Inheriting the Wrong Constructor

[AalaarDB]

struct base
{
int x, y, z;
base() {x = 0; y = 0; z = 0;};
base(int x1, int y1, int z1) {x = x1; y = y1; z = z1;};
};

struct intermediate1 : public virtual base {};

struct intermediate2 : public virtual base
{
intermediate2() : base(1,2,3) {};
};

struct derived : public virtual intermediate2, public virtual
intermediate1
{
derived() : intermediate2() {};
};

int main()
{
derived temp;
return 0;
}

OK, I know that classes don't actually inherit constructors. But it
sure seems like derived is inheriting something it shouldn't. When I
step through derived's instantiation I see that derived calls base's
default ctor and then calls intermediate2's ctor which does NOT call
base's custom ctor (I guess since it's already ctored). What I want
is derived to call intermediate2 to call base custom ctor. Why does
it do this, and how can I do what I want?

[Victor Bazarov]

Aa******@gmail.com wrote:

struct base
{
int x, y, z;
base() {x = 0; y = 0; z = 0;};
base(int x1, int y1, int z1) {x = x1; y = y1; z = z1;};

Read the FAQ. Prefer initialisation to assignment. Also, it would
seem that both constructors can be merged:

base(int x = 0, int y = 0, int z = 0) : x(x), y(y), z(z) {}

(and learn to omit the semicolons after function bodies).

};

struct intermediate1 : public virtual base {};

struct intermediate2 : public virtual base
{
intermediate2() : base(1,2,3) {};
};

struct derived : public virtual intermediate2, public virtual
intermediate1
{
derived() : intermediate2() {};
};

int main()
{
derived temp;
return 0;
}

OK, I know that classes don't actually inherit constructors. But it
sure seems like derived is inheriting something it shouldn't. When I
step through derived's instantiation I see that derived calls base's
default ctor and then calls intermediate2's ctor which does NOT call
base's custom ctor (I guess since it's already ctored). What I want
is derived to call intermediate2 to call base custom ctor. Why does
it do this, and how can I do what I want?

This has nothing to do with inheriting constructors, and everything with
the rule that the _most derived class_ is responsible for doing the
initialisation of the virtual base class[es].

'derived' has 'base' as virtual base class. That means it *itself* is
responsible for initialising the 'base' subobject. That means, in turn,
that in the initialiser list of 'derived' there *will be* the call to
'base::base()' (implicitly) if you don't specify explicitly what
constructor should be used. It is up to you to do it right, the
'derived' object cannot delegate the responsibility to any of its other
base classes. So, you ought to write

derived() : base(1,2,3) {}

if you wanted the particular values to be passed to 'base::base'.

HTH

V
--
Please remove capital 'A's when replying by e-mail
I do not respond to top-posted replies, please don't ask

[AalaarDB]

Read the FAQ. *Prefer initialisation to assignment. *Also, it would

seem that both constructors can be merged:

* * * base(int x = 0, int y = 0, int z = 0) : x(x), y(y), z(z) {}

(and learn to omit the semicolons after function bodies).
This has nothing to do with inheriting constructors, and everything with
the rule that the _most derived class_ is responsible for doing the
initialisation of the virtual base class[es].

'derived' has 'base' as virtual base class. *That means it *itself* is
responsible for initialising the 'base' subobject. *That means, in turn,
that in the initialiser list of 'derived' there *will be* the call to
'base::base()' (implicitly) if you don't specify explicitly what
constructor should be used. *It is up to you to do it right, the
'derived' object cannot delegate the responsibility to any of its other
base classes. *So, you ought to write

* * *derived() : base(1,2,3) {}

if you wanted the particular values to be passed to 'base::base'.

Thanks for the response but it's not the answer I was looking for. I
don't want the answer I can't or my design in insufficient, I want the
answer do x y z to get around it. So what is my reason to want the
base class initialized by an intermediate class rather than the
derived class? Because I'm using a policy based design. Let me
give you a more refined example, and again this is example code so no
need to worry about other critiques (I knew about all those and I read
the FAQ).

enum eShape {
circle,
triangle,
square
};

struct baseInteface
{
eShape shape;
baseInteface() {};
baseInteface(eShape shape) {base::shape = shape;};
//other stuff related to interface
};

namespace Triangle {
struct InitPolicy : public virtual baseInteface
{
InitPolicy() : baseInteface(triangle){};
};
struct SomeOtherPolicy {};
};

namespace Square {
struct InitPolicy : public virtual baseInteface
{
int height, width;
InitPolicy() : baseInteface(square){height = 0; width = 0;};
};
struct SomeOtherPolicy {};
};

template <
class shapePolicy,
class otherPolicy

>

struct Shape : public virtual shapePolicy, public virtual otherPolicy
{
Shape() : shapePolicy() {};
};

int main()
{
Shape<Triangle::InitPolicy, Triangle::SomeOtherPolicyshape;
return 0;
}
Here, your suggestion of writing derived() : baseInteface(/*policy
data*/) {} violates my encapsulation, my derived class should not have
knowledge of the implementation of any of the policies, baseInteface
is the interface and derived is just the template conglomerator.

OK so you might ask me why don't you throw out all those policies and
intermediate classes and just have class Triangle inherit from class
Polygon? Because this is an example, I have too many derivations and
a policy based design cuts that number down. Or you might ask why not
just bite the bullet and use a virtual OnInit function? Well that's
what I tried at first, but C++ cannot call virtual functions from
farther down the inheritance tree in a constructor, because the object
further down doesn't exist yet!

