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From: Eric L. <ew...@an...> - 2006-07-06 20:32:26
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What does the Binary IOp DivModU64to32 do? The inline comments in the source say (I64, I32) -> I64. I interpreted it as arg1 is 64-bit dividend, arg2 is 32-bit divisor, result is 64 bits of which high 32 bits is arg1 mod arg2 and low 32 bits is arg1 div arg2. But arg1 div arg2 could be more than 32 bits so how can it fit in only the low 32 bits? Thanks, Eric |