llvm-6502/test/Transforms/InstCombine/add.ll
Chris Lattner ad5727d173 Add new testcases
git-svn-id: https://llvm.org/svn/llvm-project/llvm/trunk@8816 91177308-0d34-0410-b5e6-96231b3b80d8
2003-10-02 15:11:09 +00:00

119 lines
2.6 KiB
LLVM

; This test makes sure that add instructions are properly eliminated.
; RUN: llvm-as < %s | opt -instcombine -die | llvm-dis | grep -v OK | not grep add
implementation
int %test1(int %A) {
%B = add int %A, 0
ret int %B
}
int %test2(int %A) {
%B = add int %A, 5
%C = add int %B, -5
ret int %C
}
int %test3(int %A) {
%B = add int %A, 5
%C = sub int %B, 5 ;; This should get converted to an add
ret int %C
}
int %test4(int %A, int %B) {
%C = sub int 0, %A
%D = add int %B, %C ; D = B + -A = B - A
ret int %D
}
int %test5(int %A, int %B) {
%C = sub int 0, %A
%D = add int %C, %B ; D = -A + B = B - A
ret int %D
}
int %test6(int %A) {
%B = mul int 7, %A
%C = add int %B, %A ; C = 7*A+A == 8*A == A << 3
ret int %C
}
int %test7(int %A) {
%B = mul int 7, %A
%C = add int %A, %B ; C = A+7*A == 8*A == A << 3
ret int %C
}
int %test8(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
%C = add int %A1, %B1
ret int %C
}
int %test9(int %A) {
%B = shl int %A, ubyte 4
%C = add int %B, %B ; === shl int %A, 5
ret int %C
}
bool %test10(ubyte %A, ubyte %b) {
%B = add ubyte %A, %b
%c = setne ubyte %B, 0 ; === A != -b
ret bool %c
}
bool %test11(ubyte %A) {
%B = add ubyte %A, 255
%c = setne ubyte %B, 0 ; === A != 1
ret bool %c
}
int %test12(int %A, int %B) {
%C_OK = add int %B, %A ; Should be transformed into shl A, 1
br label %X
X:
%D = add int %C_OK, %A
ret int %D
}
int %test13(int %A, int %B, int %C) {
%D_OK = add int %A, %B
%E_OK = add int %D_OK, %C
%F = add int %E_OK, %A ;; shl A, 1
ret int %F
}
uint %test14(uint %offset, uint %difference) {
%tmp.2 = and uint %difference, 3
%tmp.3_OK = add uint %tmp.2, %offset
%tmp.5.mask = and uint %difference, 4294967292
%tmp.8 = add uint %tmp.3_OK, %tmp.5.mask ; == add %offset, %difference
ret uint %tmp.8
}
ubyte %test15(ubyte %A) {
%B = add ubyte %A, 192 ; Does not effect result
%C = and ubyte %B, 16 ; Only one bit set
ret ubyte %C
}
ubyte %test16(ubyte %A) {
%B = add ubyte %A, 16 ; Turn this into a XOR
%C = and ubyte %B, 16 ; Only one bit set
ret ubyte %C
}
int %test17(int %A) {
%B = xor int %A, -1
%C = add int %B, 1 ; == sub int 0, %A
ret int %C
}
ubyte %test18(ubyte %A) {
%B = xor ubyte %A, 255
%C = add ubyte %B, 17 ; == sub ubyte 16, %A
ret ubyte %C
}