Function call by reference in C

Advertisements


The call by reference method of passing arguments to a function copies the address of an argument into the formal parameter. Inside the function, the address is used to access the actual argument used in the call. This means that changes made to the parameter affect the passed argument.

To pass the value by reference, argument pointers are passed to the functions just like any other value. So accordingly you need to declare the function parameters as pointer types as in the following function swap(), which exchanges the values of the two integer variables pointed to by its arguments.

/* function definition to swap the values */
void swap(int *x, int *y)
{
   int temp;
   temp = *x;    /* save the value at address x */
   *x = *y;      /* put y into x */
   *y = temp;    /* put temp into y */
  
   return;
}

To check the more detail about C - Pointers, you can check C - Pointers chapter.

For now, let us call the function swap() by passing values by reference as in the following example:

#include <stdio.h>
 
/* function declaration */
void swap(int *x, int *y);
 
int main ()
{
   /* local variable definition */
   int a = 100;
   int b = 200;
 
   printf("Before swap, value of a : %d\n", a );
   printf("Before swap, value of b : %d\n", b );
 
   /* calling a function to swap the values.
    * &a indicates pointer to a ie. address of variable a and 
    * &b indicates pointer to b ie. address of variable b.
   */
   swap(&a, &b);
 
   printf("After swap, value of a : %d\n", a );
   printf("After swap, value of b : %d\n", b );
 
   return 0;
}

Let us put above code in a single C file, compile and execute it, it will produce the following result:

Before swap, value of a :100
Before swap, value of b :200
After swap, value of a :200
After swap, value of b :100

Which shows that the change has reflected outside of the function as well unlike call by value where changes does not reflect outside of the function.



Advertisements
Advertisements