Re: [Boost-users] Switch from a regular C++ pointer to boost shared_ptr

I don't need to do 'b= NULL' in A's constructor, right?

Right.

But why this line won't compile "b = new B();" ?

The constructor you need for this conversion is declared as "explicit". Do it this way: typedef boost::shared_ptr<B> BPtr; B& A::getB() { if (!b) b = BPtr(new B()); return *b; // btw, do you really wish to return the reference? }