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475b056c9a
consider: C1 = INT_MIN C2 = -1 C1 * C2 overflows without a doubt but consider the following: %x = i32 INT_MIN This means that (%X /s C1) is 1 and (%X /s C1) /s C2 is -1. N. B. Move the unsigned version of this transform to InstSimplify, it doesn't create any new instructions. This fixes PR21243. llvm-svn: 219567
61 lines
1.1 KiB
LLVM
61 lines
1.1 KiB
LLVM
; RUN: opt < %s -instsimplify -S | FileCheck %s
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; PR8862
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; CHECK-LABEL: @shift1(
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; CHECK: ret i32 %A
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define i32 @shift1(i32 %A, i32 %B) {
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%C = lshr exact i32 %A, %B
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%D = shl nuw i32 %C, %B
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ret i32 %D
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}
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; CHECK-LABEL: @shift2(
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; CHECK: lshr
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; CHECK: ret i32 %D
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define i32 @shift2(i32 %A, i32 %B) {
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%C = lshr i32 %A, %B
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%D = shl nuw i32 %C, %B
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ret i32 %D
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}
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; CHECK-LABEL: @shift3(
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; CHECK: ret i32 %A
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define i32 @shift3(i32 %A, i32 %B) {
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%C = ashr exact i32 %A, %B
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%D = shl nuw i32 %C, %B
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ret i32 %D
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}
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; CHECK-LABEL: @shift4(
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; CHECK: ret i32 %A
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define i32 @shift4(i32 %A, i32 %B) {
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%C = shl nuw i32 %A, %B
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%D = lshr i32 %C, %B
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ret i32 %D
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}
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; CHECK-LABEL: @shift5(
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; CHECK: ret i32 %A
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define i32 @shift5(i32 %A, i32 %B) {
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%C = shl nsw i32 %A, %B
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%D = ashr i32 %C, %B
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ret i32 %D
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}
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; CHECK-LABEL: @div1(
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; CHECK: ret i32 0
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define i32 @div1(i32 %V) {
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%A = udiv i32 %V, -2147483648
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%B = udiv i32 %A, -2147483648
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ret i32 %B
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}
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; CHECK-LABEL: @div2(
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; CHECK-NOT: ret i32 0
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define i32 @div2(i32 %V) {
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%A = sdiv i32 %V, -1
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%B = sdiv i32 %A, -2147483648
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ret i32 %B
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}
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