1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
59
60
61
62
63
64
65
66
|
; This test makes sure that these instructions are properly eliminated.
;
; RUN: as < %s | opt -instcombine | dis | not grep 'xor '
implementation
bool %test5(bool %A) {
%B = xor bool %A, false
ret bool %B
}
int %test6(int %A) {
%B = xor int %A, 0
ret int %B
}
bool %test7(bool %A) {
%B = xor bool %A, %A
ret bool %B
}
int %test8(int %A) {
%B = xor int %A, %A
ret int %B
}
int %test11(int %A) { ; A ^ ~A == -1
%NotA = xor int -1, %A
%B = xor int %A, %NotA
ret int %B
}
uint %test13(uint %A) { ; (A|B)^B == A & (~B)
%t1 = or uint %A, 123
%r = xor uint %t1, 123
ret uint %r
}
ubyte %test15(ubyte %A) {
%B = xor ubyte %A, 17
%C = xor ubyte %B, 17
ret ubyte %C
}
int %test16(int %A, int %B) { ; (A & C1)^(B & C2) -> (A & C1)|(B & C2) iff C1&C2 == 0
%A1 = and int %A, 7
%B1 = and int %B, 128
%OROK = xor int %A1, %B1
ret int %OROK
}
ubyte %test18(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 %test19(ubyte %A) {
%B = xor ubyte %A, 123 ; xor can be eliminated
%C = seteq ubyte %B, 34
ret bool %C
}
|
