My question is why I cannot call protected virtual member function in derived class through a pointer to the base class unless declaring derived class as a friend of base class?
For example:
#include <iostream>
class A {
friend class C; // (1)
protected:
virtual void foo() const = 0;
};
class B : public A {
void foo() const override { std::cout << "B::foo" << std::endl; }
};
class C : public A {
friend void bar(const C &);
public:
C(A *aa) : a(aa) { }
private:
void foo() const override {
a->foo(); // (2) Compile Error if we comment out (1)
//this->foo(); // (3) Compile OK, but this is not virtual call, and will cause infinite recursion
std::cout << "C::foo" << std::endl;
}
A *a;
};
void bar(const C &c) {
c.foo();
}
int main() {
B b;
C c(&b);
bar(c);
return 0;
}
The output is
B::foo
C::foo
In the above code, I want to call virtual function foo() through member a of class C (not the static bound one through this at compile time), but if I don't make C as A's friend, the call is illegal.
I think C is inherited from A, so that it can access the protected member of A, but why is it actually not happen?
Class C can access the protected members of its own base class, but not members of any other A.
In your example, the parameter a is part of the totally unrelated class B to which C has no access rights (unless you make it a friend).
Related
Look at this excerpt of a program.
I see that cout << obj->foo(); call is not polymorphic. Actually, it is obvious, because it has no virtual specificator.
But I am confused with cout << ((B*)obj)->foo(); Why the program does not use the B's definition of virtual function and will call the third version of foo()?
#include <iostream>
using namespace std;
class A{
public:
int foo(){ return 1; }
};
class B: public A{
public:
virtual int foo(){ return 2; }
};
class C: public B{
public:
int foo(){ return 3; }
};
int main() {
A* obj = new C;
cout << obj->foo();
cout << ((B*)obj)->foo();
cout << ((C*)obj)->foo();
return 0;
}
A::foo() is not virtual. Calling foo() via an A* pointer (or A& reference) will call A::foo() directly without any polymorphic dispatch.
B::foo() is virtual. Calling foo() via a B* pointer (or B& reference) will dispatch the call to the most derived implementation of foo() that exists in the object that the B* (or B&) refers to.
C derives from B, and C::foo() overrides B::foo(), and obj points to a C object, which is why C::foo() gets called by polymorphic dispatch when foo() is called via a B* or C* pointer (or a B& or C& reference).
Because ((B*)obj)->foo(); behaves by design like B* b = (B*)obj; b->foo() and calls C::foo. You may call the base's method explicitly like ((B*)obj)->B::foo();.
#include <iostream>
using namespace std;
class A{
public:
int foo(){ return 1; }
};
class B: public A{
public:
virtual int foo(){ return 2; }
};
class C: public B{
public:
int foo() override { return 3; }
};
int main() {
A* obj = new C;
cout << obj->foo();
cout << ((B*)obj)->B::foo();
cout << ((C*)obj)->foo();
return 0;
}
Output: 123
Member function foo is virtual from class B downwards, i.e. also in C, even if it is not marked virtual or override there.
Thus, call ((B*)obj)->foo() is a virtual call, actually resulting in calling C::foo.
Can I have a virtual function in the base class and some of my derived classes do have that function and some don't have.
class A{
virtual void Dosomething();
};
class B : public A{
void Dosomething();
};
class C : public A{
//Does not have Dosomething() function.
};
From one of my c++ textbook:
Once a function is declared virtual, it remains virtual all the way down the inheritance, even if the function is not explicitly declared virtual when the derived class overrides it.
When the derived class chooses not to override it, it simply inherits its base class's virtual function.
Therefore to your question the answer is No. Class c will use Class A's virtual function.
Derived classes do not have to implement all the virtual functions, unless it is a pure virtual function. Even in this case, it will cause an error only when you try to instantiate the derived class( without implementing the pure virtual function ).
#include <iostream>
class A{
public :
virtual void foo() = 0;
};
class B: public A{
public :
void foo(){ std::cout << "foo" << std::endl;}
};
class C: public A{
void bar();
};
int main() {
//C temp; The compiler will complain only if this is initialized without
// implementing foo in the derived class C
return 0;
}
I think the closest you might get, is to change the access modifier in the derived class, as depicted below.
But, I would consider it bad practice, as it violates Liskov's substitution principle.
If you have a situation like this, you might need to reconsider your class design.
#include <iostream>
class A {
public:
virtual void doSomething() { std::cout << "A" << std::endl; }
};
class B : public A {
public:
void doSomething() override { std::cout << "B" << std::endl; };
};
class C : public A {
private:
void doSomething() override { std::cout << "C" << std::endl; };
};
int main(int argc, char **args) {
A a;
a.doSomething();
B b;
b.doSomething();
C c;
//c.doSomething(); // Not part of the public interface. Violates Liskov's substitution principle.
