1; This test makes sure that these instructions are properly eliminated. 2; 3 4; RUN: opt < %s -instcombine -S | not grep xor 5 6define i32 @test1(i32 %A) { 7 %B = xor i32 %A, -1 ; <i32> [#uses=1] 8 %C = xor i32 %B, -1 ; <i32> [#uses=1] 9 ret i32 %C 10} 11 12define i1 @test2(i32 %A, i32 %B) { 13 ; Can change into setge 14 %cond = icmp sle i32 %A, %B ; <i1> [#uses=1] 15 %Ret = xor i1 %cond, true ; <i1> [#uses=1] 16 ret i1 %Ret 17} 18 19; Test that demorgans law can be instcombined 20define i32 @test3(i32 %A, i32 %B) { 21 %a = xor i32 %A, -1 ; <i32> [#uses=1] 22 %b = xor i32 %B, -1 ; <i32> [#uses=1] 23 %c = and i32 %a, %b ; <i32> [#uses=1] 24 %d = xor i32 %c, -1 ; <i32> [#uses=1] 25 ret i32 %d 26} 27 28; Test that demorgens law can work with constants 29define i32 @test4(i32 %A, i32 %B) { 30 %a = xor i32 %A, -1 ; <i32> [#uses=1] 31 %c = and i32 %a, 5 ; <i32> [#uses=1] 32 %d = xor i32 %c, -1 ; <i32> [#uses=1] 33 ret i32 %d 34} 35 36; test the mirror of demorgans law... 37define i32 @test5(i32 %A, i32 %B) { 38 %a = xor i32 %A, -1 ; <i32> [#uses=1] 39 %b = xor i32 %B, -1 ; <i32> [#uses=1] 40 %c = or i32 %a, %b ; <i32> [#uses=1] 41 %d = xor i32 %c, -1 ; <i32> [#uses=1] 42 ret i32 %d 43} 44 45; PR2298 46define zeroext i8 @test6(i32 %a, i32 %b) nounwind { 47entry: 48 %tmp1not = xor i32 %a, -1 ; <i32> [#uses=1] 49 %tmp2not = xor i32 %b, -1 ; <i32> [#uses=1] 50 %tmp3 = icmp slt i32 %tmp1not, %tmp2not ; <i1> [#uses=1] 51 %retval67 = zext i1 %tmp3 to i8 ; <i8> [#uses=1] 52 ret i8 %retval67 53} 54 55