Comparing address of std::endl

Question

I am inspecting a piece of existing code and found out it behaves differently when compiled with Visual C++ 9 and MinGW:

inline LogMsg& LogMsg::operator<<(std::ostream& (*p_manip)(std::ostream&) )
{
    if ( p_manip == static_cast< std::ostream& (*)(std::ostream&) > ( &std::endl<char, std::char_traits<char> >) )
    {
        msg(m_output.str());
        m_output.str( "" );
    }
    else
    {
        (*p_manip) (m_output);            // or // output << p_manip;
    }
    return *this;
}

As the name suggests, this is a log class and it overloads operator<<() to strip endls from the stream.

I found out why it behaves differently: the test p_manip == static_cast... succeeds with MinGW while it fails with Visual C++ 9.

• MinGW "ignores" the cast and returns the real address of std::endl;
• Visual C++ 9 actually casts the pointer-to-endl and returns a different address.

I changed the test to if ( p_manip == std::endl ) and now it behaves as expected.

My question is: what is the rationale behind such a complicated (and, as a matter of fact, wrong) test?

---

For the sake of completness:

class LogStream
{
public:
    LogStream() {}
protected:
    std::ostringstream m_output;
};

class LogMsg : public LogStream
{
    friend LogMsg& msg() ;
    static LogMsg s_stream;
public:
    LogMsg() {}
    template <typename T>
        inline LogMsg& operator<<(T p_data);
    inline LogMsg& operator<<(std::ostream& (*p_manip)(std::ostream&) );
};

Answer 1

At a guess, I'd say the original author didn't realise they were compatible types, and did the conversion on spec (without the compiler requiring him to).

Answer 2

For information:

The statement if ( p_manip == std::endl ) does not compile on the original compiler (gcc 3.4.5, the compiler on which the code was originally developed).

That means the test was not wrong as I stated in my question.