A* c2 = &c;
c2->doSomething(); // Still possible, even though it is private! But, C::doSomething() is called!
return 0;
}
I tried this code:
class A
{
virtual void foo() = 0;
};
class B
{
virtual void foo() = 0;
};
class C : public A, public B
{
//virtual void A::foo(){}
//virtual void B::foo(){}
virtual void A::foo();
virtual void B::foo();
};
void C::A::foo(){}
void C::B::foo(){}
int main()
{
C c;
return 0;
}
It is OK when using the commented part, but when I try to write the definitions outside the class declaration, the compiler reports errors.
I am using the MSVC11 compiler, does anyone know how to write this?
I need to move the code into the cpp file.
Thank you~~
A function overrides a virtual function of a base class based on the name and parameter types (see below). Therefore, your class C has two virtual functions foo, one inherited from each A and B. But a function void C::foo() overrides both:
[class.virtual]/2
If a virtual member function vf is declared in a class Base and in a class Derived, derived directly or indirectly from Base, a member function vf with the same name, parameter-type-list, cv-qualification, and ref-qualifier (or absence of same) as Base::vf is declared, then Derived::vf is also virtual (whether or not it is so declared) and it overrides Base::vf.
As I already stated in the comments, [dcl.meaning]/1 forbids the use of a qualified-id in the declaration of a (member) function:
When the declarator-id is qualified, the declaration shall refer to a previously declared member of the class or namespace to which the qualifier refers [...]"
Therefore any virtual void X::foo(); is illegal as a declaration inside C.
The code
class C : public A, public B
{
virtual void foo();
};
is the only way AFAIK to override foo, and it will override both A::foo and B::foo. There is no way to have two different overrides for A::foo and B::foo with different behaviour other than by introducing another layer of inheritance:
#include <iostream>
struct A
{
virtual void foo() = 0;
};
struct B
{
virtual void foo() = 0;
};
struct CA : A
{
virtual void foo() { std::cout << "A" << std::endl; }
};
struct CB : B
{
virtual void foo() { std::cout << "B" << std::endl; }
};
struct C : CA, CB {};
int main() {
C c;
//c.foo(); // ambiguous
A& a = c;
a.foo();
B& b = c;
b.foo();
}
You've got just one virtual function foo:
class A {
virtual void foo() = 0;
};
class B {
virtual void foo() = 0;
};
class C : public A, public B {
virtual void foo();
};
void C::foo(){}
void C::A::foo(){}
void C::B::foo(){};
int main() {
C c;
return 0;
}
I stepped into the same problem and accidentially opened a second thread. Sorry for that. One way that worked for me was to solve it without multiple inheritance.
#include <stdio.h>
class A
{
public:
virtual void foo(void) = 0;
};
class B
{
public:
virtual void foo(void) = 0;
};
class C
{
class IA: public A
{
virtual void foo(void)
{
printf("IA::foo()\r\n");
}
};
class IB: public B
{
virtual void foo(void)
{
printf("IB::foo()\r\n");
}
};
IA m_A;
IB m_B;
public:
A* GetA(void)
{
return(&m_A);
}
B* GetB(void)
{
return(&m_B);
}
};
The trick is to define classes derived from the interfaces (A and B) as local classes (IA and IB) instead of using multiple inheritance. Furthermore this approach also opens the option to have multiple realizations of each interface if desired which would not be possible using multiple inheritance.
The local classes IA and IB can be easily given access to class C, so the implementations of both interfaces IA and IB can share data.
Access of each interface can be done as follows:
main()
{
C test;
test.GetA()->foo();
test.GetB()->foo();
}
... and there is no ambiguity regarding the foo method any more.
You can resolve this ambiguity with different function parameters.
In real-world code, such virtual functions do something, so they usually already have either:
different parameters in A and B, or
different return values in A and B that you can turn into [out] parameters for the sake of solving this inheritance problem; otherwise
you need to add some tag parameters, which the optimizer will throw away.
(In my own code I usually find myself in case (1), sometimes in (2), never so far in (3).)
Your example is case (3) and would look like this:
class A
{
public:
struct tag_a { };
virtual void foo(tag_a) = 0;
};
class B
{
public:
struct tag_b { };
virtual void foo(tag_b) = 0;
};
class C : public A, public B
{
void foo(tag_a) override;
void foo(tag_b) override;
};
A slight improvement over adigostin's solution:
#include <iostream>
struct A {
virtual void foo() = 0;
};
struct B {
virtual void foo() = 0;
};
template <class T> struct Tagger : T {
struct tag {};
void foo() final { foo({}); }
virtual void foo(tag) = 0;
};
using A2 = Tagger<A>;
using B2 = Tagger<B>;
struct C : public A2, public B2 {
void foo(A2::tag) override { std::cout << "A" << std::endl; }
void foo(B2::tag) override { std::cout << "B" << std::endl; }
};
int main() {
C c;
A* pa = &c;
B* pb = &c;
pa->foo(); // A
pb->foo(); // B
return 0;
}
Assuming that the base classes A and B are given and cannot be modified.
The compiler keeps saying 'class A' has no member named 'foo'.
I am trying to use a function from a derived class with a pointer. Here is my code:
class A{
.....
