Call by Reference | C++ Call by Reference Using pointers With Examples - Learn C++ - C++ Tutorial - C++ programming




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  • In this article, you'll learn to pass pointers as an argument to the function, and use it efficiently in your program.
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  • In C++ Functions article, you learned about passing arguments to a function. This method used is called passing by value because the actual value is passed.
  • However, there is another way of passing an argument to a function where where the actual value of the argument is not passed. Instead, only the reference to that value is passed.
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Example 1: Passing by reference without pointers

#include <iostream>
using namespace std;

// Function prototype
void swap(int&, int&);

int main()
{
    int a = 1, b = 2;
    cout << "Before swapping" << endl;
    cout << "a = " << a << endl;
    cout << "b = " << b << endl;

    swap(a, b);

    cout << "\nAfter swapping" << endl;
    cout << "a = " << a << endl;
    cout << "b = " << b << endl;

    return 0;
}

void swap(int& n1, int& n2) {
    int temp;
    temp = n1;
    n1 = n2;
    n2 = temp; 
}

Output

Before swapping 
a = 1
b = 2

After swapping
a = 2
b = 1
  • In main(), two integer variables a and b are defined. And those integers are passed to a function swap() by reference.
  • Compiler can identify this is pass by reference because function definition is void swap(int& n1, int& n2) (notice the & sign after data type).
  • Only the reference (address) of the variables a and b are received in the swap() function and swapping takes place in the original address of the variables.
  • In the swap() function, n1 and n2 are formal arguments which are actually same as variables a and b respectively.
  • There is another way of doing this same exact task using pointers.

Example 2: Passing by reference using pointers

#include <iostream>
using namespace std;

// Function prototype
void swap(int*, int*);

int main()
{
    int a = 1, b = 2;
    cout << "Before swapping" << endl;
    cout << "a = " << a << endl;
    cout << "b = " << b << endl;

    swap(&a, &b);

    cout << "\nAfter swapping" << endl;
    cout << "a = " << a << endl;
    cout << "b = " << b << endl;
    return 0;
}

void swap(int* n1, int* n2) {
    int temp;
    temp = *n1;
    *n1 = *n2;
    *n2 = temp;
}
  • The output of this example is same as before.
  • In this case, the address of variable is passed during function call rather than the variable itself.
swap(&a, &b); // &a is address of a and &b is address of b
  • Since the address is passed instead of value, dereference operator must be used to access the value stored in that address.
void swap(int* n1, int* n2) { 
 ... .. ...
}
  • The *n1 and *n2 gives the value stored at address n1 and n2 respectively.
  • Since n1 contains the address of a, anything done to *n1 changes the value of a inmain() function as well. Similarly, b will have same value as *n2.

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