What operators should be declared as friends?

Rafael S. Calsaverini picture Rafael S. Calsaverini · Jun 6, 2011 · Viewed 28.7k times · Source

In some books and often around the internet I see recommendations like "operator== should be declared as friend".

How should I understand when an operator must be declared as friend and when it should be declared as member? What are the operators that will most often need to be declared as friends besides == and <<?

Answer

Chris Frederick picture Chris Frederick · Jun 6, 2011

This really depends on whether a class is going to be on the left- or right-hand side of the call to operator== (or other operator). If a class is going to be on the right-hand side of the expression—and does not provide an implicit conversion to a type that can be compared with the left-hand side—you need to implement operator== as a separate function or as a friend of the class. If the operator needs to access private class data, it must be declared as a friend.

For example,

class Message {
    std::string content;
public:
    Message(const std::string& str);
    bool operator==(const std::string& rhs) const;
};

allows you to compare a message to a string

Message message("Test");
std::string msg("Test");
if (message == msg) {
    // do stuff...
}

but not the other way around

    if (msg == message) { // this won't compile

You need to declare a friend operator== inside the class

class Message {
    std::string content;
public:
    Message(const std::string& str);
    bool operator==(const std::string& rhs) const;
    friend bool operator==(const std::string& lhs, const Message& rhs);
};

or declare an implicit conversion operator to the appropriate type

class Message {
    std::string content;
public:
    Message(const std::string& str);
    bool operator==(const std::string& rhs) const;
    operator std::string() const;
};

or declare a separate function, which doesn't need to be a friend if it doesn't access private class data

bool operator==(const std::string& lhs, const Message& rhs);