I'm trying to swap an 8 byte quantity via a Diab macro but I am unable to figure out how to get access to the second word of the unsigned long long. Here's what I have:
asm uint64 swap8Bytes(uint64 eightBytes) { % reg eightBytes; ! "r0", "r12"
mr r12,eightBytes /* LSW */ rlwinm r0,r12, 8, 8,31 rlwimi r0,r12,24,16,23 rlwimi r0,r12,24, 0, 8 mr r4,r0
mr r12,eightBytes /* MSW? How? */ rlwinm r0,r12, 8, 8,31 rlwimi r0,r12,24,16,23 rlwimi r0,r12,24, 0, 8 mr r3,r0 }
-- - Mark -> --
Have you considered writing this in C, then looking at the assembly output Diab creates for a hint?
--Gene
Gene S. Berkowitz wrote in news: snipped-for-privacy@news.comcast.giganews.com:
[snip]Uh, yes and it is atrocious code. The compiler doesn't recognized the idiom of a 64-bit integer swap. Thus, we PowerPC developers tend to use this handy sequence of rlwiXX instruction for swapping bytes in a word. What I am trying to discern is the Diab assembler mnemonic that allows the human assembler writer to specify the MS word and LS word in a 64-bit integer. The Diab manuals do not specify how to do this.
Thanks.
-- - Mark -> --
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