Reference-Type Function Returns

Functions can be declared to return a reference type. There are two reasons to make such a declaration:

• The information being returned is a large enough object that returning a reference is more efficient than returning a copy.

• The type of the function must be an l-value.

Just as it can be more efficient to pass large objects to functions by reference, it also can be more efficient to return large objects from functions by reference. Reference-return protocol eliminates the necessity of copying the object to a temporary location prior to returning.

Reference-return types can also be useful when the function must evaluate to an l-value. Most overloaded operators fall into this category, particularly the assignment operator. Overloaded operators are covered in Overloaded Operators.

Example

Consider the Point example:

// refType_function_returns.cpp
// compile with: /EHsc

#include <iostream>
using namespace std;

class Point
{
public:
// Define "accessor" functions as
//  reference types.
unsigned& x();
unsigned& y();
private:
unsigned obj_x;
unsigned obj_y;
};

unsigned& Point :: x()
{
return obj_x;
}
unsigned& Point :: y()
{
return obj_y;
}

int main()
{
Point ThePoint;
// Use x() and y() as l-values.
ThePoint.x() = 7;
ThePoint.y() = 9;

// Use x() and y() as r-values.
cout << "x = " << ThePoint.x() << "\n"
<< "y = " << ThePoint.y() << "\n";
}

Output

x = 7
y = 9

Notice that the functions x and y are declared as returning reference types. These functions can be used on either side of an assignment statement.

Declarations of reference types must contain initializers except in the following cases:

• Explicit extern declaration

• Declaration of a class member

• Declaration within a class

• Declaration of an argument to a function or the return type for a function

See Also

Reference

References (C++)