2015-08-27 08:37:40 +08:00
|
|
|
; RUN: llc -march=x86-64 -print-machineinstrs=expand-isel-pseudos %s -o /dev/null 2>&1 | FileCheck %s
|
|
|
|
|
|
Distribute the weight on the edge from switch to default statement to edges generated in lowering switch.
Currently, when edge weights are assigned to edges that are created when lowering switch statement, the weight on the edge to default statement (let's call it "default weight" here) is not considered. We need to distribute this weight properly. However, without value profiling, we have no idea how to distribute it. In this patch, I applied the heuristic that this weight is evenly distributed to successors.
For example, given a switch statement with cases 1,2,3,5,10,11,20, and every edge from switch to each successor has weight 10. If there is a binary search tree built to test if n < 10, then its two out-edges will have weight 4x10+10/2 = 45 and 3x10 + 10/2 = 35 respectively (currently they are 40 and 30 without considering the default weight). Each distribution (which is 5 here) will be stored in each SwitchWorkListItem for further distribution.
There are some exceptions:
For a jump table header which doesn't have any edge to default statement, we don't distribute the default weight to it.
For a bit test header which covers a contiguous range and hence has no edges to default statement, we don't distribute the default weight to it.
When the branch checks a single value or a contiguous range with no edge to default statement, we don't distribute the default weight to it.
In other cases, the default weight is evenly distributed to successors.
Differential Revision: http://reviews.llvm.org/D12418
llvm-svn: 246522
2015-09-01 09:42:16 +08:00
|
|
|
declare void @foo(i32)
|
2015-08-27 08:37:40 +08:00
|
|
|
|
|
|
|
|
; CHECK-LABEL: test
|
|
|
|
|
|
|
|
|
|
define void @test(i32 %x) nounwind {
|
|
|
|
|
entry:
|
|
|
|
|
switch i32 %x, label %sw.default [
|
Distribute the weight on the edge from switch to default statement to edges generated in lowering switch.
Currently, when edge weights are assigned to edges that are created when lowering switch statement, the weight on the edge to default statement (let's call it "default weight" here) is not considered. We need to distribute this weight properly. However, without value profiling, we have no idea how to distribute it. In this patch, I applied the heuristic that this weight is evenly distributed to successors.
For example, given a switch statement with cases 1,2,3,5,10,11,20, and every edge from switch to each successor has weight 10. If there is a binary search tree built to test if n < 10, then its two out-edges will have weight 4x10+10/2 = 45 and 3x10 + 10/2 = 35 respectively (currently they are 40 and 30 without considering the default weight). Each distribution (which is 5 here) will be stored in each SwitchWorkListItem for further distribution.
There are some exceptions:
For a jump table header which doesn't have any edge to default statement, we don't distribute the default weight to it.
For a bit test header which covers a contiguous range and hence has no edges to default statement, we don't distribute the default weight to it.
When the branch checks a single value or a contiguous range with no edge to default statement, we don't distribute the default weight to it.
In other cases, the default weight is evenly distributed to successors.
Differential Revision: http://reviews.llvm.org/D12418
llvm-svn: 246522
2015-09-01 09:42:16 +08:00
|
|
|
i32 1, label %sw.bb
|
|
|
|
|
i32 155, label %sw.bb
|
|
|
|
|
i32 156, label %sw.bb
|
|
|
|
|
i32 157, label %sw.bb
|
|
|
|
|
i32 158, label %sw.bb
|
|
|
|
|
i32 159, label %sw.bb
|
|
|
|
|
i32 1134, label %sw.bb
|
|
|
|
|
i32 1140, label %sw.bb
|
2015-08-27 08:37:40 +08:00
|
|
|
], !prof !1
|
|
|
|
|
|
|
|
|
|
sw.bb:
|
|
|
|
|
call void @foo(i32 0)
|
|
|
|
|
br label %sw.epilog
|
|
|
|
|
|
|
|
|
|
sw.default:
|
|
|
|
|
call void @foo(i32 1)
|
|
|
|
|
br label %sw.epilog
|
|
|
|
|
|
|
|
|
|
sw.epilog:
|
|
|
|
|
ret void
|
|
|
|
|
|
|
|
|
|
; Check if weights are correctly assigned to edges generated from switch
|
|
|
|
|
; statement.
|
|
|
|
|
;
|
Distribute the weight on the edge from switch to default statement to edges generated in lowering switch.
Currently, when edge weights are assigned to edges that are created when lowering switch statement, the weight on the edge to default statement (let's call it "default weight" here) is not considered. We need to distribute this weight properly. However, without value profiling, we have no idea how to distribute it. In this patch, I applied the heuristic that this weight is evenly distributed to successors.
