How to inherit from std::ostream?

Question

I've been googling around and I just can't find a simple answer to this. And it should be simple, as the STL generally is.

I want to define MyOStream which inherits publicly from std::ostream. Let's say I want to call foo() each time something is written into my stream.

class MyOStream : public ostream {
public:
  ...
private:
   void foo() { ... }
}

I understand that the public interface of ostream is non-virtual, so how can it be done? I want clients to be able to use both operator<< and write() and put() on MyOStream and have use the extended ability of my class.

Answer 1

It's not a simple question, unfortunately. The classes you should derive from are the basic_ classes, such as basic_ostream. However, derivation from a stream may not be what you want, you may want to derive from a stream buffer instead, and then use this class to instantiate an existing stream class.

The whole area is complex, but there is an excellent book about it Standard C++ IOStreams and Locales, which I suggest you take a look at before going any further.

Answer 2

I was spinning my head around how to do the same thing and i found out it's actually not that hard. Basically just subclass the ostream and the streambuf objects, and construct the ostream with itself as the buffer. the virtual overflow() from std::streambuf will be called for every character sent to the stream. To fit your example i just made a foo() function and called it.

struct Bar : std::ostream, std::streambuf
{
    Bar() : std::ostream(this) {}

    int overflow(int c)
    {
        foo(c);
        return 0;
    }


    void foo(char c)
    {
        std::cout.put(c);

    }
};

void main()
{
    Bar b;
    b<<"Look a number: "<<std::hex<<29<<std::endl;
}

oh and ignore the fact that the main function is not a real main function. It's in a namespace called from elsewhere ;p

Answer 3

Another working hack to achieve a similar effect is to use template and composition

class LoggedStream {
public:
  LoggedStream(ostream& _out):out(_out){}
  template<typename T>
  const LoggedStream& operator<<(const T& v) const {log();out << v;return *this;}
protected:
  virtual void log() = 0;
  ostream& out;
};

class Logger : LoggedStream {
  void log() { std::cerr << "Printing" << std::endl;}
};

int main(int,char**) {LoggedStream(std::cout) << "log" << "Three" << "times";}

Answer 4

I don't know if this is correct solution, but I inherited from std::ostream this way. It uses a buffer inherited from std::basic_streambuf and gets 64 characters at a time (or less if flushed) and sends them to a generic putChars() method where the actual handling of data is done. It also demonstrates how to give user data.

Live Example

#include <streambuf>
#include <ostream>
#include <iostream>

//#define DEBUG

class MyData
{
    //example data class, not used
};

class MyBuffer : public std::basic_streambuf<char, std::char_traits<char> >
{

public:

    inline MyBuffer(MyData data) :
    data(data)
    {
        setp(buf, buf + BUF_SIZE);
    }

protected:

    // This is called when buffer becomes full. If
    // buffer is not used, then this is called every
    // time when characters are put to stream.
    inline virtual int overflow(int c = Traits::eof())
    {
#ifdef DEBUG
        std::cout << "(over)";
#endif
        // Handle output
        putChars(pbase(), pptr());
        if (c != Traits::eof()) {
            char c2 = c;
            // Handle the one character that didn't fit to buffer
            putChars(&c2, &c2 + 1);
        }
        // This tells that buffer is empty again
        setp(buf, buf + BUF_SIZE);

        return c;
    }

    // This function is called when stream is flushed,
    // for example when std::endl is put to stream.
    inline virtual int sync(void)
    {
        // Handle output
        putChars(pbase(), pptr());
        // This tells that buffer is empty again
        setp(buf, buf + BUF_SIZE);
        return 0;
    }

private:

    // For EOF detection
    typedef std::char_traits<char> Traits;

    // Work in buffer mode. It is also possible to work without buffer.
    static const size_t BUF_SIZE = 64;
    char buf[BUF_SIZE];

    // This is the example userdata
    MyData data;

    // In this function, the characters are parsed.
    inline void putChars(const char* begin, const char* end){
#ifdef DEBUG
        std::cout << "(putChars(" << static_cast<const void*>(begin) <<
            "," << static_cast<const void*>(end) << "))";
#endif
        //just print to stdout for now
        for (const char* c = begin; c < end; c++){
            std::cout << *c;
        }
    }

};

class MyOStream : public std::basic_ostream< char, std::char_traits< char > >
{

public:

    inline MyOStream(MyData data) :
    std::basic_ostream< char, std::char_traits< char > >(&buf),
    buf(data)
    {
    }

private:

    MyBuffer buf;

};

int main(void)
{
    MyData data;
    MyOStream o(data);

    for (int i = 0; i < 8; i++)
        o << "hello world! ";

    o << std::endl;

    return 0;
}

Answer 5

Composition, not inheritance. Your class contains, "wraps" an ostream&, and forwards to it (after calling foo()).