Modification of const int through a non-const pointer

Question

#include <stdio.h>

int main(){
    const int a = 10;
    *(int*)(&a) = 9; // modify a
    printf("%d", a);
    return 0;
}

When I run this code on Xcode, the output is 10 (not changed)
When I run this code on Visual Studio Community, the output is 9 (changed)

Why?

The better question would be: why are you trying to modify `const` value instead of learning something useful. It's against the language. — syntagma, May 08 '15 at 07:23
It's UB. Look at the generated assembly code or step through the code in disassembly mode. — Jabberwocky, May 08 '15 at 07:27

score 8 · Answer 1 · edited May 08 '15 at 09:26

8

This program would compile but exhibits undefined behavior and may output 9 or 10 or something else or may crash who knows.

When you say a is const, you promise that you won't try change the value of a directly or indirectly and compiler may make certain assumptions. If you break the promise unexpected things may happen.

edited May 08 '15 at 09:26

Gyapti Jain

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answered May 08 '15 at 07:21

Mohit Jain

30,259
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Sourav Ghosh · Accepted Answer · 2015-05-08T07:28:30.573

Q: why?

Ans: undefined behaviour.

To explain, if you try to modify a const variable value by accessing is through some non-const pointer, it invokes undefined behaviour.

As per C11 standard, chapter 6.7.3, paragraph 6.

If an attempt is made to modify an object defined with a const-qualified type through use of an lvalue with non-const-qualified type, the behavior is undefined.

Note: The recommended signature of main() is int main(void).

score 0 · Answer 3 · answered May 08 '15 at 07:28

0

const keyword is used not to change the value of variable. If its done forcefully results may be unexpected

answered May 08 '15 at 07:28

Pawan

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Modification of const int through a non-const pointer

3 Answers3

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