How this const value is getting changed..?

Question

I read that constant values cant be changed but in this code below the value of i is getting changed through the use of a pointer. May I know how?

#include <stdio.h>

int main()
{
    const int i = 10;
    int *ptr = &i;
    *ptr = 20;
    printf("%d\n", i);
    return 0;
}

The output of this code is 20 with a compiler warning.

This program is cheating -- breaking the rules -- and the warning reflects that. — Steve Summit, Oct 03 '19 at 14:40
`const` is kind of like a sticker on the back of an appliance saying "Do not open -- no user-serviceable parts inside." But if you get a screwdriver and open it up, you can try... — Steve Summit, Oct 03 '19 at 14:41
will the compiler tolerate..cheating and breaking rules..dont that throws error..? — Kannan Mrithunjai, Oct 03 '19 at 14:42
You said the compiler *did* throw a warning -- isn't that good enough for you? — Steve Summit, Oct 03 '19 at 14:42
@user8763383 most (if not all) warnings in C should be considered as errors. — Jabberwocky, Oct 03 '19 at 14:43
`const` does *not* mean, "Hey, compiler: it's super important that you put this variable in read-only memory so that, no matter what, it can't possibly be modified." It's more like, "I promise not to try to modify this variable, but I might forget, so please try to help me keep my promise, if it's not too hard." — Steve Summit, Oct 03 '19 at 14:43
Since the variable `i` in this program is a local ("stack") variable, there's typically no such thing as read-only memory to put it in, anyway. — Steve Summit, Oct 03 '19 at 14:45
"I read that constant values cant be changed" --> not quite. `const` objects should not be attempted to be changed. It might work, it might not, it might crash. It is UB. Enable a warnings and heed them. — chux - Reinstate Monica, Oct 03 '19 at 14:50
Did you check the assembly code output with compiler optimizations on ? My compiler ('Clang') completely ignored `*ptr = 20`. — rustyMagnet, Oct 03 '19 at 15:07

score 3 · Answer 1 · answered Oct 03 '19 at 14:43

3

The behaviour of your code is undefined - the language does not define the behaviour of modifying an object originally declared as const via a pointer that's had const stripped from it.

On some compilers with optimisations turned on, getting 10 as the output is reasonable.

answered Oct 03 '19 at 14:43

Bathsheba

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http://port70.net/~nsz/c/c11/n1570.html#6.7.3p6 – Eugene Sh. Oct 03 '19 at 14:46

How this const value is getting changed..?

1 Answers1