[OT] Re: Why Is Escaping Data Considered So Magical?

[Michael Torrie]

On 06/29/2010 06:25 PM, Lawrence D'Oliveiro wrote:
> I have yet to find an architecture or C compiler where it DOESN’T work.
> 
> Feel free to try and prove me wrong.

Okay, I will. Your code passes a char** when a char* is expected.  Every
compiler I know of will give you a *warning*.  Mistaking char*, char**,
and char[] is a common mistake that almost every C program makes in the
beginning.  Now for the proof:

Consider this variation where I use a dynamically allocated buffer
instead of static:

#include <stdio.h>

int main(int argc, char ** argv)
{
	char *buf = malloc(512 * sizeof(char));
	const int a = 2, b = 3;
	snprintf(&buf, sizeof buf, "%d + %d = %d\n", a, b, a + b);
	fprintf(stdout, buf);
	free(buf);
	return 0;
} /*main*/

On my machine, an immediate segfault (stack overrun).  Your code only
works because your buf is statically allocated, which means &buf==buf.
But this equivalance does not hold for any other situation.  If your
buffer was dynamically allocated on the heap, instead of passing a
pointer to the buffer (which *is* what buf itself is), you are passing a
pointer to the pointer, which is where buf is stored on the stack, but
not the buffer itself.  Instant stack corruption.