C++ function call by pointer

The call by pointer 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 pointer, 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; /* store the value at address x */

*x = *y; /* insert y into x */

*y = temp; /* insert x into y */



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

#include <iostream>

// function declaration

void swap(int *x, int *y);

int main ()


// local variable declaration:

int a = 5;

int b = 10;

cout << ” Value of a before swapping :” << a << endl;

cout << ” Value of b before swapping :” << b << endl;

/* 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);

cout << ” Value of a after swapping :” << a << endl;

cout << ” Value of b after swapping :” << b << endl;

return 0;


When the above code is put together in a file, compiled and executed, it produces following result:

Mohit Arora
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