C Pointers – Understanding ‘const’ and Double Pointers Compatibility

c++constantspointers

This questions has been addressed here.

The suggested duplicate and the currently given answers don't address why there aren't issues with the examples given first. Mainly why doesn't the reasoning:

"const int ** is a pointer to const int * which is a different thing from just int*"

also apply for:

"const int * is a pointer to const int which is a different thing from just int"

I am approaching it from a different angle to hopefully get another explanation.

The code with the examples.

#include <stdio.h>

void f_a (int const a){

    /*
     *  Can't do:
     *      a = 3;  //error: assignment of read-only parameter ‘a’
     *
     * Explanation: I can't change the value of a in the scope of the function due to the const
    */
    printf("%d\n", a);
}

void f_ptr_a_type1 (int const * ptr_a){
    /*
     * Can do this:
     *     ptr_a’ = 0x3;
     * which make dereferencig to a impossible.
     *     printf("%d\n", * ptr_a’);  -> segfault
     * But const won't forbid it.
     *
     *  Can't do:
     *      *ptr_a’ = 3;  //error: assignment of read-only parameter ‘* ptr_a’
     *
     * Explanation: I can't change the value of a by pointer dereferencing and addignment due to the int const
    */
}

void f_ptr_a_type2 (int * const ptr_a){
    /*
     * Can do this:
     *     *a = 3;
     *
     *  Can't do:
     *      ptr_a = 3;  //error: assignment of read-only parameter ‘ptr_a’
     *
     * Explanation: I can't change the value because the const is protecting the value of the pointer in the funcion scope
    */
}

void f_ptr_ptr_a (int const ** ptr_ptr_a){
    /*
     * Can do this:
     *     ptr_ptr_a = 3;
     *     * ptr_ptr_a = 0x3;
     *
     *  Can't do:
     *      ** ptr_ptr_a = 0x3;  //error: assignment of read-only parameter ‘**ptr_a’
     *
     * Explanation: Makes sense. Just follows the pattern from previous functions.
    */
}

int main()
{
    int a = 7;
    f_a(a);

    int * ptr_a = &a;
    f_ptr_a_type1(&a);
    f_ptr_a_type2(&a);

    int ** ptr_ptr_a = &ptr_a;
    f_ptr_ptr_a(ptr_ptr_a);  //warning: passing argument 1 of ‘f_ptr_ptr_a’ from incompatible pointer type [-Wincompatible-pointer-types]
}

The accepted widely accepted answer goes something like this:

int ** isn't the same as const int** and you can't safely cast it

My question is why does the function suddenly care?

It didn't complain here that int isn't int const:

int a = 7;
f_a(a);

It didn't complain here because int * isn't neither int const * nor int * const:

int * ptr_a = &a;
f_ptr_a_type1(&a);
f_ptr_a_type2(&a);

But suddenly it starts complaining in the double pointer case.

  • Looking for explanations using this terminology and example?

  • Why does the function suddenly starts worrying about write
    permissions of something that is outside of her scope?

Best Answer

A conversion from e.g. char * to const char * is always safe. Through the const char *, the data that's pointed to can't be modified, and that's it.

On the other hand, a conversion from char ** to const char ** can be unsafe, therefore it isn't allowed implicitly. Instead of explaining it, consider the following code:

void foo(const char **bar)
{
    const char *str = "test string";
    *bar = str; // perfectly legal
}

int main(void)
{
    char *teststr[] = {0};
    foo((const char **)teststr);
    // now teststr points to a `const char *`!

    *teststr[0] = 'x'; // <- attempt to modify read-only memory
                       //    ok in this line, there's no const qualifier on teststr!
}

If the conversion from char ** to const char ** when calling foo() would be implicit, you would have an implicit way of converting a const away.

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