c - allocate memory to an array of pointers to functions

Question

I want my function to create an array and allocate memory for n pointers to functions (for example, functions that have no parameters and return int) and to return a pointer to that array.

I tried doing:

void* f(int n){
   return calloc(int (*arrayName[])(void),n);
}

But i'm getting a syntax error. I'm pretty new to c and i tried to dig for an hour how to solve this issue with no success. using the man page i figured calloc is the way to go but i might be wrong.

I'm a tad curious as to why you need to allocate an array of function pointers. Are you creating code on the fly? Or are you building some sort of dynamic programming environment where you chain functions together and execute them one after another. What I'm trying to say is that this "smells like bad code". — Mats Petersson
– Mats Petersson, Commented Dec 28, 2012 at 16:21

Kerrek SB · Accepted Answer · 2012-12-28 15:52:13Z

7

Make your life easier and use typedefs:

typedef int (*fp)();

fp * f(size_t n)
{
    return calloc(n, sizeof(fp));
}

_{The hand-rolled declaration: int (*f(size_t n))()}

answered Dec 28, 2012 at 15:52

Kerrek SB

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2 Comments

user529758 Over a year ago

Bah, somebody is seriously downvoting this answer? This is uncool. +1.

Marco Over a year ago

I've downvoted the answer. I think it should say something about what's wrong with Tom's attempt. Probably say something about how to use calloc, why malloc may be also right, something about the usage of sizeof.. I mean, you just said "use typedef" instead of explaining what he was doing wrong..

user529758 · Accepted Answer · 2012-12-28 15:56:44Z

2

Or if you don't want a typedef (hint: you still do want a typedef, just for completeness): wrap the type in a sizeof():

return calloc(n, sizeof(int (*)(void)));

answered Dec 28, 2012 at 15:56

user529758

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