For example, given a switch statement with cases 1,2,3,5,10,11,20, and every edge from switch to each successor has weight 10. If there is a binary search tree built to test if n < 10, then its two out-edges will have weight 4x10+10/2 = 45 and 3x10 + 10/2 = 35 respectively (currently they are 40 and 30 without considering the default weight). Each distribution (which is 5 here) will be stored in each SwitchWorkListItem for further distribution.
There are some exceptions:
For a jump table header which doesn't have any edge to default statement, we don't distribute the default weight to it.
For a bit test header which covers a contiguous range and hence has no edges to default statement, we don't distribute the default weight to it.
When the branch checks a single value or a contiguous range with no edge to default statement, we don't distribute the default weight to it.
In other cases, the default weight is evenly distributed to successors.
Differential Revision: http://reviews.llvm.org/D12418
llvm-svn: 246522
2015-09-01 09:42:16 +08:00
|
|
|
; CHECK: BB#0:
|
|
|
|
|
; BB#0 to BB#4: [0, 1133] (65 = 60 + 5)
|
|
|
|
|
; BB#0 to BB#5: [1134, UINT32_MAX] (25 = 20 + 5)
|
2015-12-01 13:29:22 +08:00
|
|
|
; CHECK: Successors according to CFG: BB#4({{[0-9a-fx/= ]+}}72.22%) BB#5({{[0-9a-fx/= ]+}}27.78%)
|
Distribute the weight on the edge from switch to default statement to edges generated in lowering switch.
Currently, when edge weights are assigned to edges that are created when lowering switch statement, the weight on the edge to default statement (let's call it "default weight" here) is not considered. We need to distribute this weight properly. However, without value profiling, we have no idea how to distribute it. In this patch, I applied the heuristic that this weight is evenly distributed to successors.
For example, given a switch statement with cases 1,2,3,5,10,11,20, and every edge from switch to each successor has weight 10. If there is a binary search tree built to test if n < 10, then its two out-edges will have weight 4x10+10/2 = 45 and 3x10 + 10/2 = 35 respectively (currently they are 40 and 30 without considering the default weight). Each distribution (which is 5 here) will be stored in each SwitchWorkListItem for further distribution.
There are some exceptions:
For a jump table header which doesn't have any edge to default statement, we don't distribute the default weight to it.
For a bit test header which covers a contiguous range and hence has no edges to default statement, we don't distribute the default weight to it.
When the branch checks a single value or a contiguous range with no edge to default statement, we don't distribute the default weight to it.
In other cases, the default weight is evenly distributed to successors.
Differential Revision: http://reviews.llvm.org/D12418
llvm-svn: 246522
2015-09-01 09:42:16 +08:00
|
|
|
;
|
2015-08-27 08:37:40 +08:00
|
|
|
; CHECK: BB#4:
|
Distribute the weight on the edge from switch to default statement to edges generated in lowering switch.
Currently, when edge weights are assigned to edges that are created when lowering switch statement, the weight on the edge to default statement (let's call it "default weight" here) is not considered. We need to distribute this weight properly. However, without value profiling, we have no idea how to distribute it. In this patch, I applied the heuristic that this weight is evenly distributed to successors.
For example, given a switch statement with cases 1,2,3,5,10,11,20, and every edge from switch to each successor has weight 10. If there is a binary search tree built to test if n < 10, then its two out-edges will have weight 4x10+10/2 = 45 and 3x10 + 10/2 = 35 respectively (currently they are 40 and 30 without considering the default weight). Each distribution (which is 5 here) will be stored in each SwitchWorkListItem for further distribution.
There are some exceptions:
For a jump table header which doesn't have any edge to default statement, we don't distribute the default weight to it.
For a bit test header which covers a contiguous range and hence has no edges to default statement, we don't distribute the default weight to it.
When the branch checks a single value or a contiguous range with no edge to default statement, we don't distribute the default weight to it.
In other cases, the default weight is evenly distributed to successors.
Differential Revision: http://reviews.llvm.org/D12418
llvm-svn: 246522
2015-09-01 09:42:16 +08:00
|
|
|
; BB#4 to BB#1: [155, 159] (50)
|
|
|
|
|
; BB#4 to BB#5: [0, 1133] - [155, 159] (15 = 10 + 5)
|
2015-12-01 13:29:22 +08:00
|
|
|
; CHECK: Successors according to CFG: BB#1({{[0-9a-fx/= ]+}}76.92%) BB#7({{[0-9a-fx/= ]+}}23.08%)
|
Distribute the weight on the edge from switch to default statement to edges generated in lowering switch.
Currently, when edge weights are assigned to edges that are created when lowering switch statement, the weight on the edge to default statement (let's call it "default weight" here) is not considered. We need to distribute this weight properly. However, without value profiling, we have no idea how to distribute it. In this patch, I applied the heuristic that this weight is evenly distributed to successors.
