On Wed, Aug 15, 2007 at 08:31:25PM +0200, Segher Boessenkool wrote:
> >>How does the compiler know that msleep() has got barrier()s?
> >
> >Because msleep_interruptible() is in a separate compilation unit,
> >the compiler has to assume that it might modify any arbitrary global.
>
> No; compilation units have nothing to do with it, GCC can optimise
> across compilation unit boundaries just fine, if you tell it to
> compile more than one compilation unit at once.
Last I checked, the Linux kernel build system did compile each .c file
as a separate compilation unit.
> What you probably mean is that the compiler has to assume any code
> it cannot currently see can do anything (insofar as allowed by the
> relevant standards etc.)
Indeed.
> >In many cases, the compiler also has to assume that
> >msleep_interruptible()
> >might call back into a function in the current compilation unit, thus
> >possibly modifying global static variables.
>
> It most often is smart enough to see what compilation-unit-local
> variables might be modified that way, though :-)
Yep. For example, if it knows the current value of a given such local
variable, and if all code paths that would change some other variable
cannot be reached given that current value of the first variable.
At least given that gcc doesn't know about multiple threads of execution!
Thanx, Paul
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