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While simplifying branches in the MachineInstr representation, the routine BuildCondBr must preserve flags on register MachineOperands. In particular, it must preserve the <undef> flag. This fixes a bug that is unlikely to occur in any real scenario, but which bugpoint is likely to introduce. Patch By Nick Johnson! Reviewers: ahatanak, sdardis Differential Revision: https://reviews.llvm.org/D34041 llvm-svn: 305290
27 lines
993 B
LLVM
27 lines
993 B
LLVM
; RUN: llc -march=mips -mcpu=mips32 -verify-machineinstrs -o /dev/null < %s
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; Confirm that MachineInstr branch simplification preserves
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; register operand flags, such as the <undef> flag.
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define void @ham() {
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bb:
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%tmp = alloca i32, align 4
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%tmp13 = ptrtoint i32* %tmp to i32
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%tmp70 = icmp eq i32 undef, -1
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br i1 %tmp70, label %bb72, label %bb40
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bb72: ; preds = %bb72, %bb
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br i1 undef, label %bb40, label %bb72
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bb40: ; preds = %bb72, %bb
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%tmp41 = phi i32 [ %tmp13, %bb72 ], [ %tmp13, %bb ]
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%tmp55 = inttoptr i32 %tmp41 to i32*
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%tmp58 = insertelement <2 x i32*> undef, i32* %tmp55, i32 1
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br label %bb59
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bb59: ; preds = %bb59, %bb40
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%tmp60 = phi <2 x i32*> [ %tmp61, %bb59 ], [ %tmp58, %bb40 ]
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%tmp61 = getelementptr i32, <2 x i32*> %tmp60, <2 x i32> <i32 -1, i32 1>
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%tmp62 = extractelement <2 x i32*> %tmp61, i32 1
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br label %bb59
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}
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