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Fix UB in APInt::ashr 

i64 -1, whose sign bit is the 0th one, can't be left shifted without invoking UB.

https://reviews.llvm.org/D23362

llvm-svn: 278280
This commit is contained in:
Jonathan Roelofs 2016-08-10 19:50:14 +00:00
parent 9ef6b6b4f4
commit 851b79dc4d
2 changed files with 6 additions and 5 deletions
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6
llvm/lib/Support/APInt.cpp
View File

@ -1042,11 +1042,7 @@ APInt APInt::ashr(unsigned shiftAmt) const {
if (isSingleWord()) {
if (shiftAmt == BitWidth)
return APInt(BitWidth, 0); // undefined
else {
unsigned SignBit = APINT_BITS_PER_WORD - BitWidth;
return APInt(BitWidth,
(((int64_t(VAL) << SignBit) >> SignBit) >> shiftAmt));
}
return APInt(BitWidth, SignExtend64(VAL, BitWidth) >> shiftAmt);
}
// If all the bits were shifted out, the result is, technically, undefined.

5
llvm/unittests/ADT/APIntTest.cpp
View File

@ -32,6 +32,11 @@ TEST(APIntTest, ShiftLeftByZero) {
EXPECT_FALSE(Shl[1]);
}
TEST(APIntTest, i64_ArithmeticRightShiftNegative) {
const APInt neg_one(64, static_cast<uint64_t>(-1), true);
EXPECT_EQ(neg_one, neg_one.ashr(7));
}
TEST(APIntTest, i128_NegativeCount) {
APInt Minus3(128, static_cast<uint64_t>(-3), true);
EXPECT_EQ(126u, Minus3.countLeadingOnes());