHHIIIIIII
6 GGGGGGGGGGJJJJJJ
5 FFFFFFFFFFFKKKKK
4 EEEEEEEEEEEELLLL
3 DDDDDDDDDDDDDMMM
2 CCCCCCCCCCCCCCNN
1 BBBBBBBBBBBBBBBO
0 AAAAAAAAAAAAAAAA
For example, to test bit X of A, we evaluate ((bits[X] & 1) != 0), or to
test bit X of I, we evaluate ((bits[9 + X] & 0x80) != 0). This can be done
in 1-2 machine instructions on x86, or 4-6 instructions on ARM.
This uses the LPT multiprocessor scheduling algorithm to lay out the bits
efficiently.
Saves ~450KB of instructions in a recent build of Chromium.
Differential Revision: http://reviews.llvm.org/D7954
llvm-svn: 231043
2015-03-03 08:49:28 +08:00
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STATISTIC(ByteArraySizeBits, "Byte array size in bits");
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STATISTIC(ByteArraySizeBytes, "Byte array size in bytes");
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STATISTIC(NumByteArraysCreated, "Number of byte arrays created");
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2015-02-21 04:30:47 +08:00
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STATISTIC(NumBitSetCallsLowered, "Number of bitset calls lowered");
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STATISTIC(NumBitSetDisjointSets, "Number of disjoint sets of bitsets");
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2015-03-20 06:02:10 +08:00
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static cl::opt<bool> AvoidReuse(
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"lowerbitsets-avoid-reuse",
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cl::desc("Try to avoid reuse of byte array addresses using aliases"),
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cl::Hidden, cl::init(true));
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2015-02-21 04:30:47 +08:00
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bool BitSetInfo::containsGlobalOffset(uint64_t Offset) const {
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if (Offset < ByteOffset)
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return false;
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if ((Offset - ByteOffset) % (uint64_t(1) << AlignLog2) != 0)
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return false;
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uint64_t BitOffset = (Offset - ByteOffset) >> AlignLog2;
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if (BitOffset >= BitSize)
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return false;
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LowerBitSets: Use byte arrays instead of bit sets to represent in-memory bit sets.
By loading from indexed offsets into a byte array and applying a mask, a
program can test bits from the bit set with a relatively short instruction
sequence. For example, suppose we have 15 bit sets to lay out:
A (16 bits), B (15 bits), C (14 bits), D (13 bits), E (12 bits),
F (11 bits), G (10 bits), H (9 bits), I (7 bits), J (6 bits), K (5 bits),
L (4 bits), M (3 bits), N (2 bits), O (1 bit)
These bits can be laid out in a 16-byte array like this:
Byte Offset
0123456789ABCDEF
Bit
7 HHHHHHHHHIIIIIII
6 GGGGGGGGGGJJJJJJ
5 FFFFFFFFFFFKKKKK
4 EEEEEEEEEEEELLLL
3 DDDDDDDDDDDDDMMM
2 CCCCCCCCCCCCCCNN
1 BBBBBBBBBBBBBBBO
0 AAAAAAAAAAAAAAAA
For example, to test bit X of A, we evaluate ((bits[X] & 1) != 0), or to
test bit X of I, we evaluate ((bits[9 + X] & 0x80) != 0). This can be done
in 1-2 machine instructions on x86, or 4-6 instructions on ARM.
This uses the LPT multiprocessor scheduling algorithm to lay out the bits
efficiently.
Saves ~450KB of instructions in a recent build of Chromium.
Differential Revision: http://reviews.llvm.org/D7954
llvm-svn: 231043
2015-03-03 08:49:28 +08:00
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return Bits.count(BitOffset);
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2015-02-21 04:30:47 +08:00
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}
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bool BitSetInfo::containsValue(
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2015-03-10 10:37:25 +08:00
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const DataLayout &DL,
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2015-02-21 04:30:47 +08:00
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const DenseMap<GlobalVariable *, uint64_t> &GlobalLayout, Value *V,
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uint64_t COffset) const {
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if (auto GV = dyn_cast<GlobalVariable>(V)) {
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auto I = GlobalLayout.find(GV);
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if (I == GlobalLayout.end())
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return false;
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return containsGlobalOffset(I->second + COffset);
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}
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if (auto GEP = dyn_cast<GEPOperator>(V)) {
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2015-03-10 10:37:25 +08:00
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APInt APOffset(DL.getPointerSizeInBits(0), 0);
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bool Result = GEP->accumulateConstantOffset(DL, APOffset);
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2015-02-21 04:30:47 +08:00
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if (!Result)
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return false;
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COffset += APOffset.getZExtValue();
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return containsValue(DL, GlobalLayout, GEP->getPointerOperand(),
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COffset);
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}
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if (auto Op = dyn_cast<Operator>(V)) {
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if (Op->getOpcode() == Instruction::BitCast)
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return containsValue(DL, GlobalLayout, Op->getOperand(0), COffset);
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if (Op->getOpcode() == Instruction::Select)
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return containsValue(DL, GlobalLayout, Op->getOperand(1), COffset) &&
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containsValue(DL, GlobalLayout, Op->getOperand(2), COffset);
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}
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return false;
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}
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BitSetInfo BitSetBuilder::build() {
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if (Min > Max)
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Min = 0;
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// Normalize each offset against the minimum observed offset, and compute
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// the bitwise OR of each of the offsets. The number of trailing zeros
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// in the mask gives us the log2 of the alignment of all offsets, which
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// allows us to compress the bitset by only storing one bit per aligned
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// address.
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uint64_t Mask = 0;
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for (uint64_t &Offset : Offsets) {
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Offset -= Min;
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Mask |= Offset;
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}
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BitSetInfo BSI;
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BSI.ByteOffset = Min;
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BSI.AlignLog2 = 0;
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if (Mask != 0)
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BSI.AlignLog2 = countTrailingZeros(Mask, ZB_Undefined);
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// Build the compressed bitset while normalizing the offsets against the
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// computed alignment.
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BSI.BitSize = ((Max - Min) >> BSI.AlignLog2) + 1;
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for (uint64_t Offset : Offsets) {
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Offset >>= BSI.AlignLog2;
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LowerBitSets: Use byte arrays instead of bit sets to represent in-memory bit sets.
By loading from indexed offsets into a byte array and applying a mask, a
program can test bits from the bit set with a relatively short instruction
sequence. For example, suppose we have 15 bit sets to lay out:
A (16 bits), B (15 bits), C (14 bits), D (13 bits), E (12 bits),
F (11 bits), G (10 bits), H (9 bits), I (7 bits), J (6 bits), K (5 bits),
L (4 bits), M (3 bits), N (2 bits), O (1 bit)
These bits can be laid out in a 16-byte array like this:
Byte Offset
0123456789ABCDEF
Bit
7 HHHHHHHHHIIIIIII
6 GGGGGGGGGGJJJJJJ
5 FFFFFFFFFFFKKKKK
4 EEEEEEEEEEEELLLL
3 DDDDDDDDDDDDDMMM
2 CCCCCCCCCCCCCCNN
1 BBBBBBBBBBBBBBBO
0 AAAAAAAAAAAAAAAA
For example, to test bit X of A, we evaluate ((bits[X] & 1) != 0), or to
test bit X of I, we evaluate ((bits[9 + X] & 0x80) != 0). This can be done
in 1-2 machine instructions on x86, or 4-6 instructions on ARM.
This uses the LPT multiprocessor scheduling algorithm to lay out the bits
efficiently.
Saves ~450KB of instructions in a recent build of Chromium.
Differential Revision: http://reviews.llvm.org/D7954
llvm-svn: 231043
2015-03-03 08:49:28 +08:00
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BSI.Bits.insert(Offset);
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2015-02-21 04:30:47 +08:00
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}
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return BSI;
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}
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2015-02-25 07:17:02 +08:00
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void GlobalLayoutBuilder::addFragment(const std::set<uint64_t> &F) {
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// Create a new fragment to hold the layout for F.
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Fragments.emplace_back();
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std::vector<uint64_t> &Fragment = Fragments.back();
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uint64_t FragmentIndex = Fragments.size() - 1;
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for (auto ObjIndex : F) {
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uint64_t OldFragmentIndex = FragmentMap[ObjIndex];
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if (OldFragmentIndex == 0) {
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// We haven't seen this object index before, so just add it to the current
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// fragment.
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Fragment.push_back(ObjIndex);
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} else {
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// This index belongs to an existing fragment. Copy the elements of the
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// old fragment into this one and clear the old fragment. We don't update
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// the fragment map just yet, this ensures that any further references to
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// indices from the old fragment in this fragment do not insert any more
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// indices.
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std::vector<uint64_t> &OldFragment = Fragments[OldFragmentIndex];
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Fragment.insert(Fragment.end(), OldFragment.begin(), OldFragment.end());
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OldFragment.clear();
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}
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}
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// Update the fragment map to point our object indices to this fragment.
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for (uint64_t ObjIndex : Fragment)
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FragmentMap[ObjIndex] = FragmentIndex;
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}
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LowerBitSets: Use byte arrays instead of bit sets to represent in-memory bit sets.
By loading from indexed offsets into a byte array and applying a mask, a
program can test bits from the bit set with a relatively short instruction
sequence. For example, suppose we have 15 bit sets to lay out:
A (16 bits), B (15 bits), C (14 bits), D (13 bits), E (12 bits),
F (11 bits), G (10 bits), H (9 bits), I (7 bits), J (6 bits), K (5 bits),
L (4 bits), M (3 bits), N (2 bits), O (1 bit)
These bits can be laid out in a 16-byte array like this:
Byte Offset
0123456789ABCDEF
Bit
7 HHHHHHHHHIIIIIII
6 GGGGGGGGGGJJJJJJ
5 FFFFFFFFFFFKKKKK
4 EEEEEEEEEEEELLLL
3 DDDDDDDDDDDDDMMM
2 CCCCCCCCCCCCCCNN
1 BBBBBBBBBBBBBBBO
0 AAAAAAAAAAAAAAAA
For example, to test bit X of A, we evaluate ((bits[X] & 1) != 0), or to
test bit X of I, we evaluate ((bits[9 + X] & 0x80) != 0). This can be done
in 1-2 machine instructions on x86, or 4-6 instructions on ARM.
This uses the LPT multiprocessor scheduling algorithm to lay out the bits
efficiently.
Saves ~450KB of instructions in a recent build of Chromium.
Differential Revision: http://reviews.llvm.org/D7954
llvm-svn: 231043
2015-03-03 08:49:28 +08:00
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void ByteArrayBuilder::allocate(const std::set<uint64_t> &Bits,
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uint64_t BitSize, uint64_t &AllocByteOffset,
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uint8_t &AllocMask) {
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// Find the smallest current allocation.
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unsigned Bit = 0;
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for (unsigned I = 1; I != BitsPerByte; ++I)
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if (BitAllocs[I] < BitAllocs[Bit])
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Bit = I;
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AllocByteOffset = BitAllocs[Bit];
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// Add our size to it.
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unsigned ReqSize = AllocByteOffset + BitSize;
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BitAllocs[Bit] = ReqSize;
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if (Bytes.size() < ReqSize)
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Bytes.resize(ReqSize);
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// Set our bits.
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AllocMask = 1 << Bit;
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for (uint64_t B : Bits)
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Bytes[AllocByteOffset + B] |= AllocMask;
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}
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2015-02-21 04:30:47 +08:00
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namespace {
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LowerBitSets: Use byte arrays instead of bit sets to represent in-memory bit sets.
By loading from indexed offsets into a byte array and applying a mask, a
program can test bits from the bit set with a relatively short instruction
sequence. For example, suppose we have 15 bit sets to lay out:
A (16 bits), B (15 bits), C (14 bits), D (13 bits), E (12 bits),
F (11 bits), G (10 bits), H (9 bits), I (7 bits), J (6 bits), K (5 bits),
L (4 bits), M (3 bits), N (2 bits), O (1 bit)
These bits can be laid out in a 16-byte array like this:
Byte Offset
0123456789ABCDEF
Bit
7 HHHHHHHHHIIIIIII
6 GGGGGGGGGGJJJJJJ
5 FFFFFFFFFFFKKKKK
4 EEEEEEEEEEEELLLL
3 DDDDDDDDDDDDDMMM
2 CCCCCCCCCCCCCCNN
1 BBBBBBBBBBBBBBBO
0 AAAAAAAAAAAAAAAA
For example, to test bit X of A, we evaluate ((bits[X] & 1) != 0), or to
test bit X of I, we evaluate ((bits[9 + X] & 0x80) != 0). This can be done
in 1-2 machine instructions on x86, or 4-6 instructions on ARM.
