How to Handle `std::endl` in Custom Stream Operator Overloading?

Operator Overloading and Handling std::endl in Custom Stream Implementations

Consider the following code snippet where operator<< is overloaded:

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

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

While this works as expected, attempting to use my_stream << endl results in a compilation error stating "binary '<<' : no operator found which takes a left-hand operand of type 'UIStream'..."

To resolve this issue, it's important to understand that std::endl is not an object but a function. In std::cout, it's utilized by implementing operator<< to accept a function pointer with the same signature as std::endl. When called, the function is invoked and its return value is forwarded.

A custom implementation of the operator<< can be defined to handle std::endl in a similar manner:

struct MyStream
{
    template 
    MyStream& operator<< (const T& x)
    {
        std::cout << x;
        return *this;
    }

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

    // Accept function with custom signature
    MyStream& operator<< (MyStreamManipulator manip)
    {
        return manip(*this);
    }

    // Define a custom endl for this stream (matches MyStreamManipulator signature)
    static MyStream& endl(MyStream& stream)
    {
        std::cout << std::endl;
        stream << "Called MyStream::endl!" << std::endl;
        return stream;
    }
};

This custom implementation defines an operator<< that accepts two types:

The first type is a generic function pointer that takes a reference to the MyStream object and returns a reference to the same object.
The second type is a function pointer that specifically takes a reference to a MyStream object and returns a reference to the same object.

This allows for the implementation of a custom endl function that operates on the custom stream.
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