The following code:
struct interface_base
{
virtual void foo() = 0;
};
struct interface : public interface_base
{
virtual void bar() = 0;
};
struct implementation_base : public interface_base
{
void foo();
};
struct implementation : public implementation_base, public interface
{
void bar();
};
int main()
{
implementation x;
}
fails to compile with the following errors:
test.cpp: In function 'int main()':
test.cpp:23:20: error: cannot declare variable 'x' to be of abstract type 'implementation'
test.cpp:16:8: note: because the following virtual functions are pure within 'implementation':
test.cpp:3:18: note: virtual void interface_base::foo()
I have played around with it and figured out that making the 'interface -> interface_base' and 'implementation_base -> interface_base' inheritances virtual, fixes the problem, but I don't understand why. Can someone please explain what is going on?
p.s. I omitted the virtual destructors on purpose to make the code shorter. Please don't tell me to put them in, I already know :)
You have two interface_base
base classes in your inheritance tree. This means you must provide two implementations of foo()
. And calling either of them will be really awkward, requiring multiple casts to disambiguate. This usually is not what you want.
To resolve this, use virtual inheritance:
struct interface_base
{
virtual void foo() = 0;
};
struct interface : virtual public interface_base
{
virtual void bar() = 0;
};
struct implementation_base : virtual public interface_base
{
void foo();
};
struct implementation : public implementation_base, virtual public interface
{
void bar();
};
int main()
{
implementation x;
}
With virtual inheritance, only one instance of the base class in question is created in the inheritance heirarchy for all virtual mentions. Thus, there's only one foo()
, which can be satisfied by implementation_base::foo()
.
For more information, see this prior question - the answers provide some nice diagrams to make this all more clear.