This uses the LPT multiprocessor scheduling algorithm to lay out the bits
efficiently.
Saves ~450KB of instructions in a recent build of Chromium.
Differential Revision: http://reviews.llvm.org/D7954
llvm-svn: 231043
2015-03-03 08:49:28 +08:00
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struct ByteArrayInfo {
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std::set<uint64_t> Bits;
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uint64_t BitSize;
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GlobalVariable *ByteArray;
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Constant *Mask;
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};
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2015-02-21 04:30:47 +08:00
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struct LowerBitSets : public ModulePass {
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static char ID;
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LowerBitSets() : ModulePass(ID) {
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initializeLowerBitSetsPass(*PassRegistry::getPassRegistry());
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}
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LowerBitSets: Use byte arrays instead of bit sets to represent in-memory bit sets.
By loading from indexed offsets into a byte array and applying a mask, a
program can test bits from the bit set with a relatively short instruction
sequence. For example, suppose we have 15 bit sets to lay out:
A (16 bits), B (15 bits), C (14 bits), D (13 bits), E (12 bits),
F (11 bits), G (10 bits), H (9 bits), I (7 bits), J (6 bits), K (5 bits),
L (4 bits), M (3 bits), N (2 bits), O (1 bit)
These bits can be laid out in a 16-byte array like this:
Byte Offset
0123456789ABCDEF
Bit
7 HHHHHHHHHIIIIIII
6 GGGGGGGGGGJJJJJJ
5 FFFFFFFFFFFKKKKK
4 EEEEEEEEEEEELLLL
3 DDDDDDDDDDDDDMMM
2 CCCCCCCCCCCCCCNN
1 BBBBBBBBBBBBBBBO
0 AAAAAAAAAAAAAAAA
For example, to test bit X of A, we evaluate ((bits[X] & 1) != 0), or to
test bit X of I, we evaluate ((bits[9 + X] & 0x80) != 0). This can be done
in 1-2 machine instructions on x86, or 4-6 instructions on ARM.
This uses the LPT multiprocessor scheduling algorithm to lay out the bits
efficiently.
Saves ~450KB of instructions in a recent build of Chromium.
Differential Revision: http://reviews.llvm.org/D7954
llvm-svn: 231043
2015-03-03 08:49:28 +08:00
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Module *M;
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2015-03-14 08:00:49 +08:00
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bool LinkerSubsectionsViaSymbols;
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2015-02-21 04:30:47 +08:00
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IntegerType *Int1Ty;
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2015-02-26 04:42:41 +08:00
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IntegerType *Int8Ty;
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2015-02-21 04:30:47 +08:00
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IntegerType *Int32Ty;
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Type *Int32PtrTy;
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IntegerType *Int64Ty;
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Type *IntPtrTy;
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// The llvm.bitsets named metadata.
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NamedMDNode *BitSetNM;
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// Mapping from bitset mdstrings to the call sites that test them.
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DenseMap<MDString *, std::vector<CallInst *>> BitSetTestCallSites;
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LowerBitSets: Use byte arrays instead of bit sets to represent in-memory bit sets.
By loading from indexed offsets into a byte array and applying a mask, a
program can test bits from the bit set with a relatively short instruction
sequence. For example, suppose we have 15 bit sets to lay out:
A (16 bits), B (15 bits), C (14 bits), D (13 bits), E (12 bits),
F (11 bits), G (10 bits), H (9 bits), I (7 bits), J (6 bits), K (5 bits),
L (4 bits), M (3 bits), N (2 bits), O (1 bit)
These bits can be laid out in a 16-byte array like this:
Byte Offset
0123456789ABCDEF
Bit
7 HHHHHHHHHIIIIIII
6 GGGGGGGGGGJJJJJJ
5 FFFFFFFFFFFKKKKK
4 EEEEEEEEEEEELLLL
3 DDDDDDDDDDDDDMMM
2 CCCCCCCCCCCCCCNN
1 BBBBBBBBBBBBBBBO
0 AAAAAAAAAAAAAAAA
For example, to test bit X of A, we evaluate ((bits[X] & 1) != 0), or to
test bit X of I, we evaluate ((bits[9 + X] & 0x80) != 0). This can be done
in 1-2 machine instructions on x86, or 4-6 instructions on ARM.
This uses the LPT multiprocessor scheduling algorithm to lay out the bits
efficiently.
Saves ~450KB of instructions in a recent build of Chromium.
Differential Revision: http://reviews.llvm.org/D7954
llvm-svn: 231043
2015-03-03 08:49:28 +08:00
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std::vector<ByteArrayInfo> ByteArrayInfos;
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2015-02-21 04:30:47 +08:00
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BitSetInfo
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buildBitSet(MDString *BitSet,
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const DenseMap<GlobalVariable *, uint64_t> &GlobalLayout);
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LowerBitSets: Use byte arrays instead of bit sets to represent in-memory bit sets.
By loading from indexed offsets into a byte array and applying a mask, a
program can test bits from the bit set with a relatively short instruction
sequence. For example, suppose we have 15 bit sets to lay out:
A (16 bits), B (15 bits), C (14 bits), D (13 bits), E (12 bits),
F (11 bits), G (10 bits), H (9 bits), I (7 bits), J (6 bits), K (5 bits),
L (4 bits), M (3 bits), N (2 bits), O (1 bit)
These bits can be laid out in a 16-byte array like this:
Byte Offset
0123456789ABCDEF
Bit
7 HHHHHHHHHIIIIIII
6 GGGGGGGGGGJJJJJJ
5 FFFFFFFFFFFKKKKK
4 EEEEEEEEEEEELLLL
3 DDDDDDDDDDDDDMMM
2 CCCCCCCCCCCCCCNN
1 BBBBBBBBBBBBBBBO
0 AAAAAAAAAAAAAAAA
For example, to test bit X of A, we evaluate ((bits[X] & 1) != 0), or to
test bit X of I, we evaluate ((bits[9 + X] & 0x80) != 0). This can be done
in 1-2 machine instructions on x86, or 4-6 instructions on ARM.
This uses the LPT multiprocessor scheduling algorithm to lay out the bits
efficiently.
Saves ~450KB of instructions in a recent build of Chromium.
Differential Revision: http://reviews.llvm.org/D7954
llvm-svn: 231043
2015-03-03 08:49:28 +08:00
|
|
|
ByteArrayInfo *createByteArray(BitSetInfo &BSI);
|
|
|
|
void allocateByteArrays();
|
|
|
|
Value *createBitSetTest(IRBuilder<> &B, BitSetInfo &BSI, ByteArrayInfo *&BAI,
|
|
|
|
Value *BitOffset);
|
2015-02-26 04:42:41 +08:00
|
|
|
Value *
|
LowerBitSets: Use byte arrays instead of bit sets to represent in-memory bit sets.
By loading from indexed offsets into a byte array and applying a mask, a
program can test bits from the bit set with a relatively short instruction
sequence. For example, suppose we have 15 bit sets to lay out:
A (16 bits), B (15 bits), C (14 bits), D (13 bits), E (12 bits),
F (11 bits), G (10 bits), H (9 bits), I (7 bits), J (6 bits), K (5 bits),
L (4 bits), M (3 bits), N (2 bits), O (1 bit)
These bits can be laid out in a 16-byte array like this:
Byte Offset
0123456789ABCDEF
Bit
7 HHHHHHHHHIIIIIII
6 GGGGGGGGGGJJJJJJ
5 FFFFFFFFFFFKKKKK
4 EEEEEEEEEEEELLLL
3 DDDDDDDDDDDDDMMM
2 CCCCCCCCCCCCCCNN
1 BBBBBBBBBBBBBBBO
0 AAAAAAAAAAAAAAAA
For example, to test bit X of A, we evaluate ((bits[X] & 1) != 0), or to
test bit X of I, we evaluate ((bits[9 + X] & 0x80) != 0). This can be done
in 1-2 machine instructions on x86, or 4-6 instructions on ARM.
This uses the LPT multiprocessor scheduling algorithm to lay out the bits
efficiently.
Saves ~450KB of instructions in a recent build of Chromium.
Differential Revision: http://reviews.llvm.org/D7954
llvm-svn: 231043
2015-03-03 08:49:28 +08:00
|
|
|
lowerBitSetCall(CallInst *CI, BitSetInfo &BSI, ByteArrayInfo *&BAI,
|
|
|
|
GlobalVariable *CombinedGlobal,
|
2015-02-21 04:30:47 +08:00
|
|
|
const DenseMap<GlobalVariable *, uint64_t> &GlobalLayout);
|
LowerBitSets: Use byte arrays instead of bit sets to represent in-memory bit sets.
By loading from indexed offsets into a byte array and applying a mask, a
program can test bits from the bit set with a relatively short instruction
sequence. For example, suppose we have 15 bit sets to lay out:
A (16 bits), B (15 bits), C (14 bits), D (13 bits), E (12 bits),
F (11 bits), G (10 bits), H (9 bits), I (7 bits), J (6 bits), K (5 bits),
L (4 bits), M (3 bits), N (2 bits), O (1 bit)
These bits can be laid out in a 16-byte array like this:
Byte Offset
0123456789ABCDEF
Bit
7 HHHHHHHHHIIIIIII
6 GGGGGGGGGGJJJJJJ
5 FFFFFFFFFFFKKKKK
4 EEEEEEEEEEEELLLL
3 DDDDDDDDDDDDDMMM
2 CCCCCCCCCCCCCCNN
1 BBBBBBBBBBBBBBBO
0 AAAAAAAAAAAAAAAA
For example, to test bit X of A, we evaluate ((bits[X] & 1) != 0), or to
test bit X of I, we evaluate ((bits[9 + X] & 0x80) != 0). This can be done
in 1-2 machine instructions on x86, or 4-6 instructions on ARM.
This uses the LPT multiprocessor scheduling algorithm to lay out the bits
efficiently.
Saves ~450KB of instructions in a recent build of Chromium.
Differential Revision: http://reviews.llvm.org/D7954
llvm-svn: 231043
2015-03-03 08:49:28 +08:00
|
|
|
void buildBitSetsFromGlobals(const std::vector<MDString *> &BitSets,
|
2015-02-21 04:30:47 +08:00
|
|
|
const std::vector<GlobalVariable *> &Globals);
|
LowerBitSets: Use byte arrays instead of bit sets to represent in-memory bit sets.
By loading from indexed offsets into a byte array and applying a mask, a
program can test bits from the bit set with a relatively short instruction
sequence. For example, suppose we have 15 bit sets to lay out:
A (16 bits), B (15 bits), C (14 bits), D (13 bits), E (12 bits),
F (11 bits), G (10 bits), H (9 bits), I (7 bits), J (6 bits), K (5 bits),
L (4 bits), M (3 bits), N (2 bits), O (1 bit)
These bits can be laid out in a 16-byte array like this:
Byte Offset
0123456789ABCDEF
Bit
7 HHHHHHHHHIIIIIII
6 GGGGGGGGGGJJJJJJ
5 FFFFFFFFFFFKKKKK
4 EEEEEEEEEEEELLLL
3 DDDDDDDDDDDDDMMM
2 CCCCCCCCCCCCCCNN
1 BBBBBBBBBBBBBBBO
0 AAAAAAAAAAAAAAAA
For example, to test bit X of A, we evaluate ((bits[X] & 1) != 0), or to
test bit X of I, we evaluate ((bits[9 + X] & 0x80) != 0). This can be done
in 1-2 machine instructions on x86, or 4-6 instructions on ARM.
This uses the LPT multiprocessor scheduling algorithm to lay out the bits
efficiently.
