Why this code give me a warning says: passing argument 1 of "test" from incompatible pointer type? I know it's about the const before char **, but why?
void test(const int ** a)
{
}
int main()
{
int a=0;
int *b=&a;
int **c=&b;
test(c);
return 0;
}
You can't assign an int ** to a const int **, because if you did so, the latter pointer would allow you to give an int * variable the address of a const int object:
const int myconst = 10;
int *intptr;
const int **x = &intptr; /* This is the implicit conversion that isn't allowed */
*x = &myconst; /* Allowed because both *x and &myconst are const int * ... */
/* ... but now intptr points at myconst, and you could try to modify myconst through it */
const int **
is a pointer to a pointer to a const int, but you are passing a pointer to a pointer to an int
I think you may want to declare test with int ** const, which says that the pointer is const and not the value.
NOTE: and I think this should be put in every question regarding pointers in C: cdecl.org is a pretty nice way of giving a human-readable expression
The second answer to this question might be helpful:
Why can't I convert 'char**' to a 'const char* const*' in C?
Unfortunately the accepted answer is not very good and does nothing to explain the reason.
