Re: [patch][simplify-rtx] Fix 16-bit -> 64-bit multiply and accumulate

[Andrew Stubbs <ams@codesourcery.com>]

On 24/05/11 20:35, Joseph S. Myers wrote:
> On Tue, 24 May 2011, Andrew Stubbs wrote:
>
>> I've created this new, simpler patch that converts
>>
>>    (extend (mult a b))
>>
>> into
>>
>>    (mult (extend a) (extend b))
>>
>> regardless of what 'a' and 'b' might be. (These are then simplified and
>> superfluous extends removed, of course.)
>
> Are there some missing conditions here?  The two aren't equivalent in
> general - (extend:SI (mult:HI a b)) multiplies the HImode values in HImode
> (with modulo arithmetic on overflow) before extending the possibly wrapped
> result to SImode.  You'd need a and b themselves to be extended from
> narrower modes in such a way that if you interpret the extended values in
> the signedness of the outer extension, the result of the multiplication is
> exactly representable in the mode of the multiplication.  (For example, if
> both values are extended from QImode, and all extensions have the same
> signedness, that would be OK.  There are cases that are OK where not all
> extensions have the same signedness, e.g. (sign_extend:DI (mult:SI a b))
> where a and b are zero-extended from HImode or QImode, at least one from
> QImode, though there the outer extension is equivalent to a
> zero-extension.)

So, you're saying that promoting a regular multiply to a widening 
multiply isn't a valid transformation anyway? I suppose that does make 
sense. I knew something was too easy.

OK, I'll go try again. :)

Andrew