C++ : const references and initialization order

Question

I am wondering if I am using the good approach in the following :

• I want to construct a parent class (class A), this class should own an instance of a given "Foo" class
• I want the parent class to own a child class member (class B) and this member should have a reference to the foo member of the parent class.

The code below seems to works, however I am wondering whether I was just "lucky" that the compiler was sympathetic enough.

For clarity, I added comment and my question in the comments below.

Thanks !

struct Foo
{
  std::string mValue;
};

class B
{
public:
  B(const Foo & foo) : mFoo_External(foo) {}
private:
  const Foo & mFoo_External; //this is an external reference to the member 
                             //(coming from A)
};

class A
{
public:
  //Here is the big question 
  //Shall I use : 
  //  A(const Foo & foo) : mFoo(foo), mB(mFoo) {}  
  //  or the declaration below
  A(const Foo & foo) : mFoo(foo), mB(foo) {}
private:
  //According to my understanding, the declaration 
  //order here *will* be important
  //(and I feel this is ugly)
  const Foo  mFoo;
  B mB;
};



void MyTest()
{
  std::auto_ptr<Foo> foo(new Foo());
  foo->mValue = "Hello";
  A a( *foo);
  foo.release();

  //At this point (after foo.release()), "a" is still OK 
  //(i.e A.mB.mFooExternal is not broken, although foo is now invalid)
  //
  //This is under Visual Studio 2005 : 
  //was I lucky ? Or is it correct C++ ?
}

Answer 1

No, this is broken. Your mB will hold a reference to whatever you passed to the constructor of the A object, not to mFoo. Instead, you should say:

A(const Foo & foo) : mFoo(foo), mB(mFoo) { }

Note that mB is a copy of the constructor argument, and not a reference, so your MyTest function is fine.