Saves ~450KB of instructions in a recent build of Chromium.
Differential Revision: http://reviews.llvm.org/D7954
llvm-svn: 231043
2015-03-03 08:49:28 +08:00
|
|
|
bool buildBitSets();
|
|
|
|
bool eraseBitSetMetadata();
|
2015-02-21 04:30:47 +08:00
|
|
|
|
|
|
|
bool doInitialization(Module &M) override;
|
|
|
|
bool runOnModule(Module &M) override;
|
|
|
|
};
|
|
|
|
|
|
|
|
} // namespace
|
|
|
|
|
|
|
|
INITIALIZE_PASS_BEGIN(LowerBitSets, "lowerbitsets",
|
|
|
|
"Lower bitset metadata", false, false)
|
|
|
|
INITIALIZE_PASS_END(LowerBitSets, "lowerbitsets",
|
|
|
|
"Lower bitset metadata", false, false)
|
|
|
|
char LowerBitSets::ID = 0;
|
|
|
|
|
|
|
|
ModulePass *llvm::createLowerBitSetsPass() { return new LowerBitSets; }
|
|
|
|
|
LowerBitSets: Use byte arrays instead of bit sets to represent in-memory bit sets.
By loading from indexed offsets into a byte array and applying a mask, a
program can test bits from the bit set with a relatively short instruction
sequence. For example, suppose we have 15 bit sets to lay out:
A (16 bits), B (15 bits), C (14 bits), D (13 bits), E (12 bits),
F (11 bits), G (10 bits), H (9 bits), I (7 bits), J (6 bits), K (5 bits),
L (4 bits), M (3 bits), N (2 bits), O (1 bit)
These bits can be laid out in a 16-byte array like this:
Byte Offset
0123456789ABCDEF
Bit
7 HHHHHHHHHIIIIIII
6 GGGGGGGGGGJJJJJJ
5 FFFFFFFFFFFKKKKK
4 EEEEEEEEEEEELLLL
3 DDDDDDDDDDDDDMMM
2 CCCCCCCCCCCCCCNN
1 BBBBBBBBBBBBBBBO
0 AAAAAAAAAAAAAAAA
For example, to test bit X of A, we evaluate ((bits[X] & 1) != 0), or to
test bit X of I, we evaluate ((bits[9 + X] & 0x80) != 0). This can be done
in 1-2 machine instructions on x86, or 4-6 instructions on ARM.
This uses the LPT multiprocessor scheduling algorithm to lay out the bits
efficiently.
Saves ~450KB of instructions in a recent build of Chromium.
Differential Revision: http://reviews.llvm.org/D7954
llvm-svn: 231043
2015-03-03 08:49:28 +08:00
|
|
|
bool LowerBitSets::doInitialization(Module &Mod) {
|
|
|
|
M = &Mod;
|
2015-03-10 10:37:25 +08:00
|
|
|
const DataLayout &DL = Mod.getDataLayout();
|
2015-02-21 04:30:47 +08:00
|
|
|
|
2015-03-14 08:00:49 +08:00
|
|
|
Triple TargetTriple(M->getTargetTriple());
|
|
|
|
LinkerSubsectionsViaSymbols = TargetTriple.isMacOSX();
|
|
|
|
|
LowerBitSets: Use byte arrays instead of bit sets to represent in-memory bit sets.
By loading from indexed offsets into a byte array and applying a mask, a
program can test bits from the bit set with a relatively short instruction
sequence. For example, suppose we have 15 bit sets to lay out:
A (16 bits), B (15 bits), C (14 bits), D (13 bits), E (12 bits),
F (11 bits), G (10 bits), H (9 bits), I (7 bits), J (6 bits), K (5 bits),
L (4 bits), M (3 bits), N (2 bits), O (1 bit)
These bits can be laid out in a 16-byte array like this:
Byte Offset
0123456789ABCDEF
Bit
7 HHHHHHHHHIIIIIII
6 GGGGGGGGGGJJJJJJ
5 FFFFFFFFFFFKKKKK
4 EEEEEEEEEEEELLLL
3 DDDDDDDDDDDDDMMM
2 CCCCCCCCCCCCCCNN
1 BBBBBBBBBBBBBBBO
0 AAAAAAAAAAAAAAAA
For example, to test bit X of A, we evaluate ((bits[X] & 1) != 0), or to
test bit X of I, we evaluate ((bits[9 + X] & 0x80) != 0). This can be done
in 1-2 machine instructions on x86, or 4-6 instructions on ARM.
This uses the LPT multiprocessor scheduling algorithm to lay out the bits
efficiently.
Saves ~450KB of instructions in a recent build of Chromium.
Differential Revision: http://reviews.llvm.org/D7954
llvm-svn: 231043
2015-03-03 08:49:28 +08:00
|
|
|
Int1Ty = Type::getInt1Ty(M->getContext());
|
|
|
|
Int8Ty = Type::getInt8Ty(M->getContext());
|
|
|
|
Int32Ty = Type::getInt32Ty(M->getContext());
|
2015-02-21 04:30:47 +08:00
|
|
|
Int32PtrTy = PointerType::getUnqual(Int32Ty);
|
LowerBitSets: Use byte arrays instead of bit sets to represent in-memory bit sets.
By loading from indexed offsets into a byte array and applying a mask, a
program can test bits from the bit set with a relatively short instruction
sequence. For example, suppose we have 15 bit sets to lay out:
A (16 bits), B (15 bits), C (14 bits), D (13 bits), E (12 bits),
F (11 bits), G (10 bits), H (9 bits), I (7 bits), J (6 bits), K (5 bits),
L (4 bits), M (3 bits), N (2 bits), O (1 bit)
These bits can be laid out in a 16-byte array like this:
Byte Offset
0123456789ABCDEF
Bit
7 HHHHHHHHHIIIIIII
6 GGGGGGGGGGJJJJJJ
5 FFFFFFFFFFFKKKKK
4 EEEEEEEEEEEELLLL
3 DDDDDDDDDDDDDMMM
2 CCCCCCCCCCCCCCNN
1 BBBBBBBBBBBBBBBO
0 AAAAAAAAAAAAAAAA
For example, to test bit X of A, we evaluate ((bits[X] & 1) != 0), or to
test bit X of I, we evaluate ((bits[9 + X] & 0x80) != 0). This can be done
in 1-2 machine instructions on x86, or 4-6 instructions on ARM.
This uses the LPT multiprocessor scheduling algorithm to lay out the bits
efficiently.
Saves ~450KB of instructions in a recent build of Chromium.
Differential Revision: http://reviews.llvm.org/D7954
llvm-svn: 231043
2015-03-03 08:49:28 +08:00
|
|
|
Int64Ty = Type::getInt64Ty(M->getContext());
|
2015-03-10 10:37:25 +08:00
|
|
|
IntPtrTy = DL.getIntPtrType(M->getContext(), 0);
|
2015-02-21 04:30:47 +08:00
|
|
|
|
LowerBitSets: Use byte arrays instead of bit sets to represent in-memory bit sets.
By loading from indexed offsets into a byte array and applying a mask, a
program can test bits from the bit set with a relatively short instruction
sequence. For example, suppose we have 15 bit sets to lay out:
A (16 bits), B (15 bits), C (14 bits), D (13 bits), E (12 bits),
F (11 bits), G (10 bits), H (9 bits), I (7 bits), J (6 bits), K (5 bits),
L (4 bits), M (3 bits), N (2 bits), O (1 bit)
These bits can be laid out in a 16-byte array like this:
Byte Offset
0123456789ABCDEF
Bit
7 HHHHHHHHHIIIIIII
6 GGGGGGGGGGJJJJJJ
5 FFFFFFFFFFFKKKKK
4 EEEEEEEEEEEELLLL
3 DDDDDDDDDDDDDMMM
2 CCCCCCCCCCCCCCNN
1 BBBBBBBBBBBBBBBO
0 AAAAAAAAAAAAAAAA
For example, to test bit X of A, we evaluate ((bits[X] & 1) != 0), or to
test bit X of I, we evaluate ((bits[9 + X] & 0x80) != 0). This can be done
in 1-2 machine instructions on x86, or 4-6 instructions on ARM.
This uses the LPT multiprocessor scheduling algorithm to lay out the bits
efficiently.
Saves ~450KB of instructions in a recent build of Chromium.
Differential Revision: http://reviews.llvm.org/D7954
llvm-svn: 231043
2015-03-03 08:49:28 +08:00
|
|
|
BitSetNM = M->getNamedMetadata("llvm.bitsets");
|
2015-02-21 04:30:47 +08:00
|
|
|
|
|
|
|
BitSetTestCallSites.clear();
|
|
|
|
|
|
|
|
return false;
|
|
|
|
}
|
|
|
|
|
2015-02-22 17:51:42 +08:00
|
|
|
/// Build a bit set for BitSet using the object layouts in
|
|
|
|
/// GlobalLayout.
|
2015-02-21 04:30:47 +08:00
|
|
|
BitSetInfo LowerBitSets::buildBitSet(
|
|
|
|
MDString *BitSet,
|
|
|
|
const DenseMap<GlobalVariable *, uint64_t> &GlobalLayout) {
|
|
|
|
BitSetBuilder BSB;
|
|
|
|
|
|
|
|
// Compute the byte offset of each element of this bitset.
|
|
|
|
if (BitSetNM) {
|
|
|
|
for (MDNode *Op : BitSetNM->operands()) {
|
|
|
|
if (Op->getOperand(0) != BitSet || !Op->getOperand(1))
|
|
|
|
continue;
|
|
|
|
auto OpGlobal = cast<GlobalVariable>(
|
|
|
|
cast<ConstantAsMetadata>(Op->getOperand(1))->getValue());
|
|
|
|
uint64_t Offset =
|
|
|
|
cast<ConstantInt>(cast<ConstantAsMetadata>(Op->getOperand(2))
|
|
|
|
->getValue())->getZExtValue();
|
|
|
|
|
|
|
|
Offset += GlobalLayout.find(OpGlobal)->second;
|
|
|
|
|
|
|
|
BSB.addOffset(Offset);
|
|
|
|
}
|
|
|
|
}
|
|
|
|
|
|
|
|
return BSB.build();
|
|
|
|
}
|
|
|
|
|
2015-02-22 17:51:42 +08:00
|
|
|
/// Build a test that bit BitOffset mod sizeof(Bits)*8 is set in
|
|
|
|
/// Bits. This pattern matches to the bt instruction on x86.
|
2015-02-21 04:30:47 +08:00
|
|
|
static Value *createMaskedBitTest(IRBuilder<> &B, Value *Bits,
|
|
|
|
Value *BitOffset) {
|
|
|
|
auto BitsType = cast<IntegerType>(Bits->getType());
|
|
|
|
unsigned BitWidth = BitsType->getBitWidth();
|
|
|
|
|
|
|
|
BitOffset = B.CreateZExtOrTrunc(BitOffset, BitsType);
|
|
|
|
Value *BitIndex =
|
|
|
|
B.CreateAnd(BitOffset, ConstantInt::get(BitsType, BitWidth - 1));
|
|
|
|
Value *BitMask = B.CreateShl(ConstantInt::get(BitsType, 1), BitIndex);
|
|
|
|
Value *MaskedBits = B.CreateAnd(Bits, BitMask);
|
|
|
|
return B.CreateICmpNE(MaskedBits, ConstantInt::get(BitsType, 0));
|
|
|
|
}
|
|
|
|
|
LowerBitSets: Use byte arrays instead of bit sets to represent in-memory bit sets.
