[PATCH] [ConstantRange] Add a smultiply method for signed multiply

[James Molloy]

Hi both,

OK, this is clearer now. Nick's suggestion of calculating both signed and unsigned and picking the smallest range makes sense as an implementation.

Thanks both for your patience! I'll put a new patch up for review based on Nick's suggestion shortly.

Cheers,

James



> On 21 Feb 2015, at 21:06, Sanjoy Das <sanjoy at playingwithpointers.com> wrote:
>
> Is it fair to say that your question is: given that
> ConstantRange::multiply uses the unsigned and signed max of the range
> to calculate its result, how can it possibly work in a signed context?
>
> So the first half of the answer is that you're right, in general the
> result computed by using the smax and smin vs. using the umax and umin
> will be different.  For instance, given this scheme, [-1, 2) `smult`
> [-1, 2) is [-1, 2) (everything is i8 in this email) while [-1, 2)
> `umult` [-1, 2) is the full i8 range. **
>
> However, we need to look at the "contract" umult and smult have to
> follow:
>
> X `umult` Y returns Zu such that Zu is a (sometimes proper) subset of
> { x * y (mod 2^n) | x \in X, y \in Y }
>
> X `smult` Y returns Zs such that Zs is a (sometimes proper) subset
> of { x * y (mod 2^n) | x \in X, y \in Y }
>
> In other words, both of them follow the exact same contract.  You can
> see how this is true for the [-1, 2) example -- both [-1, 2) and the
> full i8 set are legitimate answers for *both* [-1, 2) `smult` [-1, 2)
> and [-1, 2) `umult` [-1, 2).  Since [-1, 2) is a more "precise"
> answer, *both* `smult` and `umult` should try to return [-1, 2).
>
> Concretely, if in ConstantRange::multiply you can produce a better
> (smaller) range
> as answer with the signed interpretation of the two ranges you're
> multiplying, you should do so.  There is no need for a separate
> ConstantRange::smultiply.
>
> ** I did not bother to verify the math here
>
> -- Sanjoy
>

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