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

[Peter Zijlstra]

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?


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