std::endl is of unknown type when overloading operator<<

Kazoom picture Kazoom · Jul 16, 2009 · Viewed 22.6k times · Source

I overloaded operator <<

template <Typename T>
UIStream& operator<<(const T);

UIStream my_stream;
my_stream << 10 << " heads";

Works but:

my_stream << endl;

Gives compilation error:

error C2678: binary '<<' : no operator found which takes a left-hand operand of type 'UIStream' (or there is no acceptable conversion)

What is the work around for making my_stream << endl work?

Answer

GManNickG picture GManNickG · Jul 16, 2009

std::endl is a function and std::cout utilizes it by implementing operator<< to take a function pointer with the same signature as std::endl.

In there, it calls the function, and forwards the return value.

Here is a code example:

#include <iostream>

struct MyStream
{
    template <typename T>
    MyStream& operator<<(const T& x)
    {
        std::cout << x;

        return *this;
    }


    // function that takes a custom stream, and returns it
    typedef MyStream& (*MyStreamManipulator)(MyStream&);

    // take in a function with the custom signature
    MyStream& operator<<(MyStreamManipulator manip)
    {
        // call the function, and return it's value
        return manip(*this);
    }

    // define the custom endl for this stream.
    // note how it matches the `MyStreamManipulator`
    // function signature
    static MyStream& endl(MyStream& stream)
    {
        // print a new line
        std::cout << std::endl;

        // do other stuff with the stream
        // std::cout, for example, will flush the stream
        stream << "Called MyStream::endl!" << std::endl;

        return stream;
    }

    // this is the type of std::cout
    typedef std::basic_ostream<char, std::char_traits<char> > CoutType;

    // this is the function signature of std::endl
    typedef CoutType& (*StandardEndLine)(CoutType&);

    // define an operator<< to take in std::endl
    MyStream& operator<<(StandardEndLine manip)
    {
        // call the function, but we cannot return it's value
        manip(std::cout);

        return *this;
    }
};

int main(void)
{
    MyStream stream;

    stream << 10 << " faces.";
    stream << MyStream::endl;
    stream << std::endl;

    return 0;
}

Hopefully this gives you a better idea of how these things work.