Why was I able to change the value of a const int here?

Question

My code:

#include <stdio.h>
main()
{
    const int x = 10;
    int *p;
    p=&x;
    *p=20;
    printf("Value at p: %d \n",*p);
    printf("Value at x: %d", x);
}

The output I get is:

Value at p: 20
Value at x: 20

Thus, the value of a constant variable is changed. Is this one of the disadvantages of using pointers?

C doesn't prevent you from doing dumb things like this. You do get a warning from the compiler though: `const.c:9: warning: assignment discards qualifiers from pointer target type` — MatthewD, May 06 '13 at 01:33
Note that depending on the compiler and optimizations enabled, you could get have in your output "Value at x: 10". — fbafelipe, May 06 '13 at 01:49
The only useful `const` in C that I can see is in parameters. — Jack, May 06 '13 at 05:01

score 2 · Answer 1 · answered May 06 '13 at 01:33

2

You used a int* to point to a const int. you should get:

 error: invalid conversion from ‘const int*’ to ‘int*’

when you do:

p = &x;

You probably needs to update your compiler, a decent compiler should have told you this error or at least gave you warning about this.

answered May 06 '13 at 01:33

taocp

23,276
10
49
62

But he didn't, because he is using a bad compiler, I guess. – Elazar May 06 '13 at 01:35
@Elazar I guess so. Or may be old versioned compilers – taocp May 06 '13 at 01:35
I use gcc! I thought gcc was good enough! – user2350631 May 06 '13 at 01:38
@user2350631 I use gcc too, what version is that? – taocp May 06 '13 at 01:38
@user2350631 gcc 4.7.3 report the error. – gongzhitaao May 06 '13 at 01:38
the version is 4.5.3! – user2350631 May 06 '13 at 01:40
@user2350631 then it is weird, since I am also using gcc 4.5.3 but reported the error – taocp May 06 '13 at 01:41
@user2350631 Did you disable warnings? – David Heffernan May 06 '13 at 01:56
Or, @user2350631: are you reviewing the warnings? Many developers assume that if it compiles, it's good. – Michael Petrotta May 06 '13 at 02:01
Oh yes! I am receiving a warning! – user2350631 May 06 '13 at 02:03

score 0 · Answer 2 · edited May 23 '17 at 12:15

0

Please check the following error message:

error: invalid conversion from ‘const int*’ to ‘int*’

const int, int const, const int *, int const *, please see this post: const int = int const?

It's just how you use it.

edited May 23 '17 at 12:15

Community

1
1

answered May 06 '13 at 01:33

gongzhitaao

6,566
3
36
44

score 0 · Answer 3 · edited May 23 '17 at 12:15

0

That's because you are using the C language in the wrong way and the compiler let you compile this code while giving only a warning

warning: assignment discards ‘const’ qualifier from pointer target type [enabled by default]

And there are answers for that on SO, such as warning: assignment discards qualifiers from pointer target type

edited May 23 '17 at 12:15

Community

1
1

answered May 06 '13 at 01:34

user2348816

532
4
9

score 0 · Answer 4 · answered May 06 '13 at 01:35

Any decent compiler will tell you that you are discarding the const qualifier.

C assumes that the programmer is always right, hence it's your choice to ignore your compiler's warnings or not. As usual, this is not a disadvantage as long as you know what you're doing!

score 0 · Answer 5 · answered May 06 '13 at 01:51

As other answers have noted, writing a program to try to modify a const qualified variable like this results in a program that has undefined behaviour. This means that your program can do anything at all - as an example of this, when I compile your program with optimisations enabled, I see this output:

Value at p: 20
Value at x: 10

..and if I add the static qualifier to the variable x then the program crashes at runtime.

Why was I able to change the value of a const int here?

5 Answers5