Question about derived class and copy constructors....

[Jeroen]

Hi all,

I wrote a little bit of code to see the behaviour of (copy) constructors
of base and derived classes. But I have a question about it. Let me
explain by the following (incomplete/illustrative) code:

class A {
// (copy) ctors goe here...
};

class B : public A {
// (copy) ctors goe here...
};

void foo(A a)
{
// function body here...
}

I can call 'foo' with an object of type B:

{
B b;
foo(b);
}

What I saw in my code is that when calling 'foo' this way, an object of
type A is created and the copy constructor for type A is called (which
apparently gets 'b' passed as parameter).

My question: is it possible that a compiler creates an object of type B
and calls the copy constructor for type B? In that case a type B object
is passed to the the function 'foo'.

I'm not sure about this because if I have a function:

void foo2(A& a){}

I can call that also with foo(b) and only a reference of 'b' is passed,
and the function 'foo' is fine with processing the object of type B...

On the other hand, I can imagine that 'foo' depends on a complete object
A on the stack so that passing an object of type B is not possible.

Or maybe I completely missed the point.

Thanx for comments and clarification,

Jeroen

[Ian Collins]

Jeroen wrote:

Hi all,

I wrote a little bit of code to see the behaviour of (copy) constructors
of base and derived classes. But I have a question about it. Let me
explain by the following (incomplete/illustrative) code:

class A {
// (copy) ctors goe here...
};

class B : public A {
// (copy) ctors goe here...
};

void foo(A a)
{
// function body here...
}

I can call 'foo' with an object of type B:

{
B b;
foo(b);
}

What I saw in my code is that when calling 'foo' this way, an object of
type A is created and the copy constructor for type A is called (which
apparently gets 'b' passed as parameter).

Correct, polymorphism doesn't work when passing by value, you get the A
part of b. This is known as slicing.

My question: is it possible that a compiler creates an object of type B
and calls the copy constructor for type B? In that case a type B object
is passed to the the function 'foo'.

No.

I'm not sure about this because if I have a function:

void foo2(A& a){}

I can call that also with foo(b) and only a reference of 'b' is passed,
and the function 'foo' is fine with processing the object of type B...

That's how it is supposed to work.

--
Ian Collins.

[Jeroen]

Ian Collins schreef:

Jeroen wrote:

>Hi all,

I wrote a little bit of code to see the behaviour of (copy) constructors
of base and derived classes. But I have a question about it. Let me
explain by the following (incomplete/illustrative) code:

class A {
// (copy) ctors goe here...
};

class B : public A {
// (copy) ctors goe here...
};

void foo(A a)
{
// function body here...
}

I can call 'foo' with an object of type B:

{
B b;
foo(b);
}

What I saw in my code is that when calling 'foo' this way, an object of
type A is created and the copy constructor for type A is called (which
apparently gets 'b' passed as parameter).

Correct, polymorphism doesn't work when passing by value, you get the A
part of b. This is known as slicing.

>My question: is it possible that a compiler creates an object of type B
and calls the copy constructor for type B? In that case a type B object
is passed to the the function 'foo'.

No.

>I'm not sure about this because if I have a function:

void foo2(A& a){}

I can call that also with foo(b) and only a reference of 'b' is passed,
and the function 'foo' is fine with processing the object of type B...

That's how it is supposed to work.

OK, thanks for the answer.