2002-04-18 17:35:39 +00:00
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; This test makes sure that add instructions are properly eliminated.
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2004-02-28 05:26:06 +00:00
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; RUN: llvm-as < %s | opt -instcombine | llvm-dis | grep -v OK | not grep add
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2002-04-18 17:35:39 +00:00
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implementation
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2003-02-18 19:43:53 +00:00
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int %test1(int %A) {
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2002-04-18 17:35:39 +00:00
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%B = add int %A, 0
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ret int %B
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2003-02-18 19:43:53 +00:00
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}
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2002-04-18 17:35:39 +00:00
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2003-02-18 19:43:53 +00:00
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int %test2(int %A) {
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2002-04-18 17:35:39 +00:00
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%B = add int %A, 5
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%C = add int %B, -5
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ret int %C
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2003-02-18 19:43:53 +00:00
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}
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2002-04-18 17:35:39 +00:00
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2003-02-18 19:43:53 +00:00
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int %test3(int %A) {
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2002-04-18 17:35:39 +00:00
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%B = add int %A, 5
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%C = sub int %B, 5 ;; This should get converted to an add
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ret int %C
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2003-02-18 19:43:53 +00:00
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}
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2002-04-18 17:35:39 +00:00
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2003-02-18 19:43:21 +00:00
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int %test4(int %A, int %B) {
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2002-05-06 16:44:53 +00:00
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%C = sub int 0, %A
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%D = add int %B, %C ; D = B + -A = B - A
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ret int %D
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}
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2003-02-18 19:43:21 +00:00
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int %test5(int %A, int %B) {
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2002-05-06 16:44:53 +00:00
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%C = sub int 0, %A
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%D = add int %C, %B ; D = -A + B = B - A
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ret int %D
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}
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2003-02-18 19:55:31 +00:00
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int %test6(int %A) {
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%B = mul int 7, %A
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%C = add int %B, %A ; C = 7*A+A == 8*A == A << 3
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ret int %C
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}
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int %test7(int %A) {
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%B = mul int 7, %A
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%C = add int %A, %B ; C = A+7*A == 8*A == A << 3
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ret int %C
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}
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2003-03-10 23:52:54 +00:00
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int %test8(int %A, int %B) { ; (A & C1)+(B & C2) -> (A & C1)|(B & C2) iff C1&C2 == 0
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%A1 = and int %A, 7
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%B1 = and int %B, 128
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%C = add int %A1, %B1
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ret int %C
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}
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2003-03-11 00:10:59 +00:00
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2003-07-24 19:44:51 +00:00
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int %test9(int %A) {
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%B = shl int %A, ubyte 4
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%C = add int %B, %B ; === shl int %A, 5
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ret int %C
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}
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2003-08-13 05:27:57 +00:00
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bool %test10(ubyte %A, ubyte %b) {
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%B = add ubyte %A, %b
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%c = setne ubyte %B, 0 ; === A != -b
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ret bool %c
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}
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bool %test11(ubyte %A) {
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%B = add ubyte %A, 255
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%c = setne ubyte %B, 0 ; === A != 1
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ret bool %c
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}
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2003-08-13 19:01:09 +00:00
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int %test12(int %A, int %B) {
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%C_OK = add int %B, %A ; Should be transformed into shl A, 1
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br label %X
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X:
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%D = add int %C_OK, %A
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ret int %D
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}
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int %test13(int %A, int %B, int %C) {
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%D_OK = add int %A, %B
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%E_OK = add int %D_OK, %C
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%F = add int %E_OK, %A ;; shl A, 1
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ret int %F
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}
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uint %test14(uint %offset, uint %difference) {
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%tmp.2 = and uint %difference, 3
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%tmp.3_OK = add uint %tmp.2, %offset
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%tmp.5.mask = and uint %difference, 4294967292
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%tmp.8 = add uint %tmp.3_OK, %tmp.5.mask ; == add %offset, %difference
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ret uint %tmp.8
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}
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2003-09-19 15:33:46 +00:00
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ubyte %test15(ubyte %A) {
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%B = add ubyte %A, 192 ; Does not effect result
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%C = and ubyte %B, 16 ; Only one bit set
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ret ubyte %C
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}
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ubyte %test16(ubyte %A) {
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%B = add ubyte %A, 16 ; Turn this into a XOR
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%C = and ubyte %B, 16 ; Only one bit set
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ret ubyte %C
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}
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2003-10-02 15:11:09 +00:00
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int %test17(int %A) {
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%B = xor int %A, -1
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%C = add int %B, 1 ; == sub int 0, %A
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ret int %C
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}
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ubyte %test18(ubyte %A) {
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%B = xor ubyte %A, 255
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%C = add ubyte %B, 17 ; == sub ubyte 16, %A
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ret ubyte %C
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}
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2004-04-09 23:47:52 +00:00
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int %test19(bool %C) {
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%A = select bool %C, int 1000, int 10
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%V = add int %A, 123
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ret int %V
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}
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2004-04-10 22:01:27 +00:00
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int %test20(int %x) {
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%tmp.2 = xor int %x, -2147483648
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;; Add of sign bit -> xor of sign bit.
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%tmp.4 = add int %tmp.2, -2147483648
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ret int %tmp.4
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}
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2004-04-09 23:47:52 +00:00
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