为什么重载运算符<<必须通过引用返回？

Question

I want to print out an object of a user-defined type, like this cout << ob1; so I want to overload operator<< and I want to return by value not by reference but it gives me an error: in two files named : iosfwd and ios_base.h

ostream operator<<( ostream& out, cat& rhs){
    out << rhs.a << ", " << rhs.b << endl;
    return out ; 
}

but when I return by reference like this :

ostream& operator<<( ostream& out, cat& rhs){
    out << rhs.a << ", " << rhs.b << endl;
    return out ;
}

it works fine.

Answer 1

In the first example, you return a copy of the stream object which is not allowed because the copy-constructors (and copy-assignment as well) of all the stream classes in C++ has been disabled by having them made private .

Since you cannot make a copy of a stream object, you're required to return it by reference , which you're doing in the second example which is why it is working fine.

You may choose to return nothing at all (ie you can make the return type void ), but if you do so, then you would not be able to chain as stream << a << b . You've to write them separately as stream <<a and then stream << b .

If you want to know why copying of stream objects is disabled, see my answer here:

• Why copying stringstream is not allowed?

Answer 2

Streams are non copyable because copying a stream doesn't really make sense, a stream is unique (you can't jump in the same river twice kind of thing). return by value is in C++03 at least copying transaction.

If there are reason you want to return by value, return by reference is the correct version.

Answer 3

It is done to support safe and reasonable operator chaining like

    cout<<a<<b;  

works because of returning a reference.