Function returning pointer to array 5 of pointer to int

Question

I read this question: what does the line int *(*(x[3])())[5]; do in C?

which have the code line:

int *(*(*x[3])())[5];

in this answer https://stackoverflow.com/a/37364752/4386427

According to http://cdecl.org/ this means

declare x as array 3 of pointer to function returning pointer to array 5 of pointer to int

Now I'm wondering about this part:

function returning pointer to array 5 of pointer to int

How does the proto-type for a function returning pointer to array 5 of pointer to int look?

I tried this:

int* g()[5]    <---- ERROR: 'g' declared as function returning an array
{
    int** t = NULL;
    // t = malloc-stuff
    return t;
}

which doesn't compile.

Then I tried

#include <stdio.h>
#include <stdlib.h>

int *(*(*x[3])())[5];

int** g()
{
    int** t = NULL;
    // t = malloc-stuff
    return t;
}

int main(void) {
    x[0] = g;
    return 0;
}

which compiles fine but now the return type is more like pointer to pointer to int. There is nothing that says pointer to array 5 of pointer to int

So my question is:

Is it at all possible to write a function which returns pointer to array 5 of pointer to int ?

If yes, how does the proto-type look?

If no, what is the purpose of 5 in the declaration of x?

int *(*(*x[3])())[5];
                  ^
                what does 5 mean here?

int* (*g())[5] More parentheses are clearly the universal antidote, as more indirection is a panacea. — user3185968
– user3185968, Commented May 21, 2016 at 16:29
@EOF - Great - it compiles fine with your suggestion. Still, I'm puzzled as I can assign x[1]=g2; even when int* (*g())[2] (i.e. 3 instead of 5). I don't get any warnings - at least not here: ideone.com/mVt8g9 — 4386427
– 4386427, Commented May 21, 2016 at 16:46

Dragoslav Marinica · Accepted Answer · 2020-10-21 12:11:28Z

an array[5] of ints would be:

int array[5]; /* read: array of 5 ints */

and a pointer to that array (not just to the first element of it, but the whole array of 5!) would be:

int(* ptr)[5] = &array; /* read: pointer to an array of 5 ints */

and a function returning such a pointer would be:

int(* g())[5]; /*read: a function returning a pointer to an array of 5 ints */

Following the same logic, an array[5] of pointers_to_int would be:

int* array_of_ptrs[5]; /* read: array of 5 pointers_to_int */

and a pointer to that array would be:

int* (* PTR)[5] = &array_of_ptrs; /* read: pointer to an array of 5 pointers_to_int */

and theeeen a function returning such a pointer would be:

int* (* g())[5] /* read: function returning a pointer to an array of 5 pointers_to_int*/
/* just like @EOF said in the comments above! */

Let's try it out:

#include <stdio.h>

int array[5] ={1, 2, 3, 4, 5};

int a = 6, b = 7, c = 8, d = 9, e = 10;
int* array_of_ptrs[5] = {&a, &b, &c, &d, &e};

int(* g())[5] 
{
    return &array;
}

int* (* gg())[5]
{
    return &array_of_ptrs; 
}

int main()
{
int(* ptr)[5]; 
ptr = g();

int* (* PTR)[5];     
PTR = gg();
    
printf
("the value of the dereferenced 1st element of array_of_ptrs is: %d", *(*PTR)[0]);  

return 0;
}

clang prog.c -Wall -Wextra -std=gnu89 output:

the value of the dereferenced 1st element of array_of_ptrs is: 6

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