问题
Given the following code (without virtual inheritance) :
class A
{
public:
virtual void f() = 0;
};
class B : public A
{
public:
virtual void f() {}
};
class C : public A
{
public:
virtual void f() {}
};
class D : public B, public C
{
/* some code */
};
int main()
{
D d;
return 0;
}
the code compile.
On the other hand , here :
class A
{
public:
virtual void f() = 0;
};
class B : virtual public A
{
virtual void f() {}
};
class C : virtual public A
{
virtual void f() {}
};
class D : public B, public C
{
/* some code */
};
int main()
{
D d;
return 0;
}
The compiler presents a compilation error:
no unique final overrider for 'virtual void A::f()' in 'D' .
Why is it different in the second code ?
回答1:
Your first scenario hierarchy corresponds to:
F() F()
A A
| |
F() B C F()
\ /
D
Where D is not abstract, because there are two A subobjects in an object of type D: One that is made concrete by B through the lattice of B, and another that is made concrete through the lattice of C.
Unless you try to invoke the function F() on object of D there will not be any ambiguity.
Your second scenario hierarchy corresponds to:
F()
A
/ \
F() B C F()
\ /
D
In this scenario, the object D has a single Base class A sub object, and it must override and provide implementation of the pure virtual function in that subobject.
Herb Sutter's articles in Guru Of The Week(GOTW) are a nice read for Multiple Inheritance:
- Multiple Inheritance Part I
- Multiple Inheritance Part II
- Multiple Inheritance Part III
回答2:
With the virtual inheritance a D
object has a single base-class A
sub-object. This single sub-object can’t have two different implementations of a virtual function. In contrast, without virtual inheritance a D
object has two distinct base-class A
sub-objects, each with its own implementation of the function (which is OK until you try to call it on a D
object, at which point you need to indicate which one you want).
Cheers & hth.
来源:https://stackoverflow.com/questions/7219941/multiple-diamond-inheritance-compiles-without-virtual-but-doesnt-with