llvm-6502/test/Transforms/InstCombine/shift.ll
2003-09-16 15:29:54 +00:00

99 lines
2.0 KiB
LLVM

; This test makes sure that these instructions are properly eliminated.
;
; RUN: llvm-as < %s | opt -instcombine | llvm-dis | not grep sh
implementation
int %test1(int %A) {
%B = shl int %A, ubyte 0
ret int %B
}
int %test2(ubyte %A) {
%B = shl int 0, ubyte %A
ret int %B
}
int %test3(int %A) {
%B = shr int %A, ubyte 0
ret int %B
}
int %test4(ubyte %A) {
%B = shr int 0, ubyte %A
ret int %B
}
uint %test5(uint %A) {
%B = shr uint %A, ubyte 32 ;; shift all bits out
ret uint %B
}
uint %test5a(uint %A) {
%B = shl uint %A, ubyte 32 ;; shift all bits out
ret uint %B
}
uint %test6(uint %A) {
%B = shl uint %A, ubyte 1 ;; convert to an mul instruction
%C = mul uint %B, 3
ret uint %C
}
int %test7(ubyte %A) {
%B = shr int -1, ubyte %A ;; Always equal to -1
ret int %B
}
ubyte %test8(ubyte %A) { ;; (A << 5) << 3 === A << 8 == 0
%B = shl ubyte %A, ubyte 5
%C = shl ubyte %B, ubyte 3
ret ubyte %C
}
ubyte %test9(ubyte %A) { ;; (A << 7) >> 7 === A & 1
%B = shl ubyte %A, ubyte 7
%C = shr ubyte %B, ubyte 7
ret ubyte %C
}
ubyte %test10(ubyte %A) { ;; (A >> 7) << 7 === A & 128
%B = shr ubyte %A, ubyte 7
%C = shl ubyte %B, ubyte 7
ret ubyte %C
}
ubyte %test11(ubyte %A) { ;; (A >> 3) << 4 === (A & 0x1F) << 1
%a = mul ubyte %A, 3
%B = shr ubyte %a, ubyte 3
%C = shl ubyte %B, ubyte 4
ret ubyte %C
}
int %test12(int %A) {
%B = shr int %A, ubyte 8 ;; (A >> 8) << 8 === A & -256
%C = shl int %B, ubyte 8
ret int %C
}
sbyte %test13(sbyte %A) { ;; (A >> 3) << 4 === (A & -8) * 2
%a = mul sbyte %A, 3
%B = shr sbyte %a, ubyte 3
%C = shl sbyte %B, ubyte 4
ret sbyte %C
}
uint %test14(uint %A) {
%B = shr uint %A, ubyte 4
%C = or uint %B, 1234
%D = shl uint %C, ubyte 4 ;; D = ((B | 1234) << 4) === ((B << 4)|(1234 << 4)
ret uint %D
}
uint %test14a(uint %A) {
%B = shl uint %A, ubyte 4
%C = and uint %B, 1234
%D = shr uint %C, ubyte 4 ;; D = ((B | 1234) << 4) === ((B << 4)|(1234 << 4)
ret uint %D
}