For example, given a switch statement with cases 1,2,3,5,10,11,20, and every edge from switch to each successor has weight 10. If there is a binary search tree built to test if n < 10, then its two out-edges will have weight 4x10+10/2 = 45 and 3x10 + 10/2 = 35 respectively (currently they are 40 and 30 without considering the default weight). Each distribution (which is 5 here) will be stored in each SwitchWorkListItem for further distribution.
There are some exceptions:
For a jump table header which doesn't have any edge to default statement, we don't distribute the default weight to it.
For a bit test header which covers a contiguous range and hence has no edges to default statement, we don't distribute the default weight to it.
When the branch checks a single value or a contiguous range with no edge to default statement, we don't distribute the default weight to it.
In other cases, the default weight is evenly distributed to successors.
Differential Revision: http://reviews.llvm.org/D12418
llvm-svn: 246522
2015-09-01 09:42:16 +08:00
|
|
|
;
|
|
|
|
|
; CHECK: BB#5:
|
|
|
|
|
; BB#5 to BB#1: {1140} (10)
|
|
|
|
|
; BB#5 to BB#6: [1134, UINT32_MAX] - {1140} (15 = 10 + 5)
|
2015-12-01 13:29:22 +08:00
|
|
|
; CHECK: Successors according to CFG: BB#1({{[0-9a-fx/= ]+}}40.00%) BB#6({{[0-9a-fx/= ]+}}60.00%)
|
Distribute the weight on the edge from switch to default statement to edges generated in lowering switch.
Currently, when edge weights are assigned to edges that are created when lowering switch statement, the weight on the edge to default statement (let's call it "default weight" here) is not considered. We need to distribute this weight properly. However, without value profiling, we have no idea how to distribute it. In this patch, I applied the heuristic that this weight is evenly distributed to successors.
For example, given a switch statement with cases 1,2,3,5,10,11,20, and every edge from switch to each successor has weight 10. If there is a binary search tree built to test if n < 10, then its two out-edges will have weight 4x10+10/2 = 45 and 3x10 + 10/2 = 35 respectively (currently they are 40 and 30 without considering the default weight). Each distribution (which is 5 here) will be stored in each SwitchWorkListItem for further distribution.
There are some exceptions:
For a jump table header which doesn't have any edge to default statement, we don't distribute the default weight to it.
For a bit test header which covers a contiguous range and hence has no edges to default statement, we don't distribute the default weight to it.
When the branch checks a single value or a contiguous range with no edge to default statement, we don't distribute the default weight to it.
In other cases, the default weight is evenly distributed to successors.
Differential Revision: http://reviews.llvm.org/D12418
llvm-svn: 246522
2015-09-01 09:42:16 +08:00
|
|
|
;
|
2015-08-27 08:37:40 +08:00
|
|
|
; CHECK: BB#6:
|
Distribute the weight on the edge from switch to default statement to edges generated in lowering switch.
Currently, when edge weights are assigned to edges that are created when lowering switch statement, the weight on the edge to default statement (let's call it "default weight" here) is not considered. We need to distribute this weight properly. However, without value profiling, we have no idea how to distribute it. In this patch, I applied the heuristic that this weight is evenly distributed to successors.
For example, given a switch statement with cases 1,2,3,5,10,11,20, and every edge from switch to each successor has weight 10. If there is a binary search tree built to test if n < 10, then its two out-edges will have weight 4x10+10/2 = 45 and 3x10 + 10/2 = 35 respectively (currently they are 40 and 30 without considering the default weight). Each distribution (which is 5 here) will be stored in each SwitchWorkListItem for further distribution.
There are some exceptions:
For a jump table header which doesn't have any edge to default statement, we don't distribute the default weight to it.
For a bit test header which covers a contiguous range and hence has no edges to default statement, we don't distribute the default weight to it.
When the branch checks a single value or a contiguous range with no edge to default statement, we don't distribute the default weight to it.
In other cases, the default weight is evenly distributed to successors.
Differential Revision: http://reviews.llvm.org/D12418
llvm-svn: 246522
2015-09-01 09:42:16 +08:00
|
|
|
; BB#6 to BB#1: {1134} (10)
|
|
|
|
|
; BB#6 to BB#2: [1134, UINT32_MAX] - {1134, 1140} (5)
|
2015-12-01 13:29:22 +08:00
|
|
|
; CHECK: Successors according to CFG: BB#1({{[0-9a-fx/= ]+}}66.67%) BB#2({{[0-9a-fx/= ]+}}33.33%)
|
2015-08-27 08:37:40 +08:00
|
|
|
}
|
|
|
|
|
|
Distribute the weight on the edge from switch to default statement to edges generated in lowering switch.