By loading from indexed offsets into a byte array and applying a mask, a
program can test bits from the bit set with a relatively short instruction
sequence. For example, suppose we have 15 bit sets to lay out:
A (16 bits), B (15 bits), C (14 bits), D (13 bits), E (12 bits),
F (11 bits), G (10 bits), H (9 bits), I (7 bits), J (6 bits), K (5 bits),
L (4 bits), M (3 bits), N (2 bits), O (1 bit)
These bits can be laid out in a 16-byte array like this:
Byte Offset
0123456789ABCDEF
Bit
7 HHHHHHHHHIIIIIII
6 GGGGGGGGGGJJJJJJ
5 FFFFFFFFFFFKKKKK
4 EEEEEEEEEEEELLLL
3 DDDDDDDDDDDDDMMM
2 CCCCCCCCCCCCCCNN
1 BBBBBBBBBBBBBBBO
0 AAAAAAAAAAAAAAAA
For example, to test bit X of A, we evaluate ((bits[X] & 1) != 0), or to
test bit X of I, we evaluate ((bits[9 + X] & 0x80) != 0). This can be done
in 1-2 machine instructions on x86, or 4-6 instructions on ARM.
This uses the LPT multiprocessor scheduling algorithm to lay out the bits
efficiently.
Saves ~450KB of instructions in a recent build of Chromium.
Differential Revision: http://reviews.llvm.org/D7954
llvm-svn: 231043
2015-03-03 08:49:28 +08:00
|
|
|
ByteArrayInfo *LowerBitSets::createByteArray(BitSetInfo &BSI) {
|
|
|
|
// Create globals to stand in for byte arrays and masks. These never actually
|
|
|
|
// get initialized, we RAUW and erase them later in allocateByteArrays() once
|
|
|
|
// we know the offset and mask to use.
|
|
|
|
auto ByteArrayGlobal = new GlobalVariable(
|
|
|
|
*M, Int8Ty, /*isConstant=*/true, GlobalValue::PrivateLinkage, nullptr);
|
|
|
|
auto MaskGlobal = new GlobalVariable(
|
|
|
|
*M, Int8Ty, /*isConstant=*/true, GlobalValue::PrivateLinkage, nullptr);
|
|
|
|
|
|
|
|
ByteArrayInfos.emplace_back();
|
|
|
|
ByteArrayInfo *BAI = &ByteArrayInfos.back();
|
|
|
|
|
|
|
|
BAI->Bits = BSI.Bits;
|
|
|
|
BAI->BitSize = BSI.BitSize;
|
|
|
|
BAI->ByteArray = ByteArrayGlobal;
|
|
|
|
BAI->Mask = ConstantExpr::getPtrToInt(MaskGlobal, Int8Ty);
|
|
|
|
return BAI;
|
|
|
|
}
|
|
|
|
|
|
|
|
void LowerBitSets::allocateByteArrays() {
|
|
|
|
std::stable_sort(ByteArrayInfos.begin(), ByteArrayInfos.end(),
|
|
|
|
[](const ByteArrayInfo &BAI1, const ByteArrayInfo &BAI2) {
|
|
|
|
return BAI1.BitSize > BAI2.BitSize;
|
|
|
|
});
|
|
|
|
|
|
|
|
std::vector<uint64_t> ByteArrayOffsets(ByteArrayInfos.size());
|
|
|
|
|
|
|
|
ByteArrayBuilder BAB;
|
|
|
|
for (unsigned I = 0; I != ByteArrayInfos.size(); ++I) {
|
|
|
|
ByteArrayInfo *BAI = &ByteArrayInfos[I];
|
|
|
|
|
|
|
|
uint8_t Mask;
|
|
|
|
BAB.allocate(BAI->Bits, BAI->BitSize, ByteArrayOffsets[I], Mask);
|
|
|
|
|
|
|
|
BAI->Mask->replaceAllUsesWith(ConstantInt::get(Int8Ty, Mask));
|
|
|
|
cast<GlobalVariable>(BAI->Mask->getOperand(0))->eraseFromParent();
|
|
|
|
}
|
|
|
|
|
|
|
|
Constant *ByteArrayConst = ConstantDataArray::get(M->getContext(), BAB.Bytes);
|
|
|
|
auto ByteArray =
|
|
|
|
new GlobalVariable(*M, ByteArrayConst->getType(), /*isConstant=*/true,
|
|
|
|
GlobalValue::PrivateLinkage, ByteArrayConst);
|
|
|
|
|
|
|
|
for (unsigned I = 0; I != ByteArrayInfos.size(); ++I) {
|
|
|
|
ByteArrayInfo *BAI = &ByteArrayInfos[I];
|
|
|
|
|
|
|
|
Constant *Idxs[] = {ConstantInt::get(IntPtrTy, 0),
|
|
|
|
ConstantInt::get(IntPtrTy, ByteArrayOffsets[I])};
|
|
|
|
Constant *GEP = ConstantExpr::getInBoundsGetElementPtr(ByteArray, Idxs);
|
|
|
|
|
|
|
|
// Create an alias instead of RAUW'ing the gep directly. On x86 this ensures
|
|
|
|
// that the pc-relative displacement is folded into the lea instead of the
|
|
|
|
// test instruction getting another displacement.
|
2015-03-17 07:36:24 +08:00
|
|
|
if (LinkerSubsectionsViaSymbols) {
|
|
|
|
BAI->ByteArray->replaceAllUsesWith(GEP);
|
|
|
|
} else {
|
|
|
|
GlobalAlias *Alias = GlobalAlias::create(
|
|
|
|
Int8Ty, 0, GlobalValue::PrivateLinkage, "bits", GEP, M);
|
|
|
|
BAI->ByteArray->replaceAllUsesWith(Alias);
|
|
|
|
}
|
LowerBitSets: Use byte arrays instead of bit sets to represent in-memory bit sets.
By loading from indexed offsets into a byte array and applying a mask, a
program can test bits from the bit set with a relatively short instruction
sequence. For example, suppose we have 15 bit sets to lay out:
A (16 bits), B (15 bits), C (14 bits), D (13 bits), E (12 bits),
F (11 bits), G (10 bits), H (9 bits), I (7 bits), J (6 bits), K (5 bits),
L (4 bits), M (3 bits), N (2 bits), O (1 bit)
These bits can be laid out in a 16-byte array like this:
Byte Offset
0123456789ABCDEF
Bit
7 HHHHHHHHHIIIIIII
6 GGGGGGGGGGJJJJJJ
5 FFFFFFFFFFFKKKKK
4 EEEEEEEEEEEELLLL
3 DDDDDDDDDDDDDMMM
2 CCCCCCCCCCCCCCNN
1 BBBBBBBBBBBBBBBO
0 AAAAAAAAAAAAAAAA
For example, to test bit X of A, we evaluate ((bits[X] & 1) != 0), or to
test bit X of I, we evaluate ((bits[9 + X] & 0x80) != 0). This can be done
in 1-2 machine instructions on x86, or 4-6 instructions on ARM.
This uses the LPT multiprocessor scheduling algorithm to lay out the bits
efficiently.
Saves ~450KB of instructions in a recent build of Chromium.
Differential Revision: http://reviews.llvm.org/D7954
llvm-svn: 231043
2015-03-03 08:49:28 +08:00
|
|
|
BAI->ByteArray->eraseFromParent();
|
|
|
|
}
|
|
|
|
|
|
|
|
ByteArraySizeBits = BAB.BitAllocs[0] + BAB.BitAllocs[1] + BAB.BitAllocs[2] +
|
|
|
|
BAB.BitAllocs[3] + BAB.BitAllocs[4] + BAB.BitAllocs[5] +
|
|
|
|
BAB.BitAllocs[6] + BAB.BitAllocs[7];
|
|
|
|
ByteArraySizeBytes = BAB.Bytes.size();
|
|
|
|
}
|
|
|
|
|
2015-02-22 17:51:42 +08:00
|
|
|
/// Build a test that bit BitOffset is set in BSI, where
|
|
|
|
/// BitSetGlobal is a global containing the bits in BSI.
|
LowerBitSets: Use byte arrays instead of bit sets to represent in-memory bit sets.
By loading from indexed offsets into a byte array and applying a mask, a
program can test bits from the bit set with a relatively short instruction
sequence. For example, suppose we have 15 bit sets to lay out:
A (16 bits), B (15 bits), C (14 bits), D (13 bits), E (12 bits),
F (11 bits), G (10 bits), H (9 bits), I (7 bits), J (6 bits), K (5 bits),
L (4 bits), M (3 bits), N (2 bits), O (1 bit)
These bits can be laid out in a 16-byte array like this:
Byte Offset
0123456789ABCDEF
Bit
7 HHHHHHHHHIIIIIII
6 GGGGGGGGGGJJJJJJ
5 FFFFFFFFFFFKKKKK
4 EEEEEEEEEEEELLLL
3 DDDDDDDDDDDDDMMM
2 CCCCCCCCCCCCCCNN
1 BBBBBBBBBBBBBBBO
0 AAAAAAAAAAAAAAAA
For example, to test bit X of A, we evaluate ((bits[X] & 1) != 0), or to
test bit X of I, we evaluate ((bits[9 + X] & 0x80) != 0). This can be done
in 1-2 machine instructions on x86, or 4-6 instructions on ARM.
This uses the LPT multiprocessor scheduling algorithm to lay out the bits
efficiently.
Saves ~450KB of instructions in a recent build of Chromium.
Differential Revision: http://reviews.llvm.org/D7954
llvm-svn: 231043
2015-03-03 08:49:28 +08:00
|
|
|
Value *LowerBitSets::createBitSetTest(IRBuilder<> &B, BitSetInfo &BSI,
|
|
|
|
ByteArrayInfo *&BAI, Value *BitOffset) {
|
|
|
|
if (BSI.BitSize <= 64) {
|
2015-02-21 04:30:47 +08:00
|
|
|
// If the bit set is sufficiently small, we can avoid a load by bit testing
|
|
|
|
// a constant.
|
|
|
|
IntegerType *BitsTy;
|
LowerBitSets: Use byte arrays instead of bit sets to represent in-memory bit sets.
By loading from indexed offsets into a byte array and applying a mask, a
program can test bits from the bit set with a relatively short instruction
sequence. For example, suppose we have 15 bit sets to lay out:
A (16 bits), B (15 bits), C (14 bits), D (13 bits), E (12 bits),
F (11 bits), G (10 bits), H (9 bits), I (7 bits), J (6 bits), K (5 bits),
L (4 bits), M (3 bits), N (2 bits), O (1 bit)
These bits can be laid out in a 16-byte array like this:
Byte Offset
0123456789ABCDEF
Bit
7 HHHHHHHHHIIIIIII
6 GGGGGGGGGGJJJJJJ
5 FFFFFFFFFFFKKKKK
4 EEEEEEEEEEEELLLL
3 DDDDDDDDDDDDDMMM
2 CCCCCCCCCCCCCCNN
1 BBBBBBBBBBBBBBBO
0 AAAAAAAAAAAAAAAA
For example, to test bit X of A, we evaluate ((bits[X] & 1) != 0), or to
test bit X of I, we evaluate ((bits[9 + X] & 0x80) != 0). This can be done
in 1-2 machine instructions on x86, or 4-6 instructions on ARM.
This uses the LPT multiprocessor scheduling algorithm to lay out the bits
efficiently.
Saves ~450KB of instructions in a recent build of Chromium.
Differential Revision: http://reviews.llvm.org/D7954
llvm-svn: 231043
2015-03-03 08:49:28 +08:00
|
|
|
if (BSI.BitSize <= 32)
|
2015-02-21 04:30:47 +08:00
|
|
|
BitsTy = Int32Ty;
|
|
|
|
else
|
|
|
|
BitsTy = Int64Ty;
|
|
|
|
|
|
|
|
uint64_t Bits = 0;
|
LowerBitSets: Use byte arrays instead of bit sets to represent in-memory bit sets.
By loading from indexed offsets into a byte array and applying a mask, a
program can test bits from the bit set with a relatively short instruction
sequence. For example, suppose we have 15 bit sets to lay out:
A (16 bits), B (15 bits), C (14 bits), D (13 bits), E (12 bits),
F (11 bits), G (10 bits), H (9 bits), I (7 bits), J (6 bits), K (5 bits),
L (4 bits), M (3 bits), N (2 bits), O (1 bit)
These bits can be laid out in a 16-byte array like this:
Byte Offset
0123456789ABCDEF
Bit
7 HHHHHHHHHIIIIIII
6 GGGGGGGGGGJJJJJJ
5 FFFFFFFFFFFKKKKK
4 EEEEEEEEEEEELLLL
3 DDDDDDDDDDDDDMMM
2 CCCCCCCCCCCCCCNN
1 BBBBBBBBBBBBBBBO
0 AAAAAAAAAAAAAAAA
For example, to test bit X of A, we evaluate ((bits[X] & 1) != 0), or to
test bit X of I, we evaluate ((bits[9 + X] & 0x80) != 0). This can be done
in 1-2 machine instructions on x86, or 4-6 instructions on ARM.
