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Date: Tue, 4 Mar 2014 09:58:27 +0400
From: Solar Designer <>
Subject: Re: [PHC] wider integer multiply on 32-bit x86

On Tue, Mar 04, 2014 at 04:04:30AM +0000, Samuel Neves wrote:
> Poly1305 used the full x86 64-bit mantissa [1, Section 4]. So did
> curve25519 [2]. According to Agner Fog, the floating-point and integer
> multipliers in the Pentium III share the same circuitry (and the
> floating-point multiplication has one extra cycle of latency), so the
> only advantage here would appear to be register usage. Register renaming
> does help there.

So if I do a MUL, immediately save EDX:EAX to other registers, and
follow that with another similar MUL, the second MUL would be able to
proceed out-of-order, before the first one completes, correct?  (As long
as there's no data dependency between the two MULs, indeed.  Only the
same ISA registers used, which a renamer might resolve.)

> By the way, in the 63x63 -> 63 case the floating-point unit would
> actually be computing the most-significant bits of the product, not the
> least-significant.

Good point!  When the full result fits in 63 bits, we can as well treat
this as the 63 least-significant bits, and in fact this is what fistpll
gives us.  However, yes, being able to obtain the most-significant bits
may be more interesting, because they're not easily available otherwise.

> This could be used to compute the xor of both product
> halves relatively quickly, with the integer multiplier computing the
> lower part, while the floating-point multiplier would compute the higher
> part. Like so (warning: not thoroughly tested, but seems to work):

This is very cool!

>             "fldcw   %4\n\t" // Only needed for setup, really
>             "fildll  %1\n\t"
>             "fildll  %2\n\t"
>             "fmulp     \n\t"
>             "fldt    %3\n\t"
>             "fmulp     \n\t"
>             "fistpll %0\n\t"
>             : "=m" (high)
>             : "m" (a), "m" (b), "m" (twom63), "m" (ctrl)
>         );
>         return high ^ ((a * b) & 0x7FFFFFFFFFFFFFFFULL);

Unfortunately, the (a * b) here corresponds to 3 MUL instructions, for a
total of 5 MULs (of various types) for this function.  Luckily, I think
those 3 can proceed in parallel, so the total latency is not bad, and
this can keep the multiplier pipeline busy.

> To be honest, though, I don't think using floating-point is worth it.



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