Declaring a function pointer returning an array

Question

For practice, I'm trying to :

Declare fubar to be a pointer to a function that takes a pointer to a char and returns a pointer to an array of 24 elements where each element is a pointer to a struct foo.

My logic is:

-fubar is a pointer to a function taking a char pointer: (*fubar)(char*)

-...returning a pointer to an array of 24 elems of where each elem is a struct foo:

(struct foo *)(*fubar)(char*)[24]

Is my logic correct?

Answer 1

After fixing the syntax error and removing the parentheses around struct foo *,

struct foo* (*fubar)(char*)[24]

...the one part that you got wrong is that it actually returns an array of pointers, not a pointer to an array. In order to declare a pointer to the array, you need an extra set of parentheses:

struct foo (*(*fubar)(char*))[24]

You can pretend that that the star belongs to an identifier (i.e., the name of the array) inside the parentheses.

Answer 2

Functions never return arrays, they may return a pointer (conventionally to the first cell of an array), since as return value or as argument an array decays to a pointer.

So, declare with a typedef the signature of your function:

typedef struct foo** funsig_t(char*);

^{notice that if you omit the typedef you would declare a function funsig_t of the desired signature.}

then declare a pointer using that typedef:

funsig_t* fubar;