This uses the LPT multiprocessor scheduling algorithm to lay out the bits
efficiently.
Saves ~450KB of instructions in a recent build of Chromium.
Differential Revision: http://reviews.llvm.org/D7954
llvm-svn: 231043
2015-03-03 08:49:28 +08:00
|
|
|
for (auto Bit : BSI.Bits)
|
|
|
|
Bits |= uint64_t(1) << Bit;
|
2015-02-21 04:30:47 +08:00
|
|
|
Constant *BitsConst = ConstantInt::get(BitsTy, Bits);
|
|
|
|
return createMaskedBitTest(B, BitsConst, BitOffset);
|
|
|
|
} else {
|
LowerBitSets: Use byte arrays instead of bit sets to represent in-memory bit sets.
By loading from indexed offsets into a byte array and applying a mask, a
program can test bits from the bit set with a relatively short instruction
sequence. For example, suppose we have 15 bit sets to lay out:
A (16 bits), B (15 bits), C (14 bits), D (13 bits), E (12 bits),
F (11 bits), G (10 bits), H (9 bits), I (7 bits), J (6 bits), K (5 bits),
L (4 bits), M (3 bits), N (2 bits), O (1 bit)
These bits can be laid out in a 16-byte array like this:
Byte Offset
0123456789ABCDEF
Bit
7 HHHHHHHHHIIIIIII
6 GGGGGGGGGGJJJJJJ
5 FFFFFFFFFFFKKKKK
4 EEEEEEEEEEEELLLL
3 DDDDDDDDDDDDDMMM
2 CCCCCCCCCCCCCCNN
1 BBBBBBBBBBBBBBBO
0 AAAAAAAAAAAAAAAA
For example, to test bit X of A, we evaluate ((bits[X] & 1) != 0), or to
test bit X of I, we evaluate ((bits[9 + X] & 0x80) != 0). This can be done
in 1-2 machine instructions on x86, or 4-6 instructions on ARM.
This uses the LPT multiprocessor scheduling algorithm to lay out the bits
efficiently.
Saves ~450KB of instructions in a recent build of Chromium.
Differential Revision: http://reviews.llvm.org/D7954
llvm-svn: 231043
2015-03-03 08:49:28 +08:00
|
|
|
if (!BAI) {
|
|
|
|
++NumByteArraysCreated;
|
|
|
|
BAI = createByteArray(BSI);
|
|
|
|
}
|
|
|
|
|
2015-03-20 06:02:10 +08:00
|
|
|
Constant *ByteArray = BAI->ByteArray;
|
|
|
|
if (!LinkerSubsectionsViaSymbols && AvoidReuse) {
|
|
|
|
// Each use of the byte array uses a different alias. This makes the
|
|
|
|
// backend less likely to reuse previously computed byte array addresses,
|
|
|
|
// improving the security of the CFI mechanism based on this pass.
|
|
|
|
ByteArray = GlobalAlias::create(
|
|
|
|
BAI->ByteArray->getType()->getElementType(), 0,
|
|
|
|
GlobalValue::PrivateLinkage, "bits_use", ByteArray, M);
|
|
|
|
}
|
|
|
|
|
|
|
|
Value *ByteAddr = B.CreateGEP(ByteArray, BitOffset);
|
LowerBitSets: Use byte arrays instead of bit sets to represent in-memory bit sets.
By loading from indexed offsets into a byte array and applying a mask, a
program can test bits from the bit set with a relatively short instruction
sequence. For example, suppose we have 15 bit sets to lay out:
A (16 bits), B (15 bits), C (14 bits), D (13 bits), E (12 bits),
F (11 bits), G (10 bits), H (9 bits), I (7 bits), J (6 bits), K (5 bits),
L (4 bits), M (3 bits), N (2 bits), O (1 bit)
These bits can be laid out in a 16-byte array like this:
Byte Offset
0123456789ABCDEF
Bit
7 HHHHHHHHHIIIIIII
6 GGGGGGGGGGJJJJJJ
5 FFFFFFFFFFFKKKKK
4 EEEEEEEEEEEELLLL
3 DDDDDDDDDDDDDMMM
2 CCCCCCCCCCCCCCNN
1 BBBBBBBBBBBBBBBO
0 AAAAAAAAAAAAAAAA
For example, to test bit X of A, we evaluate ((bits[X] & 1) != 0), or to
test bit X of I, we evaluate ((bits[9 + X] & 0x80) != 0). This can be done
in 1-2 machine instructions on x86, or 4-6 instructions on ARM.
This uses the LPT multiprocessor scheduling algorithm to lay out the bits
efficiently.
Saves ~450KB of instructions in a recent build of Chromium.
Differential Revision: http://reviews.llvm.org/D7954
llvm-svn: 231043
2015-03-03 08:49:28 +08:00
|
|
|
Value *Byte = B.CreateLoad(ByteAddr);
|
|
|
|
|
|
|
|
Value *ByteAndMask = B.CreateAnd(Byte, BAI->Mask);
|
|
|
|
return B.CreateICmpNE(ByteAndMask, ConstantInt::get(Int8Ty, 0));
|
2015-02-21 04:30:47 +08:00
|
|
|
}
|
|
|
|
}
|
|
|
|
|
2015-02-26 04:42:41 +08:00
|
|
|
/// Lower a llvm.bitset.test call to its implementation. Returns the value to
|
|
|
|
/// replace the call with.
|
|
|
|
Value *LowerBitSets::lowerBitSetCall(
|
LowerBitSets: Use byte arrays instead of bit sets to represent in-memory bit sets.
By loading from indexed offsets into a byte array and applying a mask, a
program can test bits from the bit set with a relatively short instruction
sequence. For example, suppose we have 15 bit sets to lay out:
A (16 bits), B (15 bits), C (14 bits), D (13 bits), E (12 bits),
F (11 bits), G (10 bits), H (9 bits), I (7 bits), J (6 bits), K (5 bits),
L (4 bits), M (3 bits), N (2 bits), O (1 bit)
These bits can be laid out in a 16-byte array like this:
Byte Offset
0123456789ABCDEF
Bit
7 HHHHHHHHHIIIIIII
6 GGGGGGGGGGJJJJJJ
5 FFFFFFFFFFFKKKKK
4 EEEEEEEEEEEELLLL
3 DDDDDDDDDDDDDMMM
2 CCCCCCCCCCCCCCNN
1 BBBBBBBBBBBBBBBO
0 AAAAAAAAAAAAAAAA
For example, to test bit X of A, we evaluate ((bits[X] & 1) != 0), or to
test bit X of I, we evaluate ((bits[9 + X] & 0x80) != 0). This can be done
in 1-2 machine instructions on x86, or 4-6 instructions on ARM.
This uses the LPT multiprocessor scheduling algorithm to lay out the bits
efficiently.
Saves ~450KB of instructions in a recent build of Chromium.
Differential Revision: http://reviews.llvm.org/D7954
llvm-svn: 231043
2015-03-03 08:49:28 +08:00
|
|
|
CallInst *CI, BitSetInfo &BSI, ByteArrayInfo *&BAI,
|
2015-02-21 04:30:47 +08:00
|
|
|
GlobalVariable *CombinedGlobal,
|
|
|
|
const DenseMap<GlobalVariable *, uint64_t> &GlobalLayout) {
|
|
|
|
Value *Ptr = CI->getArgOperand(0);
|
2015-03-10 10:37:25 +08:00
|
|
|
const DataLayout &DL = M->getDataLayout();
|
2015-02-21 04:30:47 +08:00
|
|
|
|
2015-02-26 04:42:41 +08:00
|
|
|
if (BSI.containsValue(DL, GlobalLayout, Ptr))
|
LowerBitSets: Use byte arrays instead of bit sets to represent in-memory bit sets.
By loading from indexed offsets into a byte array and applying a mask, a
program can test bits from the bit set with a relatively short instruction
sequence. For example, suppose we have 15 bit sets to lay out:
A (16 bits), B (15 bits), C (14 bits), D (13 bits), E (12 bits),
F (11 bits), G (10 bits), H (9 bits), I (7 bits), J (6 bits), K (5 bits),
L (4 bits), M (3 bits), N (2 bits), O (1 bit)
These bits can be laid out in a 16-byte array like this:
Byte Offset
0123456789ABCDEF
Bit
7 HHHHHHHHHIIIIIII
6 GGGGGGGGGGJJJJJJ
5 FFFFFFFFFFFKKKKK
4 EEEEEEEEEEEELLLL
3 DDDDDDDDDDDDDMMM
2 CCCCCCCCCCCCCCNN
1 BBBBBBBBBBBBBBBO
0 AAAAAAAAAAAAAAAA
For example, to test bit X of A, we evaluate ((bits[X] & 1) != 0), or to
test bit X of I, we evaluate ((bits[9 + X] & 0x80) != 0). This can be done
in 1-2 machine instructions on x86, or 4-6 instructions on ARM.
This uses the LPT multiprocessor scheduling algorithm to lay out the bits
efficiently.
Saves ~450KB of instructions in a recent build of Chromium.
Differential Revision: http://reviews.llvm.org/D7954
llvm-svn: 231043
2015-03-03 08:49:28 +08:00
|
|
|
return ConstantInt::getTrue(CombinedGlobal->getParent()->getContext());
|
2015-02-21 04:30:47 +08:00
|
|
|
|
|
|
|
Constant *GlobalAsInt = ConstantExpr::getPtrToInt(CombinedGlobal, IntPtrTy);
|
|
|
|
Constant *OffsetedGlobalAsInt = ConstantExpr::getAdd(
|
|
|
|
GlobalAsInt, ConstantInt::get(IntPtrTy, BSI.ByteOffset));
|
|
|
|
|
|
|
|
BasicBlock *InitialBB = CI->getParent();
|
|
|
|
|
|
|
|
IRBuilder<> B(CI);
|
|
|
|
|
|
|
|
Value *PtrAsInt = B.CreatePtrToInt(Ptr, IntPtrTy);
|
|
|
|
|
2015-02-26 04:42:41 +08:00
|
|
|
if (BSI.isSingleOffset())
|
|
|
|
return B.CreateICmpEQ(PtrAsInt, OffsetedGlobalAsInt);
|
2015-02-21 04:30:47 +08:00
|
|
|
|
|
|
|
Value *PtrOffset = B.CreateSub(PtrAsInt, OffsetedGlobalAsInt);
|
|
|
|
|
|
|
|
Value *BitOffset;
|
|
|
|
if (BSI.AlignLog2 == 0) {
|
|
|
|
BitOffset = PtrOffset;
|
|
|
|
} else {
|
|
|
|
// We need to check that the offset both falls within our range and is
|
|
|
|
// suitably aligned. We can check both properties at the same time by
|
|
|
|
// performing a right rotate by log2(alignment) followed by an integer
|
|
|
|
// comparison against the bitset size. The rotate will move the lower
|
|
|
|
// order bits that need to be zero into the higher order bits of the
|
|
|
|
// result, causing the comparison to fail if they are nonzero. The rotate
|
|
|
|
// also conveniently gives us a bit offset to use during the load from
|
|
|
|
// the bitset.
|
|
|
|
Value *OffsetSHR =
|
|
|
|
B.CreateLShr(PtrOffset, ConstantInt::get(IntPtrTy, BSI.AlignLog2));
|
|
|
|
Value *OffsetSHL = B.CreateShl(
|
2015-03-10 10:37:25 +08:00
|
|
|
PtrOffset,
|
|
|
|
ConstantInt::get(IntPtrTy, DL.getPointerSizeInBits(0) - BSI.AlignLog2));
|
2015-02-21 04:30:47 +08:00
|
|
|
BitOffset = B.CreateOr(OffsetSHR, OffsetSHL);
|
|
|
|
}
|
|
|
|
|
|
|
|
Constant *BitSizeConst = ConstantInt::get(IntPtrTy, BSI.BitSize);
|
|
|
|
Value *OffsetInRange = B.CreateICmpULT(BitOffset, BitSizeConst);
|
|
|
|
|
2015-02-26 04:42:41 +08:00
|
|
|
// If the bit set is all ones, testing against it is unnecessary.
|
|
|
|
if (BSI.isAllOnes())
|
|
|
|
return OffsetInRange;
|
|
|
|
|
2015-02-21 04:30:47 +08:00
|
|
|
TerminatorInst *Term = SplitBlockAndInsertIfThen(OffsetInRange, CI, false);
|
|
|
|
IRBuilder<> ThenB(Term);
|
|
|
|
|
|
|
|
// Now that we know that the offset is in range and aligned, load the
|
|
|
|
// appropriate bit from the bitset.
|
LowerBitSets: Use byte arrays instead of bit sets to represent in-memory bit sets.
