Re: [PATCH] Fix a race in pid generation that causes pids to be reused immediately.

Peter Zijlstra
2010-06-09T21:34:09Z

On Wed, 2010-06-09 at 14:21 -0700, Linus Torvalds wrote:
>
> On Wed, 9 Jun 2010, Salman wrote:
> > +/*
> > + * If we started walking pids at 'base', is 'a' seen before 'b'?
> > + *
> > + */
> > +static int pid_before(int base, int a, int b)
> > +{
> > + int a_lt_b = (a < b);
> > + int min_a_b = min(a, b);
> > + int max_a_b = max(a, b);
> > +
> > + if ((base <= min_a_b) || (base >= max_a_b))
> > + return a_lt_b;
> > +
> > + return !a_lt_b;
> > +}
>
> Ok, so that's a very confusing expression. I'm sure it gets the right
> value, but it's not exactly straightforward, is it?
>
> Wouldn't it be nicer to write it out in a more straightforward way?
> Something like
>
> /* a and b in order? base must not be between them */
> if (a <= b)
> return (base <= a || base >= b);
> /* b < a? We reach 'a' first iff base is between them */
> return base >= b && base <= a;
>
> would seem to be equivalent and easier to explain, no?
>
> And when you write it that way, it looks like the compiler should be able
> to trivially CSE the five comparisons down to just three (notice how the
> "base <= a" and "base >= b" comparisons are repeated. Which I'm sure some
> super-optimizing compiler can do from your version too, but mine seems
> more straightforward.
>
> But maybe I did that thing wrong, and I just confused myself. I have _not_
> checked the logic deeply, somebody else should definitely double-check me.

Isn't: return a - base < b - base, the natural way to express this?