Operator overloading and friend functions

Can an overloaded operator be a friend function?

If the operator can be a free function it can also be a friend.

Quote:

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Originally Posted by nuzzle

If the operator can be a free function it can also be a friend.

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By free function you mean a function that is not a data member? I was talking about a data member of a class

Quote:

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Originally Posted by maverick786us

By free function you mean a function that is not a data member? I was talking about a data member of a class

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A member function cannot be a friend function, it's either or. And a friend function is a free function (with special access priviliges in the class where it's declared).

If you declare a member function to be a friend it turns into a free function.

yes, you can friend a member function:

Code:

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class A{ public: void f(); };
class B{ friend void A::f(); void g(); };

void A::f(){ B b; b.g(); }

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in the same way, you can friend member operators.

To clarify: nuzzle's statement is true with respect to the same class, i.e., a member function of a given class cannot also be a friend function of that class. As you can see from superbonzo's code snippet, such a member function can be a friend function of another class. Since an overloaded operator is ultimately just a function, the same applies.

Quote:

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Originally Posted by superbonzo

yes, you can friend a member function:

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That doesn't compile in VS2012. It compiles only if class B makes the whole class A a friend (and not only the individual f function). But I think it should work actually so it seems like a bug.

Anyway there are two different situations to consider and the OP will have to decide which one applies.

I'm talking about friend functions/operators of a class, and you're talking about granting friendship to functions/operators of another class.

I think the first usage is most common actually, see here for example,

http://msdn.microsoft.com/en-us/libr...v=vs.110).aspx

Quote:

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Originally Posted by nuzzle

That doesn't compile in VS2012. It compiles only if class B makes the whole class A a friend (and not only the individual f function). But I think it should work actually so it seems like a bug.

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Definitely a bug:

Quote:

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Originally Posted by C++11 Clause 11.3 Paragraph 5

When a friend declaration refers to an overloaded name or operator, only the function specified by the parameter types becomes a friend. A member function of a class X can be a friend of a class Y. [ Example:

Code:

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class Y {
  friend char* X::foo(int);
  friend X::X(char); // constructors can be friends
  friend X::~X(); // destructors can be friends
};

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—end example ]

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or earlier:

Quote:

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Originally Posted by C++03 Clause 11.4 Paragraph 4

When a friend declaration refers to an overloaded name or operator, only the function specified by the parameter types becomes a friend. A member function of a class X can be a friend of a class Y. [Example:

Code:

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class Y {
    friend char* X::foo(int);
    // ...
};

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—end example]

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