By loading from indexed offsets into a byte array and applying a mask, a
program can test bits from the bit set with a relatively short instruction
sequence. For example, suppose we have 15 bit sets to lay out:
A (16 bits), B (15 bits), C (14 bits), D (13 bits), E (12 bits),
F (11 bits), G (10 bits), H (9 bits), I (7 bits), J (6 bits), K (5 bits),
L (4 bits), M (3 bits), N (2 bits), O (1 bit)
These bits can be laid out in a 16-byte array like this:
Byte Offset
0123456789ABCDEF
Bit
7 HHHHHHHHHIIIIIII
6 GGGGGGGGGGJJJJJJ
5 FFFFFFFFFFFKKKKK
4 EEEEEEEEEEEELLLL
3 DDDDDDDDDDDDDMMM
2 CCCCCCCCCCCCCCNN
1 BBBBBBBBBBBBBBBO
0 AAAAAAAAAAAAAAAA
For example, to test bit X of A, we evaluate ((bits[X] & 1) != 0), or to
test bit X of I, we evaluate ((bits[9 + X] & 0x80) != 0). This can be done
in 1-2 machine instructions on x86, or 4-6 instructions on ARM.
This uses the LPT multiprocessor scheduling algorithm to lay out the bits
efficiently.
Saves ~450KB of instructions in a recent build of Chromium.
Differential Revision: http://reviews.llvm.org/D7954
llvm-svn: 231043
2015-03-03 08:49:28 +08:00
|
|
|
Value *Bit = createBitSetTest(ThenB, BSI, BAI, BitOffset);
|
2015-02-21 04:30:47 +08:00
|
|
|
|
|
|
|
// The value we want is 0 if we came directly from the initial block
|
|
|
|
// (having failed the range or alignment checks), or the loaded bit if
|
|
|
|
// we came from the block in which we loaded it.
|
|
|
|
B.SetInsertPoint(CI);
|
|
|
|
PHINode *P = B.CreatePHI(Int1Ty, 2);
|
|
|
|
P->addIncoming(ConstantInt::get(Int1Ty, 0), InitialBB);
|
|
|
|
P->addIncoming(Bit, ThenB.GetInsertBlock());
|
2015-02-26 04:42:41 +08:00
|
|
|
return P;
|
2015-02-21 04:30:47 +08:00
|
|
|
}
|
|
|
|
|
|
|
|
/// Given a disjoint set of bitsets and globals, layout the globals, build the
|
|
|
|
/// bit sets and lower the llvm.bitset.test calls.
|
|
|
|
void LowerBitSets::buildBitSetsFromGlobals(
|
|
|
|
const std::vector<MDString *> &BitSets,
|
|
|
|
const std::vector<GlobalVariable *> &Globals) {
|
|
|
|
// Build a new global with the combined contents of the referenced globals.
|
|
|
|
std::vector<Constant *> GlobalInits;
|
2015-03-10 10:37:25 +08:00
|
|
|
const DataLayout &DL = M->getDataLayout();
|
2015-02-26 04:42:41 +08:00
|
|
|
for (GlobalVariable *G : Globals) {
|
2015-02-21 04:30:47 +08:00
|
|
|
GlobalInits.push_back(G->getInitializer());
|
2015-03-10 10:37:25 +08:00
|
|
|
uint64_t InitSize = DL.getTypeAllocSize(G->getInitializer()->getType());
|
2015-02-26 04:42:41 +08:00
|
|
|
|
|
|
|
// Compute the amount of padding required to align the next element to the
|
|
|
|
// next power of 2.
|
|
|
|
uint64_t Padding = NextPowerOf2(InitSize - 1) - InitSize;
|
|
|
|
|
|
|
|
// Cap at 128 was found experimentally to have a good data/instruction
|
|
|
|
// overhead tradeoff.
|
|
|
|
if (Padding > 128)
|
|
|
|
Padding = RoundUpToAlignment(InitSize, 128) - InitSize;
|
|
|
|
|
|
|
|
GlobalInits.push_back(
|
|
|
|
ConstantAggregateZero::get(ArrayType::get(Int8Ty, Padding)));
|
|
|
|
}
|
|
|
|
if (!GlobalInits.empty())
|
|
|
|
GlobalInits.pop_back();
|
LowerBitSets: Use byte arrays instead of bit sets to represent in-memory bit sets.
By loading from indexed offsets into a byte array and applying a mask, a
program can test bits from the bit set with a relatively short instruction
sequence. For example, suppose we have 15 bit sets to lay out:
A (16 bits), B (15 bits), C (14 bits), D (13 bits), E (12 bits),
F (11 bits), G (10 bits), H (9 bits), I (7 bits), J (6 bits), K (5 bits),
L (4 bits), M (3 bits), N (2 bits), O (1 bit)
These bits can be laid out in a 16-byte array like this:
Byte Offset
0123456789ABCDEF
Bit
7 HHHHHHHHHIIIIIII
6 GGGGGGGGGGJJJJJJ
5 FFFFFFFFFFFKKKKK
4 EEEEEEEEEEEELLLL
3 DDDDDDDDDDDDDMMM
2 CCCCCCCCCCCCCCNN
1 BBBBBBBBBBBBBBBO
0 AAAAAAAAAAAAAAAA
For example, to test bit X of A, we evaluate ((bits[X] & 1) != 0), or to
test bit X of I, we evaluate ((bits[9 + X] & 0x80) != 0). This can be done
in 1-2 machine instructions on x86, or 4-6 instructions on ARM.
This uses the LPT multiprocessor scheduling algorithm to lay out the bits
efficiently.
Saves ~450KB of instructions in a recent build of Chromium.
Differential Revision: http://reviews.llvm.org/D7954
llvm-svn: 231043
2015-03-03 08:49:28 +08:00
|
|
|
Constant *NewInit = ConstantStruct::getAnon(M->getContext(), GlobalInits);
|
2015-02-21 04:30:47 +08:00
|
|
|
auto CombinedGlobal =
|
LowerBitSets: Use byte arrays instead of bit sets to represent in-memory bit sets.
By loading from indexed offsets into a byte array and applying a mask, a
program can test bits from the bit set with a relatively short instruction
sequence. For example, suppose we have 15 bit sets to lay out:
A (16 bits), B (15 bits), C (14 bits), D (13 bits), E (12 bits),
F (11 bits), G (10 bits), H (9 bits), I (7 bits), J (6 bits), K (5 bits),
L (4 bits), M (3 bits), N (2 bits), O (1 bit)
These bits can be laid out in a 16-byte array like this:
Byte Offset
0123456789ABCDEF
Bit
7 HHHHHHHHHIIIIIII
6 GGGGGGGGGGJJJJJJ
5 FFFFFFFFFFFKKKKK
4 EEEEEEEEEEEELLLL
3 DDDDDDDDDDDDDMMM
2 CCCCCCCCCCCCCCNN
1 BBBBBBBBBBBBBBBO
0 AAAAAAAAAAAAAAAA
For example, to test bit X of A, we evaluate ((bits[X] & 1) != 0), or to
test bit X of I, we evaluate ((bits[9 + X] & 0x80) != 0). This can be done
in 1-2 machine instructions on x86, or 4-6 instructions on ARM.
This uses the LPT multiprocessor scheduling algorithm to lay out the bits
efficiently.
Saves ~450KB of instructions in a recent build of Chromium.
Differential Revision: http://reviews.llvm.org/D7954
llvm-svn: 231043
2015-03-03 08:49:28 +08:00
|
|
|
new GlobalVariable(*M, NewInit->getType(), /*isConstant=*/true,
|
2015-02-21 04:30:47 +08:00
|
|
|
GlobalValue::PrivateLinkage, NewInit);
|
|
|
|
|
|
|
|
const StructLayout *CombinedGlobalLayout =
|
2015-03-10 10:37:25 +08:00
|
|
|
DL.getStructLayout(cast<StructType>(NewInit->getType()));
|
2015-02-21 04:30:47 +08:00
|
|
|
|
|
|
|
// Compute the offsets of the original globals within the new global.
|
|
|
|
DenseMap<GlobalVariable *, uint64_t> GlobalLayout;
|
|
|
|
for (unsigned I = 0; I != Globals.size(); ++I)
|
2015-02-26 04:42:41 +08:00
|
|
|
// Multiply by 2 to account for padding elements.
|
|
|
|
GlobalLayout[Globals[I]] = CombinedGlobalLayout->getElementOffset(I * 2);
|
2015-02-21 04:30:47 +08:00
|
|
|
|
|
|
|
// For each bitset in this disjoint set...
|
|
|
|
for (MDString *BS : BitSets) {
|
|
|
|
// Build the bitset.
|
|
|
|
BitSetInfo BSI = buildBitSet(BS, GlobalLayout);
|
|
|
|
|
LowerBitSets: Use byte arrays instead of bit sets to represent in-memory bit sets.
By loading from indexed offsets into a byte array and applying a mask, a
program can test bits from the bit set with a relatively short instruction
sequence. For example, suppose we have 15 bit sets to lay out:
A (16 bits), B (15 bits), C (14 bits), D (13 bits), E (12 bits),
F (11 bits), G (10 bits), H (9 bits), I (7 bits), J (6 bits), K (5 bits),
L (4 bits), M (3 bits), N (2 bits), O (1 bit)
These bits can be laid out in a 16-byte array like this:
Byte Offset
0123456789ABCDEF
Bit
7 HHHHHHHHHIIIIIII
6 GGGGGGGGGGJJJJJJ
5 FFFFFFFFFFFKKKKK
4 EEEEEEEEEEEELLLL
3 DDDDDDDDDDDDDMMM
2 CCCCCCCCCCCCCCNN
1 BBBBBBBBBBBBBBBO
0 AAAAAAAAAAAAAAAA
For example, to test bit X of A, we evaluate ((bits[X] & 1) != 0), or to
test bit X of I, we evaluate ((bits[9 + X] & 0x80) != 0). This can be done
in 1-2 machine instructions on x86, or 4-6 instructions on ARM.
This uses the LPT multiprocessor scheduling algorithm to lay out the bits
efficiently.
Saves ~450KB of instructions in a recent build of Chromium.
Differential Revision: http://reviews.llvm.org/D7954
llvm-svn: 231043
2015-03-03 08:49:28 +08:00
|
|
|
ByteArrayInfo *BAI = 0;
|
2015-02-21 04:30:47 +08:00
|
|
|
|
|
|
|
// Lower each call to llvm.bitset.test for this bitset.
|
|
|
|
for (CallInst *CI : BitSetTestCallSites[BS]) {
|
|
|
|
++NumBitSetCallsLowered;
|
LowerBitSets: Use byte arrays instead of bit sets to represent in-memory bit sets.
