What is the significance of forward declaration in C programming?

Question

I am now learning C programming through Learn C the Hard Way by Zed A. Shaw. There is this code (taking from his website):

#include <stdio.h>
#include <ctype.h>

// forward declarations
int can_print_it(char ch);
void print_letters(char arg[]);

void print_arguments(int argc, char *argv[])
{
    int i = 0;

    for(i = 0; i < argc; i++) {
        print_letters(argv[i]);
    }
}

void print_letters(char arg[])
{
    int i = 0;

    for(i = 0; arg[i] != '\0'; i++) {
        char ch = arg[i];

        if(can_print_it(ch)) {
            printf("'%c' == %d ", ch, ch);
        }
    }

    printf("\n");
}

int can_print_it(char ch)
{
    return isalpha(ch) || isblank(ch);
}


int main(int argc, char *argv[])
{
    print_arguments(argc, argv);
    return 0;
}

Don't we just can code it like this (put the can_print_it and print_letters functions at the top and remove the need for forward declaration):

#include <stdio.h>
#include <ctype.h>

int can_print_it(char ch)
{
    return isalpha(ch) || isblank(ch);
}

void print_letters(char arg[])
{
    int i = 0;

    for(i = 0; arg[i] != '\0'; i++) {
        char ch = arg[i];

        if(can_print_it(ch)) {
            printf("'%c' == %d ", ch, ch);
        }
    }

    printf("\n");
}

Is there really times when the forward declaration is important and unavoidable?

Answer 1

Forward declarations of functions in C typically have two different uses.

Modules

The header of exported functions are declared in a header file which is included in a client module.

Mutual Recursion

In mutual recursion two functions call each other repeatedly. Without a forward declaration one of the two functions will be undeclared in the body of the other.

Example:

int Odd(int n);

int Even(int n)
{
    return (n == 0)? 1: Odd(n - 1);
}

int Odd(int n)
{
    return (n == 0)? 0: Even(n - 1);
}

With a function pointer though, we can do without a forward declaration:

int (*odd)(int n);

int Even(int n)
{
    return (n == 0)? 1: odd(n - 1);
}

int Odd(int n)
{
    return (n == 0)? 0: Even(n - 1);
}

void Init(void)
{
    odd = Odd;
    ...
}