C language - Invoking functions based on index value passed as argument - array of pointers to functions

Question

Thank you for taking the time to read this, I looked for answers before posting but I'm very new to the language. This exercise I'm trying to do is from the book "Effective C: An introduction to professional C programming".

This is my first go at learning the language, and the exercise from the 2nd chapter of the book is as follows:

Declare an array of three pointers to functions and invoke the appropriate function based on an index value passed in as an argument

I am not totally sure I understand what it's saying, but I have a piece of functioning code I think does the job. However, I'm not sure if I'm interpreting it correctly. Here's my code:

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

void f0(int x) {
  printf("I am f0 and in index location %d\n", x);
}

void f1(int x) {
  printf("I am f1 and in index location %d\n", x);
}

void f2(int x) {
  printf("I am f2 and in index location %d\n", x);
}

int main(void){

  void (*f0p)(int);
  f0p = &f0;

  void (*f1p)(int);
  f1p = &f1;

  void (*f2p)(int);
  f2p = &f2;

  void *array[3] = {f0p, f1p, f2p};

  for (int i = 0; i < 3; i++) {
    void (*program)(int);
    program = array[i];
    program(i);
  }
  
  return 0;
}

this works after compiling and returns the following:

I am f0 and in index location 0
I am f1 and in index location 1
I am f2 and in index location 2

However, am I completing the exercise correctly? I don't think I'm technically using the index as an argument and calling the function, but I'm a noob. Any validation or correction / education you provide would be extremely appreciated. I spent many hours on this today!

Looks reasonable to me. Could you do it without the local variables, and declare the array like this: void *array[3] = {&f0, &f1, &f2}; — bruceg
– bruceg, Commented Jan 30, 2021 at 2:04

Eric Postpischil · Accepted Answer · 2021-01-30 02:13:00Z

3

Pointers should not be converted between pointers-to-functions and pointers-to-objects (including void) except in special situations.

The array is better declared as an array of pointers to functions:

void (*arrray[])(int) = { f0, f1, f2 };

and the functions can be called without an intermediate variable:

array[i](i);

answered Jan 30, 2021 at 2:13

Eric Postpischil

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1 Comment

libertypointer Over a year ago

Thanks! I updated my code with your suggestion and it works great, thank you so much.

0___________ · Accepted Answer · 2021-01-30 02:27:46Z

0

You can also typedef functions and use syntax similar to "normal" pointers and arrays.

typedef void functype(int);

void f0(int x) {
  printf("I am f0 and in index location %d\n", x);
}

void f1(int x) {
  printf("I am f1 and in index location %d\n", x);
}

void f2(int x) {
  printf("I am f2 and in index location %d\n", x);
}

int main(void)
{
    functype *array[] = {f0, f1, f2};

    functype *ptr = f0;

    for(int x = 0; x < sizeof(array) / sizeof(array[0]); x++)
    {
        array[x](x);
    }

    ptr(4);
}

answered Jan 30, 2021 at 2:27

0___________

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