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[SCCP] Teach the pass how to handle div
with overdefined operands.
This can prove that: extern int f; int g() { int x = 0; for (int i = 0; i < 365; ++i) { x /= f; } return x; } always returns zero. Thanks to Sanjoy for confirming this transformation actually made sense (bugs are mine). llvm-svn: 292531
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@ -910,6 +910,12 @@ void SCCPSolver::visitBinaryOperator(Instruction &I) {
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// Otherwise, one of our operands is overdefined. Try to produce something
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// better than overdefined with some tricks.
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// If this is 0 / Y, it doesn't matter that the second operand is
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// overdefined, and we can replace it with zero.
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if (I.getOpcode() == Instruction::UDiv || I.getOpcode() == Instruction::SDiv)
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if (V1State.isConstant() && V1State.getConstant()->isNullValue())
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return markConstant(IV, &I, V1State.getConstant());
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// If this is:
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// -> AND/MUL with 0
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// -> OR with -1
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32
test/Transforms/SCCP/overdefined-div.ll
Normal file
32
test/Transforms/SCCP/overdefined-div.ll
Normal file
@ -0,0 +1,32 @@
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; RUN: opt < %s -sccp -S | FileCheck %s
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; Test that SCCP has basic knowledge of when div can nuke overdefined values.
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; 0 / X = 0 even if X is overdefined.
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; CHECK-LABEL: test1
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; CHECK-NEXT: ret i32 0
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define i32 @test1(i32 %foo) {
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%tinkywinky = udiv i32 0, %foo
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ret i32 %tinkywinky
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}
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; CHECK-LABEL: test2
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; CHECK-NEXT: ret i32 0
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define i32 @test2(i32 %foo) {
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%tinkywinky = sdiv i32 0, %foo
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ret i32 %tinkywinky
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}
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; CHECK-LABEL: test3
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; CHECK: ret i32 %tinkywinky
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define i32 @test3(i32 %foo) {
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%tinkywinky = udiv i32 %foo, 0
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ret i32 %tinkywinky
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}
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; CHECK-LABEL: test4
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; CHECK: ret i32 %tinkywinky
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define i32 @test4(i32 %foo) {
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%tinkywinky = sdiv i32 %foo, 0
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ret i32 %tinkywinky
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}
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