Thread: How come compiler lets this pass?

How come the gnu compiler lets this pass?

Code:

#include <stdio.h>
#include <stdlib.h>

int main(void) {
    char x = 1;
    char const *f = &x;

    *f=3;

    printf("The value is: %d \n", x);
    return 0;
}

$gcc -Wall mod.c -o mod
mod.c: In function `main':
mod.c:8: warning: assignment of read-only location
$./mod
The value is: 3
$

All the compiler does is give a warning, but lets this pass. Is this just how they chose to implement the c compiler?

I tested this on gcc version 3.3.3 using Suse Linux 9.1 on an standard PC and
gcc version 2.95.3 20010315 using FreeBSD 4.6 on some doped out VAX. I got the same error messages on both platforms.

So in other words, you are using a compiler on a platform that corrects this nonesense.

I don't know how to put this in quotes: I just tried using -Werror

$gcc -Werror mod.c -o mod
mod.c: In function `main':
mod.c:8: warning: assignment of read-only location
$./mod
bash: ./mod: No such file or directory
$

The compiler didn't let it pass. That is truly amazing. I didn't realize gcc had this kind of option.

Thanks. I've been enlightened.

note: these kind of thing dont happen in C++. the following program is written in C++ and compilers in GNU compiler just with -W flag

Code:

#include <iostream>

using std::cout;
using std::endl;

int main(void) {
    char x = 1;
    char const *f = &x;

    *f=3;

    cout<<"The value is: "<<x<<endl;
    return 0;
}
/* it gives u an error at first first itself
const.cpp: In function âint main()â:
const.cpp:10: error: assignment of read-only location
*/

ssharish2005

FWIW, on gcc 4.0.2, it's an error, whether -W or -Wall is specified or not.

On a plain vanilla compile, it shows up as an error.

BTW, the platform is Linux.