So any ideas on how to have some intermediate policy init a base class
without the derived class knowing about it?

[Victor Bazarov]

Aa******@gmail.com wrote:

[..]
Thanks for the response but it's not the answer I was looking for.[..]

If you're looking for a particular answer, you're in the wrong place.

Virtual base class subobjects are initialised by the most derived class.
Period. That's a requirement of the language, and you have to live
with it. If you missed it in my post, here it is again: the most
derived class initialises the virtual bases, all of them, either
explicitly (when you provide the initialiser in the initialiser list) or
implicitly (default, the compiler can do it for you). There is no way
around this *if* your base class is a *virtual base*. The reason why it
has to be that way is that if you have two paths which lead to the
single subobject of the virtual base type, and both have their own way
to initialise the base class subobject, then the compiler cannot pick
one or the other, *you* have to do it:

struct A { A(int); };

struct Intermediate1 : virtual A {
Intermediate1() : A(123) {}
};

struct Intermediate2 : virtual A {
Intermediate2() : A(456) {}
};

struct MostDerived : Intermediate1, Intermediate2
{
MostDerived() {} // what 'A' becomes? A(123) or A(456)?
};

The correct way:

struct MostDerived : Intermediate1, Intermediate2
{
MostDerived() : A(123) {} // or any other, like A(789)
};

Now that you know there is a limitation in the language with respect to
initialising virtual base classes, you need to look again at your design
or your implementation. Perhaps, just perhaps, you could do this:

struct A { A(int); };

struct Intermediate1 : virtual A {
static int initialiserForA() { return 123; }
Intermediate1() : A(initialiserForA()) {}
};

struct Intermediate2 : virtual A {
static int initialiserForA() { return 456; }
Intermediate2() : A(initialiserForA()) {}
};

struct MostDerived : Intermediate1, Intermediate2
{
MostDerived() : A(Intermediate1::initialiserForA()) {}
};

In which case the initialisation is still done in 'MostDerived', but the
actual value with which to initialised comes from some other place.

Please do not start your reply with "it's not the answer I was looking
for" again.

V
--
Please remove capital 'A's when replying by e-mail
I do not respond to top-posted replies, please don't ask

[James Kanze]

On Aug 28, 6:11 am, [e-mail] wrote:

Read the FAQ. Prefer initialisation to assignment. Also,
it would seem that both constructors can be merged:

base(int x = 0, int y = 0, int z = 0) : x(x), y(y), z(z) {}

(and learn to omit the semicolons after function bodies).

This has nothing to do with inheriting constructors, and
everything with the rule that the _most derived class_ is
responsible for doing the initialisation of the virtual base
class[es].

'derived' has 'base' as virtual base class. That means it
*itself* is responsible for initialising the 'base'
subobject. That means, in turn, that in the initialiser
list of 'derived' there *will be* the call to 'base::base()'
(implicitly) if you don't specify explicitly what
constructor should be used. It is up to you to do it right,
the 'derived' object cannot delegate the responsibility to
any of its other base classes. So, you ought to write

derived() : base(1,2,3) {}

if you wanted the particular values to be passed to
'base::base'.

Thanks for the response but it's not the answer I was looking for.

It's the correct answer, however.

I don't want the answer I can't or my design in insufficient,
I want the answer do x y z to get around it.

But that's what Victor just gave you. Specify the initializers
in the most derived class.

So what is my reason to want the base class initialized by an
intermediate class rather than the derived class?

Which intermediate class? You can have several intermediate
classes deriving from the same instance of the virtual base.

A long, long time ago, John Skaller, Fergus Henderson and I
discussed the possibility of requiring the compiler to determing
the lowest common derived class for each virtual base, and
calling the constructor there. In the end, we didn't follow
through with a proposal, because it requires a lot more
complexity in the compiler, and the cases where you would want a
non empty virtual base (requiring an initializer) are
exceedingly rare. (Note, however, that there is an example of
such in the standard library: basic_ios<>. If worse comes to
worse, you can always use something like is used in the iostream
hierarchy. Be forewarned, however, that it can be extremely
tricky to make it work without undefined behavior is some
special cases.)

Because I'm using a policy based design. Let me give you a
more refined example, and again this is example code so no
need to worry about other critiques (I knew about all those
and I read the FAQ).

enum eShape {
circle,
triangle,
square
};

struct baseInteface
{
eShape shape;
baseInteface() {};
baseInteface(eShape shape) {base::shape = shape;};
//other stuff related to interface
};

namespace Triangle {
struct InitPolicy : public virtual baseInteface
{
InitPolicy() : baseInteface(triangle){};
};
struct SomeOtherPolicy {};
};

namespace Square {
struct InitPolicy : public virtual baseInteface
{
int height, width;
InitPolicy() : baseInteface(square){height = 0; width = 0;};
};
struct SomeOtherPolicy {};
};

template <
class shapePolicy,
class otherPolicy

struct Shape : public virtual shapePolicy, public virtual otherPolicy
{
Shape() : shapePolicy() {};
};

int main()
{
Shape<Triangle::InitPolicy, Triangle::SomeOtherPolicyshape;
return 0;
}

I don't quite see why you need, or even want, virtual
inheritance here. All of your shapePolicy should inherit from
baseInteface (which is in fact BasicShapePolicy). And no other
classes should. So you really don't want virtual inheritance
here.

[...]

So any ideas on how to have some intermediate policy init a
base class without the derived class knowing about it?

In simple cases like this, you can simply use an init() member
function, calling it where ever you want.

--
James Kanze (GABI Software) email:ja*********@gmail.com
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