问题
Consider the following code:
struct A
{
virtual void f() = 0;
};
struct B
{
void f();
};
struct C : public A, public B
{};
int main()
{
A* a = new C();
B* b = new C();
C* c = new C();
// All these calls should result in B::f
a->f();
b->f();
c->f();
}
The compiler states that C
is abstract.
How can this situation be resolved?
The issue seems similar to diamond inheritance, but I fail to see the solution.
EDIT: Thanks, this is the working example:
#include "stdio.h"
struct A
{
virtual void f() = 0;
};
struct B
{
void f()
{
printf("B::f\n");
}
};
struct C : public A, public B
{
void f()
{
printf("C::f\n");
B::f();
}
};
int main()
{
A* a = new C();
B* b = new C();
C* c = new C();
printf("Calling from A\n");
a->f();
printf("Calling from B\n");
b->f();
printf("Calling from C\n");
c->f();
}
Output:
Calling from A
C::f
B::f
Calling from B
B::f
Calling from C
C::f
B::f
回答1:
The issue is that the two f()
functions are completely unrelated, even though they happen to have the same name.
If the desired behaviour is for the C
's virtual
f()
to call B::f()
, you have to do it explicitly:
struct C : public A, public B
{
void f();
};
void C::f() {
B::f();
}
回答2:
You need to override the virtual function from the base class in C
. If you want it to call B::f
, say it like this:
struct C : A, B
{
void f() { B::f(); }
};
来源:https://stackoverflow.com/questions/14176916/derived-class-defines-function-via-base-class