Currently, when edge weights are assigned to edges that are created when lowering switch statement, the weight on the edge to default statement (let's call it "default weight" here) is not considered. We need to distribute this weight properly. However, without value profiling, we have no idea how to distribute it. In this patch, I applied the heuristic that this weight is evenly distributed to successors.
For example, given a switch statement with cases 1,2,3,5,10,11,20, and every edge from switch to each successor has weight 10. If there is a binary search tree built to test if n < 10, then its two out-edges will have weight 4x10+10/2 = 45 and 3x10 + 10/2 = 35 respectively (currently they are 40 and 30 without considering the default weight). Each distribution (which is 5 here) will be stored in each SwitchWorkListItem for further distribution.
There are some exceptions:
For a jump table header which doesn't have any edge to default statement, we don't distribute the default weight to it.
For a bit test header which covers a contiguous range and hence has no edges to default statement, we don't distribute the default weight to it.
When the branch checks a single value or a contiguous range with no edge to default statement, we don't distribute the default weight to it.
In other cases, the default weight is evenly distributed to successors.
Differential Revision: http://reviews.llvm.org/D12418
llvm-svn: 246522
2015-09-01 09:42:16 +08:00
|
|
|
; CHECK-LABEL: test2
|
|
|
|
|
|
|
|
|
|
define void @test2(i32 %x) nounwind {
|
|
|
|
|
entry:
|
|
|
|
|
|
|
|
|
|
; In this switch statement, there is an edge from jump table to default
|
|
|
|
|
; statement.
|
|
|
|
|
|
|
|
|
|
switch i32 %x, label %sw.default [
|
|
|
|
|
i32 1, label %sw.bb
|
|
|
|
|
i32 10, label %sw.bb2
|
|
|
|
|
i32 11, label %sw.bb3
|
|
|
|
|
i32 12, label %sw.bb4
|
|
|
|
|
i32 13, label %sw.bb5
|
|
|
|
|
i32 14, label %sw.bb5
|
|
|
|
|
], !prof !3
|
|
|
|
|
|
|
|
|
|
sw.bb:
|
|
|
|
|
call void @foo(i32 0)
|
|
|
|
|
br label %sw.epilog
|
|
|
|
|
|
|
|
|
|
sw.bb2:
|
|
|
|
|
call void @foo(i32 2)
|
|
|
|
|
br label %sw.epilog
|
|
|
|
|
|
|
|
|
|
sw.bb3:
|
|
|
|
|
call void @foo(i32 3)
|
|
|
|
|
br label %sw.epilog
|
|
|
|
|
|
|
|
|
|
sw.bb4:
|
|
|
|
|
call void @foo(i32 4)
|
|
|
|
|
br label %sw.epilog
|
|
|
|
|
|
|
|
|
|
sw.bb5:
|
|
|
|
|
call void @foo(i32 5)
|
|
|
|
|
br label %sw.epilog
|
|
|
|
|
|
|
|
|
|
sw.default:
|
|
|
|
|
call void @foo(i32 1)
|
|
|
|
|
br label %sw.epilog
|
|
|
|
|
|
|
|
|
|
sw.epilog:
|
|
|
|
|
ret void
|
|
|
|
|
|
|
|
|
|
; Check if weights are correctly assigned to edges generated from switch
|
|
|
|
|
; statement.
|
|
|
|
|
;
|
|
|
|
|
; CHECK: BB#0:
|
|
|
|
|
; BB#0 to BB#6: {0} + [15, UINT32_MAX] (5)
|
|
|
|
|
; BB#0 to BB#8: [1, 14] (jump table) (65 = 60 + 5)
|
2015-12-01 13:29:22 +08:00
|
|
|
; CHECK: Successors according to CFG: BB#6({{[0-9a-fx/= ]+}}7.14%) BB#8({{[0-9a-fx/= ]+}}92.86%
|
Distribute the weight on the edge from switch to default statement to edges generated in lowering switch.
Currently, when edge weights are assigned to edges that are created when lowering switch statement, the weight on the edge to default statement (let's call it "default weight" here) is not considered. We need to distribute this weight properly. However, without value profiling, we have no idea how to distribute it. In this patch, I applied the heuristic that this weight is evenly distributed to successors.
For example, given a switch statement with cases 1,2,3,5,10,11,20, and every edge from switch to each successor has weight 10. If there is a binary search tree built to test if n < 10, then its two out-edges will have weight 4x10+10/2 = 45 and 3x10 + 10/2 = 35 respectively (currently they are 40 and 30 without considering the default weight). Each distribution (which is 5 here) will be stored in each SwitchWorkListItem for further distribution.
