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MIPS multi-word integer arithmetic

Started by ShmeeBegek Aug 7, 2005 at 7:43 PM 1 replies 900+ views

Original Post

Aug 7, 2005 at 7:43 PM

I'm fairly new to MIPS programming and I am wondering how one does multi-word integer arithmetic on MIPS (as there doesn't seem to be any facility like a carry bit). Obviously one could hack it in a variety of ways, but does anyone know of an elegant / standard way of doing it? In case there's any ambiguity, what I'm referring to is arithmetic on integers larger than 32 bits, perhaps 64 or 128 bits. Thanks, ~SPH

SiCrane

Aug 7, 2005 at 7:54 PM

Off the top of my head, say you had two numbers in $8 and $9, you could do

addu $10, $8, $9sltu $11, $10, $9

$10 would hold the sum, and $11 would hold 1 if there was a carry out, and 0 if there wasn't.

ShmeeBegek

Aug 7, 2005 at 8:06 PM

That makes sense for the most part, but wouldn't one use 'add' instead of 'addu' because one would want overflow?

Also this would mean that the high word would require 3 instructions to add (or 4 if there is an even higher word) right?

add  $10, $8 , $9sltu $11, $10, $9add  $10, $10, $6add  $10, $10, $7add  $10, $10, $11

Thanks, ~SPH

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