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Re: why mult generated for unsigned int multiply on mips?


Amker.Cheng wrote:

now think about two number U1, U2, the corresponding signed value are S1, S2.
S1 * S2 = (U1-2^32 *s1 ) * (U2-2^32 *s2)
             =  U1*U2 - 2^32*s2*U1 - 2^32*s1*U2 + 2^64*s1*s2
It's easy to prove that the lower 32 bit of S1*S2 is determined by the
lower part of U1*U2.

Maybe this is the reason gcc can safely use mult for unsigned
multiplication for mips.

Hope this is right and it's hard to edit equations in plain text -_-

Yes this is right, it's well known that for multiplication of 32 x 32 => 32, unsigned and signed are equivalent except for overflow considerations.

The only reason a processor would have
separate signed and unsigned single length multiplication
instructions is if there are overflow traps, or overflow
or other arithmetic flags set. In other words the same
situation as for addition/subtraction.

Note that for a double length result multplication, the results
are quite different.

Consider -1 * -1, the result is +1 signed,

so FFFF*FFFF signed = 0000_0001 but
FFFF*FFFF unsigned is FFFE_0001

Note that as we expect, the low order 32 bits is the same.

For division, even the single length quotient is different, so
there you need two instructions, or if the processor has only
unsigned division, then you have to fiddle to get a signed result.



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