There are some exceptions:
For a jump table header which doesn't have any edge to default statement, we don't distribute the default weight to it.
For a bit test header which covers a contiguous range and hence has no edges to default statement, we don't distribute the default weight to it.
When the branch checks a single value or a contiguous range with no edge to default statement, we don't distribute the default weight to it.
In other cases, the default weight is evenly distributed to successors.
Differential Revision: http://reviews.llvm.org/D12418
llvm-svn: 246522
2015-09-01 09:42:16 +08:00
|
|
|
;
|
|
|
|
|
; CHECK: BB#8:
|
|
|
|
|
; BB#8 to BB#1: {1} (10)
|
|
|
|
|
; BB#8 to BB#6: [2, 9] (5)
|
|
|
|
|
; BB#8 to BB#2: {10} (10)
|
|
|
|
|
; BB#8 to BB#3: {11} (10)
|
|
|
|
|
; BB#8 to BB#4: {12} (10)
|
|
|
|
|
; BB#8 to BB#5: {13, 14} (20)
|
2015-12-05 13:00:55 +08:00
|
|
|
; CHECK: Successors according to CFG: BB#1({{[0-9a-fx/= ]+}}15.38%) BB#6({{[0-9a-fx/= ]+}}7.69%) BB#2({{[0-9a-fx/= ]+}}15.38%) BB#3({{[0-9a-fx/= ]+}}15.38%) BB#4({{[0-9a-fx/= ]+}}15.38%) BB#5({{[0-9a-fx/= ]+}}30.77%)
|
Distribute the weight on the edge from switch to default statement to edges generated in lowering switch.
Currently, when edge weights are assigned to edges that are created when lowering switch statement, the weight on the edge to default statement (let's call it "default weight" here) is not considered. We need to distribute this weight properly. However, without value profiling, we have no idea how to distribute it. In this patch, I applied the heuristic that this weight is evenly distributed to successors.
For example, given a switch statement with cases 1,2,3,5,10,11,20, and every edge from switch to each successor has weight 10. If there is a binary search tree built to test if n < 10, then its two out-edges will have weight 4x10+10/2 = 45 and 3x10 + 10/2 = 35 respectively (currently they are 40 and 30 without considering the default weight). Each distribution (which is 5 here) will be stored in each SwitchWorkListItem for further distribution.
There are some exceptions:
For a jump table header which doesn't have any edge to default statement, we don't distribute the default weight to it.
For a bit test header which covers a contiguous range and hence has no edges to default statement, we don't distribute the default weight to it.
When the branch checks a single value or a contiguous range with no edge to default statement, we don't distribute the default weight to it.
In other cases, the default weight is evenly distributed to successors.
Differential Revision: http://reviews.llvm.org/D12418
llvm-svn: 246522
2015-09-01 09:42:16 +08:00
|
|
|
}
|
|
|
|
|
|
|
|
|
|
; CHECK-LABEL: test3
|
|
|
|
|
|
|
|
|
|
define void @test3(i32 %x) nounwind {
|
|
|
|
|
entry:
|
|
|
|
|
|
|
|
|
|
; In this switch statement, there is no edge from jump table to default
|
|
|
|
|
; statement.
|
|
|
|
|
|
|
|
|
|
switch i32 %x, label %sw.default [
|
|
|
|
|
i32 10, label %sw.bb
|
|
|
|
|
i32 11, label %sw.bb2
|
|
|
|
|
i32 12, label %sw.bb3
|
|
|
|
|
i32 13, label %sw.bb4
|
|
|
|
|
i32 14, label %sw.bb5
|
|
|
|
|
], !prof !2
|
|
|
|
|
|
|
|
|
|
sw.bb:
|
|
|
|
|
call void @foo(i32 0)
|
|
|
|
|
br label %sw.epilog
|
|
|
|
|
|
|
|
|
|
sw.bb2:
|
|
|
|
|
call void @foo(i32 2)
|
|
|
|
|
br label %sw.epilog
|
|
|
|
|
|
|
|
|
|
sw.bb3:
|
|
|
|
|
call void @foo(i32 3)
|
|
|
|
|
br label %sw.epilog
|
|
|
|
|
|
|
|
|
|
sw.bb4:
|
|
|
|
|
call void @foo(i32 4)
|
|
|
|
|
br label %sw.epilog
|
|
|
|
|
|
|
|
|
|
sw.bb5:
|
|
|
|
|
call void @foo(i32 5)
|
|
|
|
|
br label %sw.epilog
|
|
|
|
|
|
|
|
|
|
sw.default:
|
|
|
|
|
call void @foo(i32 1)
|
|
|
|
|
br label %sw.epilog
|
|
|
|
|
|
|
|
|
|
sw.epilog:
|
|
|
|
|
ret void
|
|
|
|
|
|
|
|
|
|
; Check if weights are correctly assigned to edges generated from switch
|
|
|
|
|
; statement.
