Oh, nice! Thanks.

The openness just means that a_end == b_start || b_end == a_start doesn't
mean intersection. So we just need strict inequality in your expression,
that's all.


On Fri, Jun 12, 2015 at 12:36 AM <[email protected]> wrote:

> Quick DBC...
>
>
>
> https://codereview.chromium.org/1157663007/diff/160001/src/compiler/register-allocator.h
> File src/compiler/register-allocator.h (right):
>
>
> https://codereview.chromium.org/1157663007/diff/160001/src/compiler/register-allocator.h#newcode969
> src/compiler/register-allocator.h:969
> <https://codereview.chromium.org/1157663007/diff/160001/src/compiler/register-allocator.h#newcode969src/compiler/register-allocator.h:969>:
> if (a_start < b_start) return
> a_end > b_start;
> On 2015/06/12 04:09:12, jarin wrote:
> > Out of curiosity, cannot you just leave out the "if (a_start ==
> b_start) return
> > true" line and replace "a_start < b_start" with "a_start <= b_start"?
>
> > This special casing for equality seems a bit funny.
>
> > Perhaps you can even expand the recursive call, so the body would be:
>
> > if (a_start <= b_start) return b_start < a_end;
> > return b_start <= a_start && a_start < b_end;
>
> Hmmm, this looks still too complicated, normally just a single line is
> needed:
>
>     return a_start <= b_end && b_start <= a_end;
>
> I'm not sure about the "open-ness" of our intervals, but I'm quite sure
> that this condition can be tweaked to any convention.
>
> https://codereview.chromium.org/1157663007/
>

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