llvm-6502/test/Transforms/InstCombine/phi.ll
Chris Lattner efd74a0159 The instruction combining pass removes dead instructions, there is no need
to run the die pass after it.


git-svn-id: https://llvm.org/svn/llvm-project/llvm/trunk@11942 91177308-0d34-0410-b5e6-96231b3b80d8
2004-02-28 05:26:06 +00:00

58 lines
1.2 KiB
LLVM

; This test makes sure that these instructions are properly eliminated.
;
; RUN: llvm-as < %s | opt -instcombine | llvm-dis | not grep phi
implementation
int %test1(int %A, bool %b) {
BB0: br bool %b, label %BB1, label %BB2
BB1:
%B = phi int [%A, %BB0] ; Combine away one argument PHI nodes
ret int %B
BB2:
ret int %A
}
int %test2(int %A, bool %b) {
BB0: br bool %b, label %BB1, label %BB2
BB1:
br label %BB2
BB2:
%B = phi int [%A, %BB0], [%A, %BB1] ; Combine away PHI nodes with same values
ret int %B
}
int %test3(int %A, bool %b) {
BB0: br label %Loop
Loop:
%B = phi int [%A, %BB0], [%B, %Loop] ; PHI has same value always.
br bool %b, label %Loop, label %Exit
Exit:
ret int %B
}
int %test3(bool %b) {
BB0: ret int 7 ; Loop is unreachable
Loop:
%B = phi int [%B, %L2], [%B, %Loop] ; PHI has same value always.
br bool %b, label %L2, label %Loop
L2:
br label %Loop
}
bool %test4(bool %A) {
br bool %A, label %BB1, label %BB2
BB1:
br label %Ret
BB2:
br label %Ret
Ret:
%B = phi int [1000, %BB1], [123, %BB2]
%C = cast int %B to bool
ret bool %C
}