Re: Is this the ultimate stack-smash fix?

Xavier Bestel (xavier.bestel@free.fr)
21 Feb 2001 10:30:36 +0100


Le 21 Feb 2001 01:13:03 +0100, Andreas Bombe a écrit :
> On Tue, Feb 20, 2001 at 10:09:55AM +0100, Xavier Bestel wrote:
> > Le 20 Feb 2001 02:10:12 +0100, Andreas Bombe a écrit :
> > > On Sat, Feb 17, 2001 at 09:53:48PM -0700, Eric W. Biederman wrote:
> > > > Peter Samuelson <peter@cadcamlab.org> writes:
> > > > > It also sounds like you will be
> > > > > breaking the extremely useful C postulate that, at the ABI level at
> > > > > least, arrays and pointers are equivalent. I can't see *how* you plan
> > > > > to work around that one.
> > > >
> > > > Huh? Pointers and arrays are clearly different at the ABI level.
> > > >
> > > > A pointer is a word that contains an address of something.
> > > > An array is an array.
> > >
> > > An array is a word that contains the address of the first element.
> >
> >
> > No. Exercise 3: compile and run this:
> > file a.c:
> > char array[] = "I'm really an array";
> >
> > file b.c:
> > extern char* array;
> >
> > main() { printf("array = %s\n", array); }
> >
> > ... and watch it biting the dust !
>
> Deliberately linking to the wrong symbol is not a point. Might as well
> replace file a.c with "int array = 0;". That'll also bite the dust. So?
>
> > in short: an array is NOT a pointer.
>
> In this context we were talking *function calls*, not confusing the
> linker. And whether you say "char array[];" or "char *const array;",
> array is a pointer. Even more so at the ABI = function call interface.

OK, I missed this. There are no arrays in the function call interface,
they are promoted to pointers.

Xav

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