C: Constant Is Not Constant

const does not mean that the variable is constant. In this post, we explore the true meaning of the const qualifier and its implications.

In C, const is a type qualifier that is used to specify that a variable’s value cannot be modified after it has been initialized. The const type of variable is a read-only variable, meaning that it can be read but cannot be changed.

This understanding of the const keyword is misleading!

The reassignment of c to X in the following code is not allowed!

Experiment #1

#include <stdio.h>

int main() {
    volatile const char c = 'A';

    c = 'X';

    printf(" &c: %p |    c: %c\n", &c, c);

    return 0;
}

❯ gcc main.c
main.c:6:7: error: cannot assign to variable 'c' with const-qualified type 'const volatile char'
    c = 'X';
    ~ ^
main.c:4:25: note: variable 'c' declared const here
    volatile const char c = 'A';
    ~~~~~~~~~~~~~~~~~~~~^~~~~~~
1 error generated.

So, a const variable cannot be modified?

Now try this

What do you think the output of the following code will be?

Experiment #2

#include <stdio.h>

int main() {
    volatile const char c = 'A';
    char *ptr = (char *)&c;

    printf(" &c: %p |    c: %c\n", &c, c);

    *ptr = 'X';

    printf(" &c: %p |    c: %c\n", &c, c);
    printf("ptr: %p | *ptr: %c\n", ptr, *ptr);

    return 0;
}

This code will compile with no problems. When I compile and run it, I see the following -

❯ gcc main.c
❯ ./a.out
 &c: 0x16f82701b |    c: A
 &c: 0x16f82701b |    c: X
ptr: 0x16f82701b | *ptr: X

c was supposed to be A always! No? What happened to the c being a const? How did it go from being A to being X?

What happened here?

Notice that in Experiment #2 we declared a pointer variable to point to the c variable - char *ptr = (char *)&c; and then we manipulated the memory directly - *ptr = 'X';.

In my particular case, the location at which c was stored happened to be 0x16f82701b (this varies from machine to machine and will be different for you).

All we did was go to the location and sneakily modify it!

const does not mean (or guarantee) that the variable is placed in a read-only memory region and is a constant!

We used the power of pointers to manipulate the memory directly. The location at which c was stored(0x16f82701b) happened to be writable. Because of this, writing to the location was no problem! Had that not been the case, we should have seen some runtime error or crash (like segmentation fault).

What const really means

The difference between Experiment #1 and Experiment #2 is that in the former we asked the compiler to modify the variable and in the latter, we modified it ourselves (using a pointer)!

✅ const is a way to tell the compiler to only generate load instructions to interact with the variable. The compiler refuses to generate store instructions that are needed to be able to modify the variable.

By this logic, the variable is supposed to be read-only and thus behave as if it is a constant!

Again, when we go the pointer way, the compiler has no problem generating instructions for the pointer dereference. Accessing the memory, fetching the values, and issuing writes to it is no problem!

❗ Pointers can be used to attempt a modification of const variable!

Conclusion

const qualifier instructs the compiler to not generate store instructions for the location of a variable as a result of the variable reassignment.

It does not mean in any way that the variable’s value will be constant! It is not guaranteed!

More food for thought

Remove the volatile qualifier from the code in Experiment #2, compile, and rerun.

#include <stdio.h>

int main() {
    const char c = 'A';
    char *ptr = (char *)&c;

    printf(" &c: %p |    c: %c\n", &c, c);

    *ptr = 'X';

    printf(" &c: %p |    c: %c\n", &c, c);
    printf("ptr: %p | *ptr: %c\n", ptr, *ptr);

    return 0;
}

Why is the same memory location having two different values in the output below?

❯ gcc main.c
❯ ./a.out
 &c: 0x16cf6f37b |    c: A
 &c: 0x16cf6f37b |    c: A
ptr: 0x16cf6f37b | *ptr: X