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.
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.
-- Steve
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