; This test makes sure that these instructions are properly eliminated.
;
-; RUN: if as < %s | opt -instcombine -die | dis | grep xor
-; RUN: then exit 1
-; RUN: else exit 0
-; RUN: fi
+; RUN: opt < %s -instcombine -S | not grep xor
-implementation
+define i32 @test1(i32 %A) {
+ %B = xor i32 %A, -1 ; <i32> [#uses=1]
+ %C = xor i32 %B, -1 ; <i32> [#uses=1]
+ ret i32 %C
+}
-int %test1(int %A) {
- %B = xor int %A, -1
- %C = xor int %B, -1
- ret int %C
+define i1 @test2(i32 %A, i32 %B) {
+ ; Can change into setge
+ %cond = icmp sle i32 %A, %B ; <i1> [#uses=1]
+ %Ret = xor i1 %cond, true ; <i1> [#uses=1]
+ ret i1 %Ret
}
-bool %test2(int %A, int %B) {
- %cond = setle int %A, %B ; Can change into setge
- %Ret = xor bool %cond, true
- ret bool %Ret
+; Test that demorgans law can be instcombined
+define i32 @test3(i32 %A, i32 %B) {
+ %a = xor i32 %A, -1 ; <i32> [#uses=1]
+ %b = xor i32 %B, -1 ; <i32> [#uses=1]
+ %c = and i32 %a, %b ; <i32> [#uses=1]
+ %d = xor i32 %c, -1 ; <i32> [#uses=1]
+ ret i32 %d
}
+; Test that demorgens law can work with constants
+define i32 @test4(i32 %A, i32 %B) {
+ %a = xor i32 %A, -1 ; <i32> [#uses=1]
+ %c = and i32 %a, 5 ; <i32> [#uses=1]
+ %d = xor i32 %c, -1 ; <i32> [#uses=1]
+ ret i32 %d
+}
-; Test that demorgans law can be instcombined
-int %test3(int %A, int %B) {
- %a = xor int %A, -1
- %b = xor int %B, -1
- %c = and int %a, %b
- %d = xor int %c, -1
- ret int %d
+; test the mirror of demorgans law...
+define i32 @test5(i32 %A, i32 %B) {
+ %a = xor i32 %A, -1 ; <i32> [#uses=1]
+ %b = xor i32 %B, -1 ; <i32> [#uses=1]
+ %c = or i32 %a, %b ; <i32> [#uses=1]
+ %d = xor i32 %c, -1 ; <i32> [#uses=1]
+ ret i32 %d
}
+
+; PR2298
+define zeroext i8 @test6(i32 %a, i32 %b) nounwind {
+entry:
+ %tmp1not = xor i32 %a, -1 ; <i32> [#uses=1]
+ %tmp2not = xor i32 %b, -1 ; <i32> [#uses=1]
+ %tmp3 = icmp slt i32 %tmp1not, %tmp2not ; <i1> [#uses=1]
+ %retval67 = zext i1 %tmp3 to i8 ; <i8> [#uses=1]
+ ret i8 %retval67
+}
+