access of memory beyond allocation, not causing segmentation fault...

[Chris Dollin]

One reason he gave was that ANSI permitted the implict cast. OK. I
surrender, _conversion_, reluctantly, and knowing that they were driving a
coach and horses through the type-checking system.

/Casts/ (of pointer types) give you a coach-and-horses -- nay, a veritable
battleship -- to drive with. Allowing an implicit conversion doesn't seem
to make a significant difference and, by removing the necessity to cast
where `void*` is involved, make it, I think, less likely that casts between
non-void-* pointers will appear as a path of less resistance.

 

[Peter 'Shaggy' Haywood]

Groovy hepcat (e-mail address removed) was jivin' on Tue, 08 May 2007 22:19:16
-0700. It's a cool scene! Dig it.

#include <stdlib.h>

int main()
{
char *p = (char*)malloc(100);
if(p==NULL)
{
printf("...null...\n");
}
else
{
printf("...success...\n");
printf("1.%d\n",p[10]);
memset(p,0,101);
printf("2.%d\n",p[10]);
free(p);
printf("3.%d\n",p[101]);

All these printf() calls cause undefined behaviour, since you have
neglected to include stdio.h.

}

A C90 compiler is required to issue a diagnostic for a non-void
function without a return statement. In C99 it is permissible to leave
out a return statement in main(). However, there are apparently very few,
if any, C99 compilers. So chances are you don't have one. In either case,
it's always been a good idea to put at least one return statement in
main(). Return 0 or EXIT_SUCCESS to indicate successful execution and
EXIT_FAILURE to indicate termination due to an error. Those macros are
defined in stdlib.h, which you have included.

}

I used gcc on Solaris OS and i get the following output,

[Snip.]

You needn't get any output at all. Your program needn't even
successfully compile.

--
Dig the sig!



----------- Peter 'Shaggy' Haywood ------------

Ain't I'm a dawg!!