On Wed, 9 Nov 2005, Steven Rostedt wrote:
> On Wed, 2005-11-09 at 15:40 -0800, Vadim Lobanov wrote:
> > On Thu, 10 Nov 2005, Andreas Schwab wrote:
> >
> > > Vadim Lobanov <[email protected]> writes:
> > >
> > > > However, if the code is as follows:
> > > > void foo (void) {
> > > > int myvar = 0;
> > > > printf("%d\n", myvar);
> > > > bar(&myvar);
> > > > printf("%d\n", myvar);
> > > > }
> > > > If bar is declared in _another_ file as
> > > > void bar (const int * var);
> > > > then I think the compiler can validly cache the value of 'myvar' for the
> > > > second printf without re-reading it. Correct/incorrect?
> > >
> > > Incorrect. bar() may cast away const. In C const does not mean readonly.
> >
> > In that case, I stand corrected.
> >
> > Is there any real reason to apply const to pointer targets, aside from
> > giving yourself a warning in the case you try to write the pointer
> > target directly? Seems to be a missed opportunity for optimizations
> > where the coder designates that it's okay to do so.
>
> Actually, where are you going to cache it? In a register? but calling
> bar() may use that register, so it would be stored on the stack anyway.
May, but not necessarily will.
> I doubt that this is a problem with the compiler, since if bar _is_
> small, then myvar is most likely already in the processor's cache to
> begin with, so it wouldn't need to go back out to memory, unless it was
> modified.
You're right, however. There's very few cases where such an optimization
would be useful, due to register constraints.
> -- Steve
>
>
-Vadim Lobanov
-
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