llvm-6502/test/Transforms/InstCombine/xor.ll
Chris Lattner 9716eb7676 New testcase for the other side
git-svn-id: https://llvm.org/svn/llvm-project/llvm/trunk@9714 91177308-0d34-0410-b5e6-96231b3b80d8
2003-11-05 01:05:22 +00:00

125 lines
2.3 KiB
LLVM

; This test makes sure that these instructions are properly eliminated.
;
; RUN: llvm-as < %s | opt -instcombine | llvm-dis | not grep 'xor '
implementation
bool %test0(bool %A) {
%B = xor bool %A, false
ret bool %B
}
int %test1(int %A) {
%B = xor int %A, 0
ret int %B
}
bool %test2(bool %A) {
%B = xor bool %A, %A
ret bool %B
}
int %test3(int %A) {
%B = xor int %A, %A
ret int %B
}
int %test4(int %A) { ; A ^ ~A == -1
%NotA = xor int -1, %A
%B = xor int %A, %NotA
ret int %B
}
uint %test5(uint %A) { ; (A|B)^B == A & (~B)
%t1 = or uint %A, 123
%r = xor uint %t1, 123
ret uint %r
}
ubyte %test6(ubyte %A) {
%B = xor ubyte %A, 17
%C = xor ubyte %B, 17
ret ubyte %C
}
; (A & C1)^(B & C2) -> (A & C1)|(B & C2) iff C1&C2 == 0
int %test7(int %A, int %B) {
%A1 = and int %A, 7
%B1 = and int %B, 128
%C1 = xor int %A1, %B1
ret int %C1
}
ubyte %test8(bool %c) {
%d = xor bool %c, true ; invert the condition
br bool %d, label %True, label %False
True:
ret ubyte 1
False:
ret ubyte 3
}
bool %test9(ubyte %A) {
%B = xor ubyte %A, 123 ; xor can be eliminated
%C = seteq ubyte %B, 34
ret bool %C
}
ubyte %test10(ubyte %A) {
%B = and ubyte %A, 3
%C = xor ubyte %B, 4 ; transform into an OR
ret ubyte %C
}
ubyte %test11(ubyte %A) {
%B = or ubyte %A, 12
%C = xor ubyte %B, 4 ; transform into an AND
ret ubyte %C
}
bool %test12(ubyte %A) {
%B = xor ubyte %A, 4
%c = setne ubyte %B, 0
ret bool %c
}
bool %test13(ubyte %A, ubyte %B) {
%C = setlt ubyte %A, %B
%D = setgt ubyte %A, %B
%E = xor bool %C, %D ; E = setne %A, %B
ret bool %E
}
bool %test14(ubyte %A, ubyte %B) {
%C = seteq ubyte %A, %B
%D = setne ubyte %B, %A
%E = xor bool %C, %D ; E = true
ret bool %E
}
uint %test15(uint %A) { ; ~(X-1) == -X
%B = add uint %A, 4294967295
%C = xor uint %B, 4294967295
ret uint %C
}
uint %test16(uint %A) { ; ~(X+c) == (-c-1)-X
%B = add uint %A, 123 ; A generalization of the previous case
%C = xor uint %B, 4294967295
ret uint %C
}
uint %test17(uint %A) { ; ~(c-X) == X-(c-1) == X+(-c+1)
%B = sub uint 123, %A
%C = xor uint %B, 4294967295
ret uint %C
}
uint %test18(uint %A) { ; C - ~X == X + (1+C)
%B = xor uint %A, 4294967295; -~X == 0 - ~X == X+1
%C = sub uint 123, %B
ret uint %C
}