|
|
|
|
|
;
|
|
|
|
|
; CHECK: BB#0:
|
|
|
|
|
; BB#0 to BB#6: [0, 9] + [15, UINT32_MAX] {10}
|
|
|
|
|
; BB#0 to BB#8: [10, 14] (jump table) (50)
|
2015-12-01 13:29:22 +08:00
|
|
|
; CHECK: Successors according to CFG: BB#6({{[0-9a-fx/= ]+}}16.67%) BB#8({{[0-9a-fx/= ]+}}83.33%)
|
Distribute the weight on the edge from switch to default statement to edges generated in lowering switch.
Currently, when edge weights are assigned to edges that are created when lowering switch statement, the weight on the edge to default statement (let's call it "default weight" here) is not considered. We need to distribute this weight properly. However, without value profiling, we have no idea how to distribute it. In this patch, I applied the heuristic that this weight is evenly distributed to successors.
For example, given a switch statement with cases 1,2,3,5,10,11,20, and every edge from switch to each successor has weight 10. If there is a binary search tree built to test if n < 10, then its two out-edges will have weight 4x10+10/2 = 45 and 3x10 + 10/2 = 35 respectively (currently they are 40 and 30 without considering the default weight). Each distribution (which is 5 here) will be stored in each SwitchWorkListItem for further distribution.
There are some exceptions:
For a jump table header which doesn't have any edge to default statement, we don't distribute the default weight to it.
For a bit test header which covers a contiguous range and hence has no edges to default statement, we don't distribute the default weight to it.
When the branch checks a single value or a contiguous range with no edge to default statement, we don't distribute the default weight to it.
In other cases, the default weight is evenly distributed to successors.
Differential Revision: http://reviews.llvm.org/D12418
llvm-svn: 246522
2015-09-01 09:42:16 +08:00
|
|
|
;
|
|
|
|
|
; CHECK: BB#8:
|
|
|
|
|
; BB#8 to BB#1: {10} (10)
|
|
|
|
|
; BB#8 to BB#2: {11} (10)
|
|
|
|
|
; BB#8 to BB#3: {12} (10)
|
|
|
|
|
; BB#8 to BB#4: {13} (10)
|
|
|
|
|
; BB#8 to BB#5: {14} (10)
|
2015-12-01 13:29:22 +08:00
|
|
|
; CHECK: Successors according to CFG: BB#1({{[0-9a-fx/= ]+}}20.00%) BB#2({{[0-9a-fx/= ]+}}20.00%) BB#3({{[0-9a-fx/= ]+}}20.00%) BB#4({{[0-9a-fx/= ]+}}20.00%) BB#5({{[0-9a-fx/= ]+}}20.00%)
|
Distribute the weight on the edge from switch to default statement to edges generated in lowering switch.
Currently, when edge weights are assigned to edges that are created when lowering switch statement, the weight on the edge to default statement (let's call it "default weight" here) is not considered. We need to distribute this weight properly. However, without value profiling, we have no idea how to distribute it. In this patch, I applied the heuristic that this weight is evenly distributed to successors.
For example, given a switch statement with cases 1,2,3,5,10,11,20, and every edge from switch to each successor has weight 10. If there is a binary search tree built to test if n < 10, then its two out-edges will have weight 4x10+10/2 = 45 and 3x10 + 10/2 = 35 respectively (currently they are 40 and 30 without considering the default weight). Each distribution (which is 5 here) will be stored in each SwitchWorkListItem for further distribution.
There are some exceptions:
For a jump table header which doesn't have any edge to default statement, we don't distribute the default weight to it.
For a bit test header which covers a contiguous range and hence has no edges to default statement, we don't distribute the default weight to it.
When the branch checks a single value or a contiguous range with no edge to default statement, we don't distribute the default weight to it.
In other cases, the default weight is evenly distributed to successors.
