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e0134d95cc
We can prove that a 'sub' can be a 'sub nsw' under certain conditions: - The sign bits of the operands is the same. - Both operands have more than 1 sign bit. The subtraction cannot be a signed overflow in either case. git-svn-id: https://llvm.org/svn/llvm-project/llvm/trunk@216037 91177308-0d34-0410-b5e6-96231b3b80d8
48 lines
991 B
LLVM
48 lines
991 B
LLVM
; RUN: opt -instcombine -S < %s | FileCheck %s
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define i32 @test1(i32 %x) nounwind {
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%and = and i32 %x, 31
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%sub = sub i32 63, %and
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ret i32 %sub
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; CHECK-LABEL: @test1(
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; CHECK-NEXT: and i32 %x, 31
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; CHECK-NEXT: xor i32 %and, 63
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; CHECK-NEXT: ret
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}
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declare i32 @llvm.ctlz.i32(i32, i1) nounwind readnone
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define i32 @test2(i32 %x) nounwind {
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%count = tail call i32 @llvm.ctlz.i32(i32 %x, i1 true) nounwind readnone
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%sub = sub i32 31, %count
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ret i32 %sub
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; CHECK-LABEL: @test2(
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; CHECK-NEXT: ctlz
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; CHECK-NEXT: xor i32 %count, 31
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; CHECK-NEXT: ret
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}
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define i32 @test3(i32 %x) nounwind {
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%and = and i32 %x, 31
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%sub = xor i32 31, %and
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%add = add i32 %sub, 42
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ret i32 %add
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; CHECK-LABEL: @test3(
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; CHECK-NEXT: and i32 %x, 31
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; CHECK-NEXT: sub nsw i32 73, %and
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; CHECK-NEXT: ret
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}
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define i32 @test4(i32 %x) nounwind {
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%sub = xor i32 %x, 2147483648
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%add = add i32 %sub, 42
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ret i32 %add
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; CHECK-LABEL: @test4(
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; CHECK-NEXT: add i32 %x, -2147483606
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; CHECK-NEXT: ret
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}
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