How can I use an array of function pointers?

Question

How should I use array of function pointers in C?

How can I initialize them?

Answer 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

Answer 2

This "answer" is more of an addendum to VonC's answer; just noting that the syntax can be simplified via a typedef, and aggregate initialization can be used:

typedef int FUNC(int, int);

FUNC sum, subtract, mul, div;
FUNC *p[4] = { sum, subtract, mul, div };

int main(void)
{
    int result;
    int i = 2, j = 3, op = 2;  // 2: mul

    result = p[op](i, j);   // = 6
}

// maybe even in another file
int sum(int a, int b) { return a+b; }
int subtract(int a, int b) { return a-b; }
int mul(int a, int b) { return a*b; }
int div(int a, int b) { return a/b; }

Answer 3

This should be a short & simple copy & paste piece of code example of the above responses. Hopefully this helps.

#include <iostream>
using namespace std;

#define DBG_PRINT(x) do { std::printf("Line:%-4d" "  %15s = %-10d\n", __LINE__, #x, x); } while(0);

void F0(){ printf("Print F%d\n", 0); }
void F1(){ printf("Print F%d\n", 1); }
void F2(){ printf("Print F%d\n", 2); }
void F3(){ printf("Print F%d\n", 3); }
void F4(){ printf("Print F%d\n", 4); }
void (*fArrVoid[N_FUNC])() = {F0, F1, F2, F3, F4};

int Sum(int a, int b){ return(a+b); }
int Sub(int a, int b){ return(a-b); }
int Mul(int a, int b){ return(a*b); }
int Div(int a, int b){ return(a/b); }
int (*fArrArgs[4])(int a, int b) = {Sum, Sub, Mul, Div};

int main(){
    for(int i = 0; i < 5; i++)  (*fArrVoid[i])();
    printf("\n");

    DBG_PRINT((*fArrArgs[0])(3,2))
    DBG_PRINT((*fArrArgs[1])(3,2))
    DBG_PRINT((*fArrArgs[2])(3,2))
    DBG_PRINT((*fArrArgs[3])(3,2))

    return(0);
}

Answer 4

The simplest solution is to give the address of the final vector you want , and modify it inside the function.

void calculation(double result[] ){  //do the calculation on result

   result[0] = 10+5;
   result[1] = 10 +6;
   .....
}

int main(){

    double result[10] = {0}; //this is the vector of the results

    calculation(result);  //this will modify result
}

Answer 5

This simple example for multidimensional array with function pointers":

void one( int a, int b){    printf(" \n[ ONE ]  a =  %d   b = %d",a,b);}
void two( int a, int b){    printf(" \n[ TWO ]  a =  %d   b = %d",a,b);}
void three( int a, int b){    printf("\n [ THREE ]  a =  %d   b = %d",a,b);}
void four( int a, int b){    printf(" \n[ FOUR ]  a =  %d   b = %d",a,b);}
void five( int a, int b){    printf(" \n [ FIVE ]  a =  %d   b = %d",a,b);}
void(*p[2][2])(int,int)   ;
int main()
{
    int i,j;
    printf("multidimensional array with function pointers\n");

    p[0][0] = one;    p[0][1] = two;    p[1][0] = three;    p[1][1] = four;
    for (  i  = 1 ; i >=0; i--)
        for (  j  = 0 ; j <2; j++)
            (*p[i][j])( (i, i*j);
    return 0;
}

Answer 6

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.