C++ stream as a parameter when overloading operator<<

Question

I'm trying to write my own logging class and use it as a stream:

logger L;
L << "whatever" << std::endl;

This is the code I started with:

#include <iostream>

using namespace std;


class logger{
public:
    template <typename T>
    friend logger& operator <<(logger& log, const T& value);
};

template <typename T>
logger& operator <<(logger& log, T const & value) {
    // Here I'd output the values to a file and stdout, etc.
    cout << value;
    return log;
}

int main(int argc, char *argv[])
{
    logger L;
    L << "hello" << '\n' ; // This works
    L << "bye" << "alo" << endl; // This doesn't work
    return 0;
}

But I was getting an error when trying to compile, saying that there was no definition for operator<< (when using std::endl):

pruebaLog.cpp:31: error: no match for ‘operator<<’ in ‘operator<< [with T = char [4]](((logger&)((logger*)operator<< [with T = char [4]](((logger&)(& L)), ((const char (&)[4])"bye")))), ((const char (&)[4])"alo")) << std::endl’

So, I've been trying to overload operator<< to accept this kind of streams, but it's driving me mad. I don't know how to do it. I've been loking at, for instance, the definition of std::endl at the ostream header file and written a function with this header:

logger& operator <<(logger& log, const basic_ostream<char,char_traits<char> >& (*s)(basic_ostream<char,char_traits<char> >&))

But no luck. I've tried the same using templates instead of directly using char, and also tried simply using "const ostream& os", and nothing.

Another thing that bugs me is that, in the error output, the first argument for operator<< changes, sometimes it's a reference to a pointer, sometimes looks like a double reference...

Answer 1

In C++ it is the stream buffer that encapsulates the underlying I/O mechanisim. The stream itself only encapsulates the conversions to string, and the I/O direction.

Thus you should be using one of the predefined stream classes, rather than making your own. If you have a new target you want your I/O to go to (like a system log), what you should be creating is your own stream buffer (derived from std::streambuf).

Answer 2

endl is an IO manipulator, which is a functor that accepts a stream by reference, performs some operation on it, and returns that stream, also by reference. cout << endl is equivalent to cout << '\n' << flush, where flush is a manipulator that flushes the output buffer.

In your class, you just need to write an overload for this operator:

logger& operator<<(logger&(*function)(logger&)) {
    return function(*this);
}

Where logger&(*)(logger&) is the type of a function accepting and returning a logger by reference. To write your own manipulators, just write a function that matches that signature, and have it perform some operation on the stream:

logger& newline(logger& L) {
    return L << '\n';
}

Answer 3

endl is a strange beast. It isn't a constant value. It's actually, of all things, a function. You need a special override to handle the application of endl:

logger& operator<< (logger& log, ostream& (*pf) (ostream&))
{
  cout << pf;
  return log;
}

This accepts insertion of a function that takes an ostream reference and returns an ostream reference. That's what endl is.

Edit: In response to FranticPedantic's interesting question of "why can't the compiler deduce this automatically?". The reason is that if you delve yet deeper, endl is actually itself a template function. It's defined as:

template <class charT, class traits>
  basic_ostream<charT,traits>& endl ( basic_ostream<charT,traits>& os );

That is, it can take any sort of ostream as its input and output. So the problem isn't that the compiler can't deduce that T const & could be a function pointer, but that it can't figure out which endl you meant to pass in. The templated version of operator<< presented in the question would accept a pointer to any function as its second argument, but at the same time, the endl template represents an infinite set of potential functions, so the compiler can't do anything meaningful there.

Providing the special overload of the operator<< whose second argument matches a specific instantiation of the endl template allows the call to resolve.

Answer 4

I believe that the problem is your stream doesn't overload operator<< to accept a function that has the same type as std::endl as illustrated in this answer: http://stackoverflow.com/questions/1134388/stdendl-is-of-unknown-type-when-overloading-operator