Differential Revision: http://reviews.llvm.org/D12418
llvm-svn: 246522
2015-09-01 09:42:16 +08:00
|
|
|
}
|
|
|
|
|
|
|
|
|
|
; CHECK-LABEL: test4
|
|
|
|
|
|
|
|
|
|
define void @test4(i32 %x) nounwind {
|
|
|
|
|
entry:
|
|
|
|
|
|
|
|
|
|
; In this switch statement, there is no edge from bit test to default basic
|
|
|
|
|
; block.
|
|
|
|
|
|
|
|
|
|
switch i32 %x, label %sw.default [
|
|
|
|
|
i32 1, label %sw.bb
|
|
|
|
|
i32 111, label %sw.bb2
|
|
|
|
|
i32 112, label %sw.bb3
|
|
|
|
|
i32 113, label %sw.bb3
|
|
|
|
|
i32 114, label %sw.bb2
|
|
|
|
|
i32 115, label %sw.bb2
|
|
|
|
|
], !prof !3
|
|
|
|
|
|
|
|
|
|
sw.bb:
|
|
|
|
|
call void @foo(i32 0)
|
|
|
|
|
br label %sw.epilog
|
|
|
|
|
|
|
|
|
|
sw.bb2:
|
|
|
|
|
call void @foo(i32 2)
|
|
|
|
|
br label %sw.epilog
|
|
|
|
|
|
|
|
|
|
sw.bb3:
|
|
|
|
|
call void @foo(i32 3)
|
|
|
|
|
br label %sw.epilog
|
|
|
|
|
|
|
|
|
|
sw.default:
|
|
|
|
|
call void @foo(i32 1)
|
|
|
|
|
br label %sw.epilog
|
|
|
|
|
|
|
|
|
|
sw.epilog:
|
|
|
|
|
ret void
|
|
|
|
|
|
|
|
|
|
; Check if weights are correctly assigned to edges generated from switch
|
|
|
|
|
; statement.
|
|
|
|
|
;
|
|
|
|
|
; CHECK: BB#0:
|
|
|
|
|
; BB#0 to BB#6: [0, 110] + [116, UINT32_MAX] (20)
|
|
|
|
|
; BB#0 to BB#7: [111, 115] (bit test) (50)
|
2015-12-01 13:29:22 +08:00
|
|
|
; CHECK: Successors according to CFG: BB#6({{[0-9a-fx/= ]+}}28.57%) BB#7({{[0-9a-fx/= ]+}}71.43%)
|
Distribute the weight on the edge from switch to default statement to edges generated in lowering switch.
Currently, when edge weights are assigned to edges that are created when lowering switch statement, the weight on the edge to default statement (let's call it "default weight" here) is not considered. We need to distribute this weight properly. However, without value profiling, we have no idea how to distribute it. In this patch, I applied the heuristic that this weight is evenly distributed to successors.
For example, given a switch statement with cases 1,2,3,5,10,11,20, and every edge from switch to each successor has weight 10. If there is a binary search tree built to test if n < 10, then its two out-edges will have weight 4x10+10/2 = 45 and 3x10 + 10/2 = 35 respectively (currently they are 40 and 30 without considering the default weight). Each distribution (which is 5 here) will be stored in each SwitchWorkListItem for further distribution.
There are some exceptions:
For a jump table header which doesn't have any edge to default statement, we don't distribute the default weight to it.
For a bit test header which covers a contiguous range and hence has no edges to default statement, we don't distribute the default weight to it.
When the branch checks a single value or a contiguous range with no edge to default statement, we don't distribute the default weight to it.
In other cases, the default weight is evenly distributed to successors.
Differential Revision: http://reviews.llvm.org/D12418
llvm-svn: 246522
2015-09-01 09:42:16 +08:00
|
|
|
;
|
|
|
|
|
; CHECK: BB#7:
|
|
|
|
|
; BB#7 to BB#2: {111, 114, 115} (30)
|
|
|
|
|
; BB#7 to BB#3: {112, 113} (20)
|
2015-12-01 13:29:22 +08:00
|
|
|
; CHECK: Successors according to CFG: BB#2({{[0-9a-fx/= ]+}}60.00%) BB#3({{[0-9a-fx/= ]+}}40.00%)
|
Distribute the weight on the edge from switch to default statement to edges generated in lowering switch.
Currently, when edge weights are assigned to edges that are created when lowering switch statement, the weight on the edge to default statement (let's call it "default weight" here) is not considered. We need to distribute this weight properly. However, without value profiling, we have no idea how to distribute it. In this patch, I applied the heuristic that this weight is evenly distributed to successors.
For example, given a switch statement with cases 1,2,3,5,10,11,20, and every edge from switch to each successor has weight 10. If there is a binary search tree built to test if n < 10, then its two out-edges will have weight 4x10+10/2 = 45 and 3x10 + 10/2 = 35 respectively (currently they are 40 and 30 without considering the default weight). Each distribution (which is 5 here) will be stored in each SwitchWorkListItem for further distribution.
There are some exceptions:
For a jump table header which doesn't have any edge to default statement, we don't distribute the default weight to it.
For a bit test header which covers a contiguous range and hence has no edges to default statement, we don't distribute the default weight to it.
When the branch checks a single value or a contiguous range with no edge to default statement, we don't distribute the default weight to it.
In other cases, the default weight is evenly distributed to successors.
