Triple stars: What's the difference between char* (*arr)[] and char*** arr (in C)?

arrays, c, pointers, syntax

Solution

As always, http://cdecl.org is your friend:

- `char * (*arr)[]` - "declare arr as pointer to array of pointer to char"

- `char *** arr` - "declare arr as pointer to pointer to pointer to char"

These are not the same. For a start, the first is an incomplete type (in order to use a pointer to an array, the compiler needs to know the array size).

Your aim isn't entirely clear. I'm guessing that really all you want to do is modify the underlying data in your array of `char *`. If so, then you can just pass a pointer to the first element:

void my_func(char **pointers) { 
    pointers[3] = NULL;  // Modify an element in the array
}

char *array_of_pointers[10];

// The following two lines are equivalent
my_func(&array_of_pointers[0]);
my_func(array_of_pointers);

If you really want to pass a pointer to an array, then something like this would work:

void my_func(char *(*ptr)[10]) {
    (*ptr)[3] = NULL;  // Modify an element in the array
}

char *array_of_pointers[10];

// Note how this is different to either of the calls in the first example
my_func(&array_of_pointers);

For more info on the important difference between arrays and pointers, see the dedicated chapter of the C FAQ: http://c-faq.com/aryptr/index.html.

Problem

Original source

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• Equivalence of p[0] and *p for incomplete array types