By loading from indexed offsets into a byte array and applying a mask, a
program can test bits from the bit set with a relatively short instruction
sequence. For example, suppose we have 15 bit sets to lay out:
A (16 bits), B (15 bits), C (14 bits), D (13 bits), E (12 bits),
F (11 bits), G (10 bits), H (9 bits), I (7 bits), J (6 bits), K (5 bits),
L (4 bits), M (3 bits), N (2 bits), O (1 bit)
These bits can be laid out in a 16-byte array like this:
Byte Offset
0123456789ABCDEF
Bit
7 HHHHHHHHHIIIIIII
6 GGGGGGGGGGJJJJJJ
5 FFFFFFFFFFFKKKKK
4 EEEEEEEEEEEELLLL
3 DDDDDDDDDDDDDMMM
2 CCCCCCCCCCCCCCNN
1 BBBBBBBBBBBBBBBO
0 AAAAAAAAAAAAAAAA
For example, to test bit X of A, we evaluate ((bits[X] & 1) != 0), or to
test bit X of I, we evaluate ((bits[9 + X] & 0x80) != 0). This can be done
in 1-2 machine instructions on x86, or 4-6 instructions on ARM.
This uses the LPT multiprocessor scheduling algorithm to lay out the bits
efficiently.
Saves ~450KB of instructions in a recent build of Chromium.
Differential Revision: http://reviews.llvm.org/D7954
llvm-svn: 231043
2015-03-03 08:49:28 +08:00
|
|
|
Value *Lowered = lowerBitSetCall(CI, BSI, BAI, CombinedGlobal, GlobalLayout);
|
2015-02-26 04:42:41 +08:00
|
|
|
CI->replaceAllUsesWith(Lowered);
|
|
|
|
CI->eraseFromParent();
|
2015-02-21 04:30:47 +08:00
|
|
|
}
|
|
|
|
}
|
|
|
|
|
|
|
|
// Build aliases pointing to offsets into the combined global for each
|
|
|
|
// global from which we built the combined global, and replace references
|
|
|
|
// to the original globals with references to the aliases.
|
|
|
|
for (unsigned I = 0; I != Globals.size(); ++I) {
|
2015-02-26 04:42:41 +08:00
|
|
|
// Multiply by 2 to account for padding elements.
|
2015-02-21 04:30:47 +08:00
|
|
|
Constant *CombinedGlobalIdxs[] = {ConstantInt::get(Int32Ty, 0),
|
2015-02-26 04:42:41 +08:00
|
|
|
ConstantInt::get(Int32Ty, I * 2)};
|
2015-02-21 04:30:47 +08:00
|
|
|
Constant *CombinedGlobalElemPtr =
|
|
|
|
ConstantExpr::getGetElementPtr(CombinedGlobal, CombinedGlobalIdxs);
|
2015-03-17 07:36:24 +08:00
|
|
|
if (LinkerSubsectionsViaSymbols) {
|
|
|
|
Globals[I]->replaceAllUsesWith(CombinedGlobalElemPtr);
|
|
|
|
} else {
|
|
|
|
GlobalAlias *GAlias = GlobalAlias::create(
|
|
|
|
Globals[I]->getType()->getElementType(),
|
|
|
|
Globals[I]->getType()->getAddressSpace(), Globals[I]->getLinkage(),
|
|
|
|
"", CombinedGlobalElemPtr, M);
|
|
|
|
GAlias->takeName(Globals[I]);
|
|
|
|
Globals[I]->replaceAllUsesWith(GAlias);
|
|
|
|
}
|
2015-02-21 04:30:47 +08:00
|
|
|
Globals[I]->eraseFromParent();
|
|
|
|
}
|
|
|
|
}
|
|
|
|
|
|
|
|
/// Lower all bit sets in this module.
|
LowerBitSets: Use byte arrays instead of bit sets to represent in-memory bit sets.
By loading from indexed offsets into a byte array and applying a mask, a
program can test bits from the bit set with a relatively short instruction
sequence. For example, suppose we have 15 bit sets to lay out:
A (16 bits), B (15 bits), C (14 bits), D (13 bits), E (12 bits),
F (11 bits), G (10 bits), H (9 bits), I (7 bits), J (6 bits), K (5 bits),
L (4 bits), M (3 bits), N (2 bits), O (1 bit)
These bits can be laid out in a 16-byte array like this:
Byte Offset
0123456789ABCDEF
Bit
7 HHHHHHHHHIIIIIII
6 GGGGGGGGGGJJJJJJ
5 FFFFFFFFFFFKKKKK
4 EEEEEEEEEEEELLLL
3 DDDDDDDDDDDDDMMM
2 CCCCCCCCCCCCCCNN
1 BBBBBBBBBBBBBBBO
0 AAAAAAAAAAAAAAAA
For example, to test bit X of A, we evaluate ((bits[X] & 1) != 0), or to
test bit X of I, we evaluate ((bits[9 + X] & 0x80) != 0). This can be done
in 1-2 machine instructions on x86, or 4-6 instructions on ARM.
This uses the LPT multiprocessor scheduling algorithm to lay out the bits
efficiently.
Saves ~450KB of instructions in a recent build of Chromium.
Differential Revision: http://reviews.llvm.org/D7954
llvm-svn: 231043
2015-03-03 08:49:28 +08:00
|
|
|
bool LowerBitSets::buildBitSets() {
|
2015-02-21 04:30:47 +08:00
|
|
|
Function *BitSetTestFunc =
|
LowerBitSets: Use byte arrays instead of bit sets to represent in-memory bit sets.
By loading from indexed offsets into a byte array and applying a mask, a
program can test bits from the bit set with a relatively short instruction
sequence. For example, suppose we have 15 bit sets to lay out:
A (16 bits), B (15 bits), C (14 bits), D (13 bits), E (12 bits),
F (11 bits), G (10 bits), H (9 bits), I (7 bits), J (6 bits), K (5 bits),
L (4 bits), M (3 bits), N (2 bits), O (1 bit)
These bits can be laid out in a 16-byte array like this:
Byte Offset
0123456789ABCDEF
Bit
7 HHHHHHHHHIIIIIII
6 GGGGGGGGGGJJJJJJ
5 FFFFFFFFFFFKKKKK
4 EEEEEEEEEEEELLLL
3 DDDDDDDDDDDDDMMM
2 CCCCCCCCCCCCCCNN
1 BBBBBBBBBBBBBBBO
0 AAAAAAAAAAAAAAAA
For example, to test bit X of A, we evaluate ((bits[X] & 1) != 0), or to
test bit X of I, we evaluate ((bits[9 + X] & 0x80) != 0). This can be done
in 1-2 machine instructions on x86, or 4-6 instructions on ARM.
This uses the LPT multiprocessor scheduling algorithm to lay out the bits
efficiently.
Saves ~450KB of instructions in a recent build of Chromium.
Differential Revision: http://reviews.llvm.org/D7954
llvm-svn: 231043
2015-03-03 08:49:28 +08:00
|
|
|
M->getFunction(Intrinsic::getName(Intrinsic::bitset_test));
|
2015-02-21 04:30:47 +08:00
|
|
|
if (!BitSetTestFunc)
|
|
|
|
return false;
|
|
|
|
|
|
|
|
// Equivalence class set containing bitsets and the globals they reference.
|
|
|
|
// This is used to partition the set of bitsets in the module into disjoint
|
|
|
|
// sets.
|
|
|
|
typedef EquivalenceClasses<PointerUnion<GlobalVariable *, MDString *>>
|
|
|
|
GlobalClassesTy;
|
|
|
|
GlobalClassesTy GlobalClasses;
|
|
|
|
|
|
|
|
for (const Use &U : BitSetTestFunc->uses()) {
|
|
|
|
auto CI = cast<CallInst>(U.getUser());
|
|
|
|
|
|
|
|
auto BitSetMDVal = dyn_cast<MetadataAsValue>(CI->getArgOperand(1));
|
|
|
|
if (!BitSetMDVal || !isa<MDString>(BitSetMDVal->getMetadata()))
|
|
|
|
report_fatal_error(
|
|
|
|
"Second argument of llvm.bitset.test must be metadata string");
|
|
|
|
auto BitSet = cast<MDString>(BitSetMDVal->getMetadata());
|
|
|
|
|
|
|
|
// Add the call site to the list of call sites for this bit set. We also use
|
|
|
|
// BitSetTestCallSites to keep track of whether we have seen this bit set
|
|
|
|
// before. If we have, we don't need to re-add the referenced globals to the
|
|
|
|
// equivalence class.
|
|
|
|
std::pair<DenseMap<MDString *, std::vector<CallInst *>>::iterator,
|
|
|
|
bool> Ins =
|
|
|
|
BitSetTestCallSites.insert(
|
|
|
|
std::make_pair(BitSet, std::vector<CallInst *>()));
|
|
|
|
Ins.first->second.push_back(CI);
|
|
|
|
if (!Ins.second)
|
|
|
|
continue;
|
|
|
|
|
|
|
|
// Add the bitset to the equivalence class.
|
|
|
|
GlobalClassesTy::iterator GCI = GlobalClasses.insert(BitSet);
|
|
|
|
GlobalClassesTy::member_iterator CurSet = GlobalClasses.findLeader(GCI);
|
|
|
|
|
|
|
|
if (!BitSetNM)
|
|
|
|
continue;
|
|
|
|
|
|
|
|
// Verify the bitset metadata and add the referenced globals to the bitset's
|
|
|
|
// equivalence class.
|
|
|
|
for (MDNode *Op : BitSetNM->operands()) {
|
|
|
|
if (Op->getNumOperands() != 3)
|
|
|
|
report_fatal_error(
|
|
|
|
"All operands of llvm.bitsets metadata must have 3 elements");
|
|
|
|
|
|
|
|
if (Op->getOperand(0) != BitSet || !Op->getOperand(1))
|
|
|
|
continue;
|
|
|
|
|
|
|
|
auto OpConstMD = dyn_cast<ConstantAsMetadata>(Op->getOperand(1));
|
|
|
|
if (!OpConstMD)
|
|
|
|
report_fatal_error("Bit set element must be a constant");
|
|
|
|
auto OpGlobal = dyn_cast<GlobalVariable>(OpConstMD->getValue());
|
|
|
|
if (!OpGlobal)
|
|
|
|
report_fatal_error("Bit set element must refer to global");
|
|
|
|
|
|
|
|
auto OffsetConstMD = dyn_cast<ConstantAsMetadata>(Op->getOperand(2));
|
|
|
|
if (!OffsetConstMD)
|
|
|
|
report_fatal_error("Bit set element offset must be a constant");
|
|
|
|
auto OffsetInt = dyn_cast<ConstantInt>(OffsetConstMD->getValue());
|
|
|
|
if (!OffsetInt)
|
|
|
|
report_fatal_error(
|
|
|
|
"Bit set element offset must be an integer constant");
|
|
|
|
|
|
|
|
CurSet = GlobalClasses.unionSets(
|
|
|
|
CurSet, GlobalClasses.findLeader(GlobalClasses.insert(OpGlobal)));
|
|
|
|
}
|
|
|
|
}
|
|
|
|
|
|
|
|
if (GlobalClasses.empty())
|
|
|
|
return false;
|
|
|
|
|
|
|
|
// For each disjoint set we found...
|
|
|
|
for (GlobalClassesTy::iterator I = GlobalClasses.begin(),
|
|
|
|
E = GlobalClasses.end();
|
|
|
|
I != E; ++I) {
|
|
|
|
if (!I->isLeader()) continue;
|
|
|
|
|
|
|
|
++NumBitSetDisjointSets;
|
|
|
|
|
|
|
|
// Build the list of bitsets and referenced globals in this disjoint set.
|
|
|
|
std::vector<MDString *> BitSets;
|
|
|
|
std::vector<GlobalVariable *> Globals;
|
2015-02-25 07:17:02 +08:00
|
|
|
llvm::DenseMap<MDString *, uint64_t> BitSetIndices;
|
|
|
|
llvm::DenseMap<GlobalVariable *, uint64_t> GlobalIndices;
|
2015-02-21 04:30:47 +08:00
|
|
|
for (GlobalClassesTy::member_iterator MI = GlobalClasses.member_begin(I);
|
|
|
|
MI != GlobalClasses.member_end(); ++MI) {
|
2015-02-25 07:17:02 +08:00
|
|
|
if ((*MI).is<MDString *>()) {
|
|
|
|
BitSetIndices[MI->get<MDString *>()] = BitSets.size();
|
2015-02-21 04:30:47 +08:00
|
|
|
BitSets.push_back(MI->get<MDString *>());
|
2015-02-25 07:17:02 +08:00
|
|
|
} else {
|
|
|
|
GlobalIndices[MI->get<GlobalVariable *>()] = Globals.size();
|
2015-02-21 04:30:47 +08:00
|
|
|
Globals.push_back(MI->get<GlobalVariable *>());
|
2015-02-25 07:17:02 +08:00
|
|
|
}
|
|
|
|
}
|
|
|
|
|
|
|
|
// For each bitset, build a set of indices that refer to globals referenced
|
|
|
|
// by the bitset.
