Proper return statement and function type

Question

I am experimenting with malloc and I am trying to create a dynamic 2D array, I know the size of the second dimension and I am allocating memory like so:

int (*arr)[SIZE] = NULL;
arr = malloc(sizeof(arr[SIZE]) * 10);

My question is, is it possible place the above code into a function, and return the address of the allocated memory, if so what should the type be?

Edit: I am aware of the other method of allocating and return a pointer to a pointer.

Thanks in advance.

Answer 1

For starters the argument of malloc is confusing

arr = malloc(sizeof(arr[SIZE]) * 10);

It seems you mean

arr = malloc(sizeof( *arr ) * 10);

That is you are trying to allocate dynamically an array of the type int[10][SIZE].

More precisely the record used as an argument in the call of malloc

arr = malloc(sizeof(arr[SIZE]) * 10);

is correct but very confusing. It is better not to use such a record for the sizeof operator.

The function declaration can look like

int ( *allocation( size_t n ) )[SIZE];

Or you can introduce a typedef name like

typedef int ( *Array2D )[SIZE];

and then declare the function like

Array2D allocation( size_t n );

where n corresponds to the used by you value 10. That is using the parameter you can specify any number for the array dimension apart from 10.