quoting mess? More complaints about my top posting? terrible. :|
Thanks,
Tyler Littlefield
http://tds-solutions.net
Twitter: sorressean
On Feb 25, 2010, at 7:56 AM, Paul Herring wrote:
> On Thu, Feb 25, 2010 at 2:47 PM, Benjamin Scott
> <benjamin_cail_sc...@yahoo.com> wrote:
> > --- On Wed, 2/24/10, Tyler Littlefield <ty...@tysdomain.com> wrote:
> >> Benjamin Scott <benjamin_cail_sc...@yahoo.com> wrote:
> >>
> >> > Just curious, but why would the compiler
> >> > knowing the value of a pointer be pointless?
> >
> >> It doesn't know the value *at* the
> >> pointer, nor the value it points to, unless your using a
> >> statement to assign/check. The pointer points to an address,
> >> so the compiler doesn't really know the value unless it's
> >> assigned, or unless it's a constant in some cases.
> >
> > Okay, I got it: Since the compilier is not
> > Intelligent, then it can't know anything.
> > Now how does that apply to my first statement?
>
> How does it not? (presuming I've managed to cull the correct quote
> from the mess that is people's inability to quote properly on here)
>
> > If a Pointer has the value of 0, then it SHOWS
> > the compilier that it points to nothing.
> >
> > Isn't KNOWS and SHOWS different? I thought
> > SHOWS doesn't mean understanding, but knows
> > definely does.
>
> 0 is one of the few _exceptions_ (as are explicit constants used in,
> for example, placement new and/or hardware addresses.)
>
> Given the following program:
>
> #include <stdio.h>
> int main(void){
> int i;
> printf("%p\n", &i);
> }
>
> How can the compiler possibly know, at compile time, what the value
> printed will be? And even if it did, what use could the compiler make
> of it?
>
> --
> PJH
>
> http://shabbleland.myminicity.com/env
> http://www.chavgangs.com/register.php?referer=9375
>
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