allocating memory to a function pointer

Question

I have a func to add two nos and return (a+b). Then I created a func pointer to the func. Want to allocate memory for an array of that function pointer and access them. code is below.

My question is on the following line using malloc:

pp = (add_2nos*)malloc(5 * sizeof(add_2nos*))

sizeof(add_2nos*) and sizeof(add_2nos) does not make any difference while compiling. What is the difference if there is any ?? Also if type casting is necessary while I am allocating memory of the same type...?

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

int add(int a, int b) {
        return (a+b);
}

typedef int (*add_2nos)(int, int);     

int main() {

        add_2nos *pp; 
        
        // Defining an array of pointers to function add and accessing them
        pp = (add_2nos*)malloc(5 * sizeof(add_2nos*));
        pp[0] = add;
        pp[1] = add;
        printf("\n\nAdding two nos -- (14, 15): %d ", pp1[0](14, 15));
        printf("\nAdding two nos -- (16, 16): %d \n\n", pp1[1](16, 16));
}

In C the function name is the same of a pointer to function. Try `printf("sizeof fnc=%ld, sizeof fnc pointer=%ld\n", sizeof(add), sizeof(*add));` — Frankie_C, Dec 13 '20 at 18:01
Trying to apply `sizeof` to a function type is undefined behavior, per 6.5.3.4 in the C spec. So you might get the same size as a function pointer or you might get an error, or anything else. — Chris Dodd, Dec 13 '20 at 19:56
@ChrisDodd you're right. It is an UB. My example isn't correct. But the concept of function designator conversion from 'function returning type' to 'pointer to function returning type' is correct, as stated in §6.3.2.1 Lvalues, arrays, and function designators comma 4: . It also enforce the UB for the `sizeof`. — Frankie_C, Dec 13 '20 at 20:53
Thanks for responding to my query. I have another thought ... when we do sizeof() to a predefined data type pointer such as int *p or char *c, we do sizeof(*p) but when we use the data type itself we place the dereference symbol afterwards such as sizeof(int *) or same in case of user defined data type such as sizeof(add_2nos*) in the code above. Why it is like this way ?? Can someone point it out...? — Niranjan Das, Dec 15 '20 at 13:36

score 2 · Accepted Answer · answered Dec 13 '20 at 23:59

sizeof(add_2nos*) and sizeof(add_2nos) does not make any difference while compiling. What is the difference if there is any ?

add_2nos is void (*)(int, int) - it's a pointer to a function.

add_2nos* is a void (**)(int, int) - it's a pointer, to a pointer to a function.

Because on most architectures all pointers have the same size, including function pointers and pointers to pointers, and malloc() just takes a number, it doesn't make a difference. Anyway you want to allocate space for 5 function pointers (ie. 5 * sizeof(add_2nos)), not for 5 pointers to function pointers.

Don't think about it and let the compiler figure it out: pp = malloc(5 * sizeof(*pp)); is a common pattern.

Also if type casting is necessary while I am allocating memory of the same type...?

See do we cast the result of malloc in C.

Chris Dodd · Answer 2 · 2020-12-14T04:03:27.870

This

pp = (add_2nos*)malloc(5 * sizeof(add_2nos*));

is giving you an allocation based on the size of a pointer to a function pointer. That's probably the same as a function pointer, but maybe not.

pp = malloc(5 * sizeof(add_2nos));

or

pp = malloc(5 * sizeof *pp);

You've typedef's add_2nos to be a typename for "pointer to function with a particular signature", so to allocate an array of such pointers, you just want to use sizeof on it directly (no dereference), or use sizeof on the dereference of the array pointer.

Karol Adamiak · Answer 3 · 2020-12-15T06:26:16.330

-1

Pointer determines memory address. The easiest way to explain this is by this example:

printf("%ld\n",sizeof(unsigned char));
printf("%ld\n",sizeof(unsigned char *));

In the first line, the size of an unsigned char is printed, which is 1 byte. In the second line, the size of a pointer which keeps the information, where unsigned char is stored in the memory. This may vary between compilers/platforms. For example, on https://www.onlinegdb.com/online_c_compiler size of 8 bytes is printed.

edited Dec 15 '20 at 06:26

answered Dec 13 '20 at 18:07

Karol Adamiak

141
1
9

1

This causes undefined behaviour, `%d` is incorrect format specifier for the result of `sizeof` – M.M Dec 14 '20 at 04:07

allocating memory to a function pointer

3 Answers3