Function call by reference in C

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 functionswap(), 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 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.

Function call by value in C

The call by value method of passing arguments to a function copies the actual value of an argument into the formal parameter of the function. In this case, changes made to the parameter inside the function have no effect on the argument.

By default, C programming language uses call by value method to pass arguments. In general, this means that code within a function cannot alter the arguments used to call the function. Consider the function swap() definition as follows.

/* function definition to swap the values */
void swap(int x, int y)
{
   int temp;

   temp = x; /* save the value of x */
   x = y;    /* put y into x */
   y = temp; /* put temp into y */
  
   return;
}

Now let us call the function swap() by passing actual values 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 */
   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 following result:

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

Which shows that there is no change in the values though they had been changed inside the function.