Re: Is this the ultimate stack-smash fix?

[Xavier Bestel]

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 <[e-mail]> 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