Base-from-Member Idiom in C++

The following code is from here:

#include <streambuf>  // for std::streambuf
#include <ostream>    // for std::ostream

class fdoutbuf
    : public std::streambuf
{
public:
    explicit fdoutbuf( int fd );
    //...
};

class fdostream
    : public std::ostream
{
protected:
    f开发者_如何学Pythondoutbuf buf;
public:
    explicit fdostream( int fd ) 
        : buf( fd ), std::ostream( &buf ) // This is not allowed. 
                                          // buf can't be initialized before std::ostream.
        {}
    //...
};

I didn't really understand the comment. Why "buf can't be initialized before std::ostream"? Can I use some help understanding this?

The order of initialization is determined by the order of declaring your class members, and inherited classes come before all of that. Take a simple example that illustrate the basic problem without referring to inheritance :

class C
{
  int a, b;
public:
  C() : b(1), a(b) {} // a is initialized before b!
};

The code doesn't do what you think! a is initialized first, then b is initialized to one. So it depends on the order of the declaration not the order in the initialization list:

int a, b;

Now, the same idea applies to base classes which are initialized before the derived class members. To solve this problem, you create a class that you inherent which contains the member you want to initialize from a base class. Of course, that helper class must come before the one you are actually deriving from.

You have to call the base class constructor before you initialize your member variables, but you pass a pointer to buf (a member variable which is undefined at this point) to this constructor.