Differential Revision: http://reviews.llvm.org/D12418
llvm-svn: 246522
2015-09-01 09:42:16 +08:00
|
|
|
}
|
2015-08-27 08:37:40 +08:00
|
|
|
|
2015-09-23 08:34:56 +08:00
|
|
|
; CHECK-LABEL: test5
|
|
|
|
|
|
|
|
|
|
define void @test5(i32 %x) nounwind {
|
|
|
|
|
entry:
|
|
|
|
|
|
|
|
|
|
; In this switch statement, there is an edge from jump table to default basic
|
|
|
|
|
; block.
|
|
|
|
|
|
|
|
|
|
switch i32 %x, label %sw.default [
|
|
|
|
|
i32 1, label %sw.bb
|
|
|
|
|
i32 5, label %sw.bb2
|
|
|
|
|
i32 7, label %sw.bb3
|
|
|
|
|
i32 9, label %sw.bb4
|
|
|
|
|
i32 31, label %sw.bb5
|
|
|
|
|
], !prof !2
|
|
|
|
|
|
|
|
|
|
sw.bb:
|
|
|
|
|
call void @foo(i32 0)
|
|
|
|
|
br label %sw.epilog
|
|
|
|
|
|
|
|
|
|
sw.bb2:
|
|
|
|
|
call void @foo(i32 1)
|
|
|
|
|
br label %sw.epilog
|
|
|
|
|
|
|
|
|
|
sw.bb3:
|
|
|
|
|
call void @foo(i32 2)
|
|
|
|
|
br label %sw.epilog
|
|
|
|
|
|
|
|
|
|
sw.bb4:
|
|
|
|
|
call void @foo(i32 3)
|
|
|
|
|
br label %sw.epilog
|
|
|
|
|
|
|
|
|
|
sw.bb5:
|
|
|
|
|
call void @foo(i32 4)
|
|
|
|
|
br label %sw.epilog
|
|
|
|
|
|
|
|
|
|
sw.default:
|
|
|
|
|
call void @foo(i32 5)
|
|
|
|
|
br label %sw.epilog
|
|
|
|
|
|
|
|
|
|
sw.epilog:
|
|
|
|
|
ret void
|
|
|
|
|
|
|
|
|
|
; Check if weights are correctly assigned to edges generated from switch
|
|
|
|
|
; statement.
|
|
|
|
|
;
|
|
|
|
|
; CHECK: BB#0:
|
|
|
|
|
; BB#0 to BB#6: [10, UINT32_MAX] (15)
|
|
|
|
|
; BB#0 to BB#8: [1, 5, 7, 9] (jump table) (45)
|
2015-12-01 13:29:22 +08:00
|
|
|
; CHECK: Successors according to CFG: BB#8({{[0-9a-fx/= ]+}}25.00%) BB#9({{[0-9a-fx/= ]+}}75.00%)
|
2015-09-23 08:34:56 +08:00
|
|
|
}
|
|
|
|
|
|
2015-08-27 08:37:40 +08:00
|
|
|
!1 = !{!"branch_weights", i32 10, i32 10, i32 10, i32 10, i32 10, i32 10, i32 10, i32 10, i32 10}
|
Distribute the weight on the edge from switch to default statement to edges generated in lowering switch.
Currently, when edge weights are assigned to edges that are created when lowering switch statement, the weight on the edge to default statement (let's call it "default weight" here) is not considered. We need to distribute this weight properly. However, without value profiling, we have no idea how to distribute it. In this patch, I applied the heuristic that this weight is evenly distributed to successors.
For example, given a switch statement with cases 1,2,3,5,10,11,20, and every edge from switch to each successor has weight 10. If there is a binary search tree built to test if n < 10, then its two out-edges will have weight 4x10+10/2 = 45 and 3x10 + 10/2 = 35 respectively (currently they are 40 and 30 without considering the default weight). Each distribution (which is 5 here) will be stored in each SwitchWorkListItem for further distribution.
There are some exceptions:
For a jump table header which doesn't have any edge to default statement, we don't distribute the default weight to it.
For a bit test header which covers a contiguous range and hence has no edges to default statement, we don't distribute the default weight to it.
When the branch checks a single value or a contiguous range with no edge to default statement, we don't distribute the default weight to it.
In other cases, the default weight is evenly distributed to successors.
Differential Revision: http://reviews.llvm.org/D12418
llvm-svn: 246522
2015-09-01 09:42:16 +08:00
|
|
|
!2 = !{!"branch_weights", i32 10, i32 10, i32 10, i32 10, i32 10, i32 10}
|
|
|
|
|
!3 = !{!"branch_weights", i32 10, i32 10, i32 10, i32 10, i32 10, i32 10, i32 10}
|