How can I use an array of function pointers?

How should I use array of function pointers in C?

How can I initialize them?

Solution 1:

You have a good example here (Array of Function pointers), with the syntax detailed.

int sum(int a, int b);
int subtract(int a, int b);
int mul(int a, int b);
int div(int a, int b);

int (*p[4]) (int x, int y);

int main(void)
{
  int result;
  int i, j, op;

  p[0] = sum; /* address of sum() */
  p[1] = subtract; /* address of subtract() */
  p[2] = mul; /* address of mul() */
  p[3] = div; /* address of div() */
[...]

To call one of those function pointers:

result = (*p[op]) (i, j); // op being the index of one of the four functions

Solution 2:

The above answers may help you but you may also want to know how to use array of function pointers.

void fun1()
{

}

void fun2()
{

}

void fun3()
{

}

void (*func_ptr[3])() = {fun1, fun2, fun3};

main()
{
    int option;


    printf("\nEnter function number you want");
    printf("\nYou should not enter other than 0 , 1, 2"); /* because we have only 3 functions */
    scanf("%d",&option);

    if((option>=0)&&(option<=2))
    { 
        (*func_ptr[option])();
    }

    return 0;
}

You can only assign the addresses of functions with the same return type and same argument types and no of arguments to a single function pointer array.

You can also pass arguments like below if all the above functions are having the same number of arguments of same type.

  (*func_ptr[option])(argu1);

Note: here in the array the numbering of the function pointers will be starting from 0 same as in general arrays. So in above example fun1 can be called if option=0, fun2 can be called if option=1 and fun3 can be called if option=2.