Why is it not OK to pass `char **` to a function that takes a `const char **` in C? [duplicate]

Question

Possible Duplicate:
Why can’t I convert ‘char**’ to a ‘const char* const*’ in C?

I am curious, why can't I pass a char ** to const char ** function? Where as it is OK to pass char * to a const char * function it seems not to be OK to do it with double pointers. I thought it was always ok to add constness (but not ok to drop constness) but now it seems I have been wrong.

Gcc compiler is giving me the errror:

note: expected ‘const char **’ but argument is of type ‘char **’

Here is the code snippet:

int f(const char **a) { }

int main() {
    char *a;
    f(&a);
}

Any ideas?

Answer 1

Because the compiler can't guarantee safety.

See Q11.10 from the comp.lang.c FAQ: Why can't I pass a char ** to a function which expects a const char **?

suppose you performed the following more complicated series of assignments:

const char c = 'x';    /* 1 */
char *p1;              /* 2 */
const char **p2 = &p1; /* 3 */
*p2 = &c;              /* 4 */
*p1 = 'X';             /* 5 */

In line 3, we assign a char ** to a const char **. (The compiler should complain.) In line 4, we assign a const char * to a const char *; this is clearly legal. In line 5, we modify what a char * points to--this is supposed to be legal. However, p1 ends up pointing to c, which is const. This came about in line 4, because *p2 was really p1. This was set up in line 3, which is an assignment of a form that is disallowed, and this is exactly why line 3 is disallowed.

Assigning a char ** to a const char ** (as in line 3, and in the original question) is not immediately dangerous. But it sets up a situation in which p2's promise--that the ultimately-pointed-to value won't be modified--cannot be kept.