|
|
|
|
std::vector<std::set<uint64_t>> BitSetMembers(BitSets.size());
|
|
|
|
if (BitSetNM) {
|
|
|
|
for (MDNode *Op : BitSetNM->operands()) {
|
|
|
|
// Op = { bitset name, global, offset }
|
|
|
|
if (!Op->getOperand(1))
|
|
|
|
continue;
|
|
|
|
auto I = BitSetIndices.find(cast<MDString>(Op->getOperand(0)));
|
|
|
|
if (I == BitSetIndices.end())
|
|
|
|
continue;
|
|
|
|
|
|
|
|
auto OpGlobal = cast<GlobalVariable>(
|
|
|
|
cast<ConstantAsMetadata>(Op->getOperand(1))->getValue());
|
|
|
|
BitSetMembers[I->second].insert(GlobalIndices[OpGlobal]);
|
|
|
|
}
|
2015-02-21 04:30:47 +08:00
|
|
|
}
|
|
|
|
|
2015-02-25 07:17:02 +08:00
|
|
|
// Order the sets of indices by size. The GlobalLayoutBuilder works best
|
|
|
|
// when given small index sets first.
|
|
|
|
std::stable_sort(
|
|
|
|
BitSetMembers.begin(), BitSetMembers.end(),
|
|
|
|
[](const std::set<uint64_t> &O1, const std::set<uint64_t> &O2) {
|
|
|
|
return O1.size() < O2.size();
|
|
|
|
});
|
|
|
|
|
|
|
|
// Create a GlobalLayoutBuilder and provide it with index sets as layout
|
|
|
|
// fragments. The GlobalLayoutBuilder tries to lay out members of fragments
|
|
|
|
// as close together as possible.
|
|
|
|
GlobalLayoutBuilder GLB(Globals.size());
|
|
|
|
for (auto &&MemSet : BitSetMembers)
|
|
|
|
GLB.addFragment(MemSet);
|
|
|
|
|
|
|
|
// Build a vector of globals with the computed layout.
|
|
|
|
std::vector<GlobalVariable *> OrderedGlobals(Globals.size());
|
|
|
|
auto OGI = OrderedGlobals.begin();
|
|
|
|
for (auto &&F : GLB.Fragments)
|
|
|
|
for (auto &&Offset : F)
|
|
|
|
*OGI++ = Globals[Offset];
|
|
|
|
|
|
|
|
// Order bitsets by name for determinism.
|
2015-02-21 04:30:47 +08:00
|
|
|
std::sort(BitSets.begin(), BitSets.end(), [](MDString *S1, MDString *S2) {
|
|
|
|
return S1->getString() < S2->getString();
|
|
|
|
});
|
|
|
|
|
|
|
|
// Build the bitsets from this disjoint set.
|
LowerBitSets: Use byte arrays instead of bit sets to represent in-memory bit sets.
By loading from indexed offsets into a byte array and applying a mask, a
program can test bits from the bit set with a relatively short instruction
sequence. For example, suppose we have 15 bit sets to lay out:
A (16 bits), B (15 bits), C (14 bits), D (13 bits), E (12 bits),
F (11 bits), G (10 bits), H (9 bits), I (7 bits), J (6 bits), K (5 bits),
L (4 bits), M (3 bits), N (2 bits), O (1 bit)
These bits can be laid out in a 16-byte array like this:
Byte Offset
0123456789ABCDEF
Bit
7 HHHHHHHHHIIIIIII
6 GGGGGGGGGGJJJJJJ
5 FFFFFFFFFFFKKKKK
4 EEEEEEEEEEEELLLL
3 DDDDDDDDDDDDDMMM
2 CCCCCCCCCCCCCCNN
1 BBBBBBBBBBBBBBBO
0 AAAAAAAAAAAAAAAA
For example, to test bit X of A, we evaluate ((bits[X] & 1) != 0), or to
test bit X of I, we evaluate ((bits[9 + X] & 0x80) != 0). This can be done
in 1-2 machine instructions on x86, or 4-6 instructions on ARM.
This uses the LPT multiprocessor scheduling algorithm to lay out the bits
efficiently.
Saves ~450KB of instructions in a recent build of Chromium.
Differential Revision: http://reviews.llvm.org/D7954
llvm-svn: 231043
2015-03-03 08:49:28 +08:00
|
|
|
buildBitSetsFromGlobals(BitSets, OrderedGlobals);
|
2015-02-21 04:30:47 +08:00
|
|
|
}
|
|
|
|
|
LowerBitSets: Use byte arrays instead of bit sets to represent in-memory bit sets.
By loading from indexed offsets into a byte array and applying a mask, a
program can test bits from the bit set with a relatively short instruction
sequence. For example, suppose we have 15 bit sets to lay out:
A (16 bits), B (15 bits), C (14 bits), D (13 bits), E (12 bits),
F (11 bits), G (10 bits), H (9 bits), I (7 bits), J (6 bits), K (5 bits),
L (4 bits), M (3 bits), N (2 bits), O (1 bit)
These bits can be laid out in a 16-byte array like this:
Byte Offset
0123456789ABCDEF
Bit
7 HHHHHHHHHIIIIIII
6 GGGGGGGGGGJJJJJJ
5 FFFFFFFFFFFKKKKK
4 EEEEEEEEEEEELLLL
3 DDDDDDDDDDDDDMMM
2 CCCCCCCCCCCCCCNN
1 BBBBBBBBBBBBBBBO
0 AAAAAAAAAAAAAAAA
For example, to test bit X of A, we evaluate ((bits[X] & 1) != 0), or to
test bit X of I, we evaluate ((bits[9 + X] & 0x80) != 0). This can be done
in 1-2 machine instructions on x86, or 4-6 instructions on ARM.
This uses the LPT multiprocessor scheduling algorithm to lay out the bits
efficiently.
Saves ~450KB of instructions in a recent build of Chromium.
Differential Revision: http://reviews.llvm.org/D7954
llvm-svn: 231043
2015-03-03 08:49:28 +08:00
|
|
|
allocateByteArrays();
|
|
|
|
|
2015-02-21 04:30:47 +08:00
|
|
|
return true;
|
|
|
|
}
|
|
|
|
|
LowerBitSets: Use byte arrays instead of bit sets to represent in-memory bit sets.
By loading from indexed offsets into a byte array and applying a mask, a
program can test bits from the bit set with a relatively short instruction
sequence. For example, suppose we have 15 bit sets to lay out:
A (16 bits), B (15 bits), C (14 bits), D (13 bits), E (12 bits),
F (11 bits), G (10 bits), H (9 bits), I (7 bits), J (6 bits), K (5 bits),
L (4 bits), M (3 bits), N (2 bits), O (1 bit)
These bits can be laid out in a 16-byte array like this:
Byte Offset
0123456789ABCDEF
Bit
7 HHHHHHHHHIIIIIII
6 GGGGGGGGGGJJJJJJ
5 FFFFFFFFFFFKKKKK
4 EEEEEEEEEEEELLLL
3 DDDDDDDDDDDDDMMM
2 CCCCCCCCCCCCCCNN
1 BBBBBBBBBBBBBBBO
0 AAAAAAAAAAAAAAAA
For example, to test bit X of A, we evaluate ((bits[X] & 1) != 0), or to
test bit X of I, we evaluate ((bits[9 + X] & 0x80) != 0). This can be done
in 1-2 machine instructions on x86, or 4-6 instructions on ARM.
This uses the LPT multiprocessor scheduling algorithm to lay out the bits
efficiently.
Saves ~450KB of instructions in a recent build of Chromium.
Differential Revision: http://reviews.llvm.org/D7954
llvm-svn: 231043
2015-03-03 08:49:28 +08:00
|
|
|
bool LowerBitSets::eraseBitSetMetadata() {
|
2015-02-21 04:30:47 +08:00
|
|
|
if (!BitSetNM)
|
|
|
|
return false;
|
|
|
|
|
LowerBitSets: Use byte arrays instead of bit sets to represent in-memory bit sets.
By loading from indexed offsets into a byte array and applying a mask, a
program can test bits from the bit set with a relatively short instruction
sequence. For example, suppose we have 15 bit sets to lay out:
A (16 bits), B (15 bits), C (14 bits), D (13 bits), E (12 bits),
F (11 bits), G (10 bits), H (9 bits), I (7 bits), J (6 bits), K (5 bits),
L (4 bits), M (3 bits), N (2 bits), O (1 bit)
These bits can be laid out in a 16-byte array like this:
Byte Offset
0123456789ABCDEF
Bit
7 HHHHHHHHHIIIIIII
6 GGGGGGGGGGJJJJJJ
5 FFFFFFFFFFFKKKKK
4 EEEEEEEEEEEELLLL
3 DDDDDDDDDDDDDMMM
2 CCCCCCCCCCCCCCNN
1 BBBBBBBBBBBBBBBO
0 AAAAAAAAAAAAAAAA
For example, to test bit X of A, we evaluate ((bits[X] & 1) != 0), or to
test bit X of I, we evaluate ((bits[9 + X] & 0x80) != 0). This can be done
in 1-2 machine instructions on x86, or 4-6 instructions on ARM.
This uses the LPT multiprocessor scheduling algorithm to lay out the bits
efficiently.
Saves ~450KB of instructions in a recent build of Chromium.
Differential Revision: http://reviews.llvm.org/D7954
llvm-svn: 231043
2015-03-03 08:49:28 +08:00
|
|
|
M->eraseNamedMetadata(BitSetNM);
|
2015-02-21 04:30:47 +08:00
|
|
|
return true;
|
|
|
|
}
|
|
|
|
|
|
|
|
bool LowerBitSets::runOnModule(Module &M) {
|
LowerBitSets: Use byte arrays instead of bit sets to represent in-memory bit sets.
By loading from indexed offsets into a byte array and applying a mask, a
program can test bits from the bit set with a relatively short instruction
sequence. For example, suppose we have 15 bit sets to lay out:
A (16 bits), B (15 bits), C (14 bits), D (13 bits), E (12 bits),
F (11 bits), G (10 bits), H (9 bits), I (7 bits), J (6 bits), K (5 bits),
L (4 bits), M (3 bits), N (2 bits), O (1 bit)
These bits can be laid out in a 16-byte array like this:
Byte Offset
0123456789ABCDEF
Bit
7 HHHHHHHHHIIIIIII
6 GGGGGGGGGGJJJJJJ
5 FFFFFFFFFFFKKKKK
4 EEEEEEEEEEEELLLL
3 DDDDDDDDDDDDDMMM
2 CCCCCCCCCCCCCCNN
1 BBBBBBBBBBBBBBBO
0 AAAAAAAAAAAAAAAA
For example, to test bit X of A, we evaluate ((bits[X] & 1) != 0), or to
test bit X of I, we evaluate ((bits[9 + X] & 0x80) != 0). This can be done
in 1-2 machine instructions on x86, or 4-6 instructions on ARM.
This uses the LPT multiprocessor scheduling algorithm to lay out the bits
efficiently.
Saves ~450KB of instructions in a recent build of Chromium.
Differential Revision: http://reviews.llvm.org/D7954
llvm-svn: 231043
2015-03-03 08:49:28 +08:00
|
|
|
bool Changed = buildBitSets();
|
|
|
|
Changed |= eraseBitSetMetadata();
|
2015-02-21 04:30:47 +08:00
|
|
|
return Changed;
|
|
|
|
}
|