Syntax for returning array pointers in C

Question

EDIT: the correct syntax for declaring a function with the return type of array pointer in the example I wrote further below in my code example section should be as follows:

int (*my_function(void))[10]
{
    ...
}

Important note for future readers who might come across the same question: don't typecast malloc! Just don't. Thanks to user3386109 for that one. Further reading on the why.

Thanks to Andrew Henle and Eric Postpischil for their answers. Here's a really good explanation if anyone wants to read further on the the topic. Original thread below:

What I want to know

How to return a pointer to array in C.

Why I want to know that

Because it (is?) should be possible to do so.

For what I want it

To return multidimensional arrays from functions.

"Have you tried structs?"

Yes, but this isn't about structs. The point (no pun intended) of this thread is doing it with pointers.

Code example

My example is wrong, thus the reason I'm asking here, but here it go:

int (*)[10]my_function(void) // Should be returning a type of array pointer to 10-elements array. Sure enough it's not.
{
    int (*ptr)[10] = (int (*)[10]) malloc(10 * sizeof(int));
    return ptr;
}

int main(void)
{
    printf("Address of ptr is: %d", my_function());
    return 0;
}

As you can see, I'm not sure about how to typecast malloc to type array pointer to 10-elements array as well. If you know how, please let me know.

IMPORTANT NOTE

As far as I know, using double pointers (pointer to pointer) in this case is wrong. Code example:

int **my_function(void); // WRONG, afaik

Rationale: this video, at 16:22.

I already tried searching online and in some books, they have examples using structs, pointer to array as function arguments, but not this.

If you have some knowledge about this topic, please reply, I'll be grateful.

Answer 1

Arrays can't be passed by value, unless you embed them in a structure. When you pass an array to or from a function, it decays to a pointer to the first element. So the way to return a dynamically-allocated array is as a single pointer.

int *my_function(void) {
    int *ptr = malloc(10 * sizeof(int));
    return ptr;
}

There's no need to specify the dimension in the type, as this will be ignored.

This is not a multi-dimensional array, it's just a 1-d array of 10 int.

The type declaration int (*ptr)[10] is for an array of 10 pointers to integers, not a pointer to an array of 10 integers.

Answer 2

int (*p)[10] declares a pointer to an array of 10 int.

int (*)[10] is a type that is a pointer to an array of 10 int. However, note that you do not need to cast the result of malloc in C. As a void * used as an initializer or right operand of assignment, it will be automatically converted to the target type.

int (*foo(void))[10] declares a function taking no parameters that returns a pointer to an array of 10 int.

Note that int (*foo(void))[10] can be derived from int (*p)[10] simply by replacing p with foo(void). Generally, to get any return type for a function, write a declaration for an object of that type, then replace the object name with the function name followed by its usual parameter declarations.