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# Returning values by reference in C++

A C++ program can be made easier to read and maintain by using references rather than pointers. A C++ function can return a reference in a similar way as it returns a pointer.

When a function returns a reference, it returns an implicit pointer to its return value. This way, a function can be used on the left side of an assignment statement. For example, consider this simple program:

    #include <iostream>
    #include <ctime>

    using namespace std;

    double vals[] = {10.1, 12.6, 33.1, 24.1, 50.0};

    double& setValues( int i )
    {
      return vals[i];   // return a reference to the ith element
    }

    // main function to call above defined function.
    int main ()
    {

       cout << "Value before change" << endl;
       for ( int i = 0; i < 5; i++ )
       {
           cout << "vals[" << i << "] = ";
           cout << vals[i] << endl;
       }

       setValues(1) = 20.23; // change 2nd element
       setValues(3) = 70.8;  // change 4th element

       cout << "Value after change" << endl;
       for ( int i = 0; i < 5; i++ )
       {
           cout << "vals[" << i << "] = ";
           cout << vals[i] << endl;
       }
       return 0;
    }

When the above code is compiled together and executed, it produces following result:

    Value before change
    vals[0] = 10.1
    vals[1] = 12.6
    vals[2] = 33.1
    vals[3] = 24.1
    vals[4] = 50
    Value after change
    vals[0] = 10.1
    vals[1] = 20.23
    vals[2] = 33.1
    vals[3] = 70.8
    vals[4] = 50

When returning a reference, be careful that the object being referred to does not go out of scope. So it is not legal to return a reference to local var. But you can always return a reference on a static variable.

    int& func() {
       int q;
       //! return q; // Compile time error
       static int x;
       return x;     // Safe, x lives outside this scope
    }

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