Lvalue Reference Declarator: &

Holds the address of an object but behaves syntactically like an object.

type-id & cast-expression

Remarks

You can think of an lvalue reference as another name for an object. An lvalue reference declaration consists of an optional list of specifiers followed by a reference declarator. A reference must be initialized and cannot be changed.

Any object whose address can be converted to a given pointer type can also be converted to the similar reference type. For example, any object whose address can be converted to type char * can also be converted to type char &.

Do not confuse reference declarations with use of the address-of operator. When the & identifier is preceded by a type, such as int or char, identifier is declared as a reference to the type. When & identifier is not preceded by a type, the usage is that of the address-of operator.

Example

The following example demonstrates the reference declarator by declaring a Person object and a reference to that object. Because rFriend is a reference to myFriend, updating either variable changes the same object.

// reference_declarator.cpp
// compile with: /EHsc
// Demonstrates the reference declarator.
#include <iostream>
using namespace std;
struct Person
{
 char* Name;
 short Age;
};
int main()
{
 // Declare a Person object.
 Person myFriend;
 // Declare a reference to the Person object.
 Person& rFriend = myFriend;
 // Set the fields of the Person object.
 // Updating either variable changes the same object.
 myFriend.Name = "Bill";
 rFriend.Age = 40;
 // Print the fields of the Person object to the console.
 cout << rFriend.Name << " is " << myFriend.Age << endl;
}

Bill is 40

See Also

Reference

References (C++)

Reference-Type Function Arguments

Reference-Type Function Returns

References to Pointers