Applying “using” keyword on C++ pure virtual function
The Class B is overriding the pure Virtual Function "print()" of class A. Class C is inheriting Class B as well as having a "using A::print" statement.
Now why Class C is not an abstract class?
class A {
public :
virtual void print() =0;
};
class B:public A {
public:
void print();
};
void B :: print() {
cout << "nClass B print ()";
}
class C : public B {
public:
using A::print;
};
void funca (A *a) {
// a->print(1);
}
void funcb (B *b) {
b->print();
}
void funcc (C *c) {
c->print();
}
int main() {
B b;
C c;
funca(&c);
funcb(&c);
funcc(&c);
return 0;
}
Output:
Class B print ()
Class B print ()
c++ using virtual-functions
asked 1 hour ago
Gtrex
411
add a comment |
The Class B is overriding the pure Virtual Function "print()" of class A. Class C is inheriting Class B as well as having a "using A::print" statement.
Now why Class C is not an abstract class?
class A {
public :
virtual void print() =0;
};
class B:public A {
public:
void print();
};
void B :: print() {
cout << "nClass B print ()";
}
class C : public B {
public:
using A::print;
};
void funca (A *a) {
// a->print(1);
}
void funcb (B *b) {
b->print();
}
void funcc (C *c) {
c->print();
}
int main() {
B b;
C c;
funca(&c);
funcb(&c);
funcc(&c);
return 0;
}
Output:
Class B print ()
Class B print ()
c++ using virtual-functions
asked 1 hour ago
Gtrex
411
Kinda related, since a using declaration is not an overrider.
– StoryTeller
1 hour ago
add a comment |
The Class B is overriding the pure Virtual Function "print()" of class A. Class C is inheriting Class B as well as having a "using A::print" statement.
Now why Class C is not an abstract class?
class A {
public :
virtual void print() =0;
};
class B:public A {
public:
void print();
};
void B :: print() {
cout << "nClass B print ()";
}
class C : public B {
public:
using A::print;
};
void funca (A *a) {
// a->print(1);
}
void funcb (B *b) {
b->print();
}
void funcc (C *c) {
c->print();
}
int main() {
B b;
C c;
funca(&c);
funcb(&c);
funcc(&c);
return 0;
}
Output:
Class B print ()
Class B print ()
c++ using virtual-functions
asked 1 hour ago
Gtrex
411
The Class B is overriding the pure Virtual Function "print()" of class A. Class C is inheriting Class B as well as having a "using A::print" statement.
Now why Class C is not an abstract class?
class A {
public :
virtual void print() =0;
};
class B:public A {
public:
void print();
};
void B :: print() {
cout << "nClass B print ()";
}
class C : public B {
public:
using A::print;
};
void funca (A *a) {
// a->print(1);
}
void funcb (B *b) {
b->print();
}
void funcc (C *c) {
c->print();
}
int main() {
B b;
C c;
funca(&c);
funcb(&c);
funcc(&c);
return 0;
}
Output:
Class B print ()
Class B print ()
c++ using virtual-functions
c++ using virtual-functions
asked 1 hour ago
Gtrex
411
asked 1 hour ago
Gtrex
411
asked 1 hour ago
Gtrex
411
asked 1 hour ago
Gtrex
411
asked 1 hour ago
Gtrex
411
411
Kinda related, since a using declaration is not an overrider.
– StoryTeller
1 hour ago
add a comment |
Kinda related, since a using declaration is not an overrider.
– StoryTeller
1 hour ago
Kinda related, since a using declaration is not an overrider.
– StoryTeller
1 hour ago
Kinda related, since a using declaration is not an overrider.
– StoryTeller
1 hour ago
add a comment |
1 Answer
1
active
oldest
votes
From cppreference we read that
Using-declaration introduces a member of a base class into the derived class definition [...] If the derived class already has a member with the same name, parameter list, and qualifications, the derived class member hides or overrides (doesn't conflict with) the member that is introduced from the base class.
As C inherits from B, there is already a virtual print member function, which is non-pure. This one hides/overrides the member function brought in by using A::print, and hence C is not an abstract class.
answered 1 hour ago
lubgr
10.3k21745
That is just wrong,derived class already has a member, means a member first declared in this class. See demo here: godbolt.org/z/ff5cEb
– Oliv
56 mins ago
@Oliv Not sure if I get your point, why does it behave differently then based on the virtualness of the member function in question? If hiding or overriding both leads to an exclusion of the set of declarations introduced by the using declaration, shouldn't it exhibit identical behavior?
– lubgr
31 mins ago
Your right this does not apply too to this case. Neither is the standard paragraph you site...
– Oliv
5 mins ago
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
From cppreference we read that
Using-declaration introduces a member of a base class into the derived class definition [...] If the derived class already has a member with the same name, parameter list, and qualifications, the derived class member hides or overrides (doesn't conflict with) the member that is introduced from the base class.
