pointer to function in c? - c

Question

I m learning c (programming in ANSI c -> fifth edition) and facing the below issue:
I'm implementing one program with pointers to functions
#include<stdio.h>
//void swap (int *a, int *b); //function declaration
int main()
{
int m = 25;
int n = 100;
printf("m is %d, n is %d\n", m, n);
swap(&m, &n); //calling a function
printf("m is %d, n is %d\n", m, n);
return 0;
}
void swap (int *a, int *b) //function implementation
{
int temp;
temp = *a;
*a = *b;
*b = temp;
}
in the above program comment the line void swap (int *a, int *b); then program run file but give one suggestion here facing the issue why here give the suggestion
main.c:10:3: warning: implicit declaration of function ‘swap’ [-Wimplicit-function-declaration]
main.c:16:6: warning: conflicting types for ‘swap’
main.c:10:3: note: previous implicit declaration of ‘swap’ was here
m is 25, n is 100
m is 100, n is 25
when I m uncomment this line void swap (int *a, int *b); then program run fine
#include<stdio.h>
void swap (int *a, int *b); //function declaration
int main()
{
int m = 25;
int n = 100;
printf("m is %d, n is %d\n", m, n);
swap(&m, &n); //calling a function
printf("m is %d, n is %d\n", m, n);
return 0;
}
void swap (int *a, int *b) //function implementation
{
int temp;
temp = *a;
*a = *b;
*b = temp;
}
problem is why not allowed this one in the above program
void swap (int, int); //function declaration
and give the error
error:
conflicting types for ‘swap’
void swap (int *a, int *b) //function implementation
but in book example allowed

Answer 1

void swap (int *a, int *b);
This here before main() is called function signature that should match function declaration
below.
void swap (int *a, int *b){
...
}
But the same function signature can be written as
void swap (int *, int *);
And not
void swap (int , int ); //This will result in type mismatch
I guess this is what you are asking.
NOTE: Forward declaration is important here which lets you declare first and define later helps maintain code readability.