Re: Is this the ultimate stack-smash fix?

[Andreas Bombe <andreas.bombe@munich.netsurf.de>]

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.

Another try:  Assume that
	#include "secret.h"
	int main() { printf("array = %s\n", array); return 0; }
is correct code.

Is the variable array a pointer to a char or an array of chars?

Oh well, who cares.

-- 
 Andreas E. Bombe <andreas.bombe@munich.netsurf.de>    DSA key 0x04880A44
http://home.pages.de/~andreas.bombe/    http://linux1394.sourceforge.net/