As C inherits from B, there is already a virtual print member function, which is non-pure. This one hides/overrides the member function brought in by using A::print, and hence C is not an abstract class.
answered 1 hour ago
lubgr
10.3k21745
That is just wrong,derived class already has a member, means a member first declared in this class. See demo here: godbolt.org/z/ff5cEb
– Oliv
56 mins ago
@Oliv Not sure if I get your point, why does it behave differently then based on the virtualness of the member function in question? If hiding or overriding both leads to an exclusion of the set of declarations introduced by the using declaration, shouldn't it exhibit identical behavior?
– lubgr
31 mins ago
Your right this does not apply too to this case. Neither is the standard paragraph you site...
– Oliv
5 mins ago
add a comment |
From cppreference we read that
Using-declaration introduces a member of a base class into the derived class definition [...] If the derived class already has a member with the same name, parameter list, and qualifications, the derived class member hides or overrides (doesn't conflict with) the member that is introduced from the base class.
As C inherits from B, there is already a virtual print member function, which is non-pure. This one hides/overrides the member function brought in by using A::print, and hence C is not an abstract class.
answered 1 hour ago
lubgr
10.3k21745
That is just wrong,derived class already has a member, means a member first declared in this class. See demo here: godbolt.org/z/ff5cEb
– Oliv
56 mins ago
@Oliv Not sure if I get your point, why does it behave differently then based on the virtualness of the member function in question? If hiding or overriding both leads to an exclusion of the set of declarations introduced by the using declaration, shouldn't it exhibit identical behavior?
– lubgr
31 mins ago
Your right this does not apply too to this case. Neither is the standard paragraph you site...
– Oliv
5 mins ago
add a comment |
From cppreference we read that
Using-declaration introduces a member of a base class into the derived class definition [...] If the derived class already has a member with the same name, parameter list, and qualifications, the derived class member hides or overrides (doesn't conflict with) the member that is introduced from the base class.
As C inherits from B, there is already a virtual print member function, which is non-pure. This one hides/overrides the member function brought in by using A::print, and hence C is not an abstract class.
answered 1 hour ago
lubgr
10.3k21745
From cppreference we read that
Using-declaration introduces a member of a base class into the derived class definition [...] If the derived class already has a member with the same name, parameter list, and qualifications, the derived class member hides or overrides (doesn't conflict with) the member that is introduced from the base class.
As C inherits from B, there is already a virtual print member function, which is non-pure. This one hides/overrides the member function brought in by using A::print, and hence C is not an abstract class.
answered 1 hour ago
lubgr
10.3k21745
answered 1 hour ago
lubgr
10.3k21745
answered 1 hour ago
lubgr
10.3k21745
answered 1 hour ago
lubgr
10.3k21745
10.3k21745
That is just wrong,derived class already has a member, means a member first declared in this class. See demo here: godbolt.org/z/ff5cEb
– Oliv
56 mins ago
@Oliv Not sure if I get your point, why does it behave differently then based on the virtualness of the member function in question? If hiding or overriding both leads to an exclusion of the set of declarations introduced by the using declaration, shouldn't it exhibit identical behavior?
– lubgr
31 mins ago
Your right this does not apply too to this case. Neither is the standard paragraph you site...
– Oliv
5 mins ago
add a comment |
That is just wrong,derived class already has a member, means a member first declared in this class. See demo here: godbolt.org/z/ff5cEb
– Oliv
56 mins ago
@Oliv Not sure if I get your point, why does it behave differently then based on the virtualness of the member function in question? If hiding or overriding both leads to an exclusion of the set of declarations introduced by the using declaration, shouldn't it exhibit identical behavior?
– lubgr
31 mins ago
Your right this does not apply too to this case. Neither is the standard paragraph you site...
– Oliv
5 mins ago
That is just wrong,
derived class already has a member, means a member first declared in this class. See demo here: godbolt.org/z/ff5cEb– Oliv
56 mins ago
That is just wrong,
derived class already has a member, means a member first declared in this class. See demo here: godbolt.org/z/ff5cEb– Oliv
56 mins ago
@Oliv Not sure if I get your point, why does it behave differently then based on the virtualness of the member function in question? If hiding or overriding both leads to an exclusion of the set of declarations introduced by the using declaration, shouldn't it exhibit identical behavior?
– lubgr
31 mins ago
@Oliv Not sure if I get your point, why does it behave differently then based on the virtualness of the member function in question? If hiding or overriding both leads to an exclusion of the set of declarations introduced by the using declaration, shouldn't it exhibit identical behavior?
– lubgr
31 mins ago
Your right this does not apply too to this case. Neither is the standard paragraph you site...
– Oliv
5 mins ago
Your right this does not apply too to this case. Neither is the standard paragraph you site...
– Oliv
5 mins ago
add a comment |
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Kinda related, since a using declaration is not an overrider.
– StoryTeller
1 hour ago