b = a - b
a = a - b
I think this works too, assuming a+b doesn't overflow.
> I think this works too, assuming a+b doesn't overflow.
Well, in two's complement arithmetic (as is used on most architectures), the intermediate overflow can be ignored, and it will work just fine.
> I think this works too, assuming a+b doesn't overflow.
Well, in two's complement arithmetic (as is used on most architectures), the intermediate overflow can be ignored, and it will work just fine.