diamond inheritance hides non-default constructor

[Tom]

Hello, I've searched groups and the std unsuccessfully for an explanation
of the following - I'd appreciate any comments you may have.

Given the following diamond multiple inheritance pattern, with the base
class having 2 constructors, only the default constructor is called. Is
there a way to require the other constructor to be used?

First, the output:

Base default constructor
Derived1:
Derived2:
Join:
Main:

class Base
{
public:
Base ()
{ cout << "Base default constructor" << endl; }
Base (string const & useMsg)
: msg (useMsg)
{ cout << "Base: " << msg << endl; }
virtual ~Base() {}
string msg;
};

class Derived1 : virtual public Base
{
public:
Derived1 (string const & useMsg)
: Base (useMsg)
{ cout << "Derived1: " << msg << endl; }
};

class Derived2 : virtual public Base
{
public:
Derived2 ()
: Base ()
{ cout << "Derived2: " << msg << endl; }
};

class Join : public Derived1, public Derived2
{
public:
Join ()
: Derived1 ( string ("Hello") ), Derived2 ()
{ cout << "Join: " << msg << endl; }
};

int main ()
{
Join join;
cout << "Main: " << join.msg << endl;
return 0;
}

[JKop]

class Join : public Derived1, public Derived2
{
public:
Join ()
: Derived1 ( string ("Hello") ), Derived2 ()
{ cout << "Join: " << msg << endl; }
};

Join() : Derived1( "Hello" ), Derived2(), Base("this is what I was
missing!")
{

}
-JKop

[Victor Bazarov]

Tom wrote:

Hello, I've searched groups and the std unsuccessfully for an explanation
of the following - I'd appreciate any comments you may have.

Given the following diamond multiple inheritance pattern, with the base
class having 2 constructors, only the default constructor is called. Is
there a way to require the other constructor to be used?

First, the output:

Base default constructor
Derived1:
Derived2:
Join:
Main:

class Base
{
public:
Base ()
{ cout << "Base default constructor" << endl; }
Base (string const & useMsg)
: msg (useMsg)
{ cout << "Base: " << msg << endl; }
virtual ~Base() {}
string msg;
};

class Derived1 : virtual public Base
{
public:
Derived1 (string const & useMsg)
: Base (useMsg)
{ cout << "Derived1: " << msg << endl; }
};

class Derived2 : virtual public Base
{
public:
Derived2 ()
: Base ()
{ cout << "Derived2: " << msg << endl; }
};

class Join : public Derived1, public Derived2
{
public:
Join ()
: Derived1 ( string ("Hello") ), Derived2 ()
{ cout << "Join: " << msg << endl; }
};

int main ()
{
Join join;
cout << "Main: " << join.msg << endl;
return 0;
}

Virtual base class initialisation is done at the level of the most derived
class. In your case the 'Join' c-tor should have

Base(???)

in its initialiser list (along with other things). Replace the ??? with
the appropriate expression you want to be evaluated and passed to the
Base's c-tor. I suspect it will be

Base("Hello")

Victor

[Tom]

Thanks for your answers.
I finally found your quote in the std doc under initialization (section 12),
not in the section on inheritance (section 10). The explanation is in the
sentence following your quote of it:

All sub-objects representing virtual base classes are initialized by
the constructor of the most derived class. If the constructor of
the most derived class does not specify a mem-initializer for a
virtual base class V, then V's default constructor is called to
initialize the virtual base class subobject.

Tom
"Victor Bazarov" <v.********@comAcast.net> wrote in message
news:Zl***************@newsread1.dllstx09.us.to.ve rio.net...

Virtual base class initialisation is done at the level of the most derived
class. In your case the 'Join' c-tor should have

Base(???)

in its initialiser list (along with other things). Replace the ??? with
the appropriate expression you want to be evaluated and passed to the
Base's c-tor. I suspect it will be

Base("Hello")

Victor