Question

All,

Why does the following code fail to compile for ‘std::endl’, but it’s fine for all of the other inserted types?

#include <sstream> // ostringstream

/// @brief A class that does streamed, formatted output via 'operator<<'.
class My_Stream
{
public:
    /// @brief A member method that manipulates the underlying stream.
    void foo()
    {
        m_oss << "foo_was_here; ";
    }

private:
    /// @brief The underlying stream.
    std::ostringstream m_oss;

    /// @brief 'operator<<' is a friend.
    template< typename T >
    friend My_Stream& operator<<( My_Stream& a_r_my_stream,
                                  const T& a_r_value );
};

/// @brief A manipulator that calls a class method.
My_Stream& manipulator_foo( My_Stream& a_r_my_stream )
{
    a_r_my_stream.foo();
    return a_r_my_stream;
}

/// @brief The generic insertion operator.
template< typename T >
My_Stream& operator<<( My_Stream& a_r_my_stream,
                       const T& a_r_value )
{
    a_r_my_stream.m_oss << a_r_value;
    return a_r_my_stream;
}

/// @brief Define an iostream-like manipulator for my-stream.
typedef My_Stream& ( * my_stream_manipulator ) ( My_Stream& );

/// @brief The specialized 'my_stream_manipulator' insertion operator.
template<>
My_Stream& operator<<( My_Stream& a_r_my_stream,
                       const my_stream_manipulator& a_r_manipulator )
{
    return a_r_manipulator( a_r_my_stream );
}

int main( int argc, char* argv[] )
{
    My_Stream my_stream;

    my_stream << 'c'; // char
    my_stream << "string"; // c-string
    my_stream << 1u; // unsigned int
    my_stream << -1; // signed int
    my_stream << 5.3f; // float
    my_stream << -23.345; // double
    my_stream << std::boolalpha; // std::ios_base manipulator
    my_stream << std::endl; // std::ostream manipulator
    my_stream << manipulator_foo; // my_stream manipulator

    return 0;
}

I get the following G++ 4.5 error:

willo:~/test_cpp$ g++ -Wall test_overloaded_insertion_manipulators.cpp
test_overloaded_insertion_manipulators.cpp: In function ‘int main(int, char**)’:
test_overloaded_insertion_manipulators.cpp:60: error: no match for ‘operator<<’ in ‘my_stream << std::endl’

I expect the code to instantiate a ‘operator<<‘ for std::endl, just like it did for the primitives, std::ios_base and my custom manipulator.

For context, I’m trying to create a light-API IOStream-like class that works with current IOStream manipulators, as well as one or two more custom manipulators.

Answer 1

Because endl is a function template:

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

So the identifier itself is not a value. It only becomes a value (function pointer) when it’s instantiated. But your operator<< is itself a template, and there is no type information available to the compiler to decide which types to instantiate endl with.

In contrast, e.g. boolalpha is:

ios_base& boolalpha(ios_base& str);

Hence why it works.

endl works for basic_ostream, because that one defines operator<< overloads as member functions taking function pointers; in particular:

basic_ostream<charT,traits>& operator<<(basic_ostream<charT,traits>& (*pf)(basic_ostream<charT,traits>&));

So in a call like stream << endl, it would know charT and traits from type of this (i.e. left side of operator), and that would give it exact type of function pointer to expect on the right side – which it would then use to instantiate the corresponding version of endl. You can do the same for your class.