};
class B:public A{
virtual void foo() = 0;
};
class C:public B{
....
public:
void foo(){
....
}
};
I have a table of A pointers named Table and when trying
Table[j]->foo()
I am getting the compiler error.
What should I do except cast?
You have a compilation error because function foo is not declared in class A.
You can declare it as pure virtual in A like this:
class A {
virtual void foo() = 0;
};
In derived classes you don't have to declare foo as explicitly virtual. If only C is a concrete class, you don't have to declare foo in class B at all.
If your example if you know that in your array of pointers to A you have only instances of class C you can explicitly cast to pointer to C but it is a sign of poor design and I don't recommend it:
static_cast<C*>(Table[j])->foo()
If you have a pointer to a base and want to access a member of the derived type you have to cast so you have a pointer to the derived type, either that or as in your case add foo() as a virtual member to the base.
class A
{ ... };
class B : public A:{
{
public:
virtual foo() { std::cout << "B::foo" << std::endl; }
};
class C : public B
{
public:
void foo() { std::cout << "C::foo" << std::endl;
};
...
A * arr[8];
for(int i = 0; 8>i; ++i)
arr[i] = new C;
((B*)arr[0])->foo(); // will print "C::foo\n"
But if you added virtual void foo() = 0; to A then you could just do arr[0]->foo() and it would still print C::foo
I have stumbled on a problem while trying to re-use code from different classes. I post it here in hope that some of you might be able to help me.
I have a set of classes (B,C) deriving from the same class (A) which forces the implementation of some methods (foo, run). Class B implements these method, and both B and C provide other methods:
#include<iostream>
template<class I, class O>
class A {
public:
A() {}
virtual ~A() {}
virtual void foo() const = 0; // force implementation of this function
virtual void run() const = 0; // force implementation of this function
};
template<class I, class O>
class B : public A<I,O> {
public:
B() {}
virtual ~B() {}
virtual void foo() const { // implementation for the Base class
std::cout << "B's implementation of foo" << std::endl;
}
virtual void run() const { // implementation for the Base class
std::cout << "B's implementation of run" << std::endl;
}
virtual void foobar() const { // some other function provided by this class
std::cout << "B's implementation of foobar" << std::endl;
}
};
template<class I, class O, class M>
class C : public A<I,O> {
public:
C() {}
virtual ~C() {}
virtual void bar(M m) const { // some other function provided by this class
std::cout << "C's implementation of bar with: " << m << std::endl;
}
};
Now, what I am trying to do is inherit from both B and C so that I can have the extra methods (foobar, bar), but also not have to implement the method from class A (foo) because it is already defined in B:
template<class I, class O>
class D : public B<I,O>, public C<I,O,int> {
public:
D() {}
void run() const {
this->bar(123);
this->foo();
this->foobar();
}
};
But for some reason the compiler gives me this error:
test.cpp: In function ‘int main(int, char**)’:
test.cpp:68:35: error: cannot allocate an object of abstract type ‘D<float, double>’
A<float, double> *d = new D<float, double>(); // what I need to do
test.cpp:48:11: note: because the following virtual functions are pure within ‘D<float, double>’:
class D : public B<I,O>, public C<I,O,int> {
^
test.cpp:9:22: note: void A<I, O>::foo() const [with I = float; O = double]
virtual void foo() const = 0; // force implementation of this function
This is the code I use to run it:
int main(int argc, char **argv)
{
A<float, double> *b = new B<float, double>();
b->foo(); // prints "B's implementation of foo"
b->run(); // prints "B's implementation of run"
//A<float, double> *c = new C<float, double, int>(); // obviously fails because C does not implement any of A's functions
//A<float, double> *d = new D<float, double>; // line 68: what I need to do
//d->run(); // ***throws the abstract class error
return 0;
}
I want to use the 'run' function of an object of class D from a pointer to a A. As all the functions are virtual I expect to execute implementation of each function defined in the lowest inheritance point, meaning that 'B::run' will be discarded. As 'D::run' uses functions from both B and C I need to inherit from both classes.
I hope I have described it enough and not confused anybody.
Thanks for the help!
If you change B and C to virtually inherit from the A template class, they will share a single base instance when combined by D and this error will go away:
template<class I, class O>
class B : virtual public A<I,O> {
// ...
template<class I, class O, class M>
class C : virtual public A<I,O> {
However, this pattern (known as the diamond inheritance (anti-)pattern) can be very difficult to reason about and I would strongly suggest avoiding it if possible. You are likely to run into even more obscure problems later.
Here is a sample of this technique working, but showing some results that may not be expected at first glance:
class A {
public:
virtual void foo() = 0;
};
class B : virtual public A {
public:
virtual void foo() override;
};
void B::foo()
{
std::cout << "B::foo()" << std::endl;
}
class C : virtual public A { };
class D : public B, public C { };
int main() {
D d;
C & c = d;
c.foo();
return 0;
}
Note that even though you are calling C::foo(), which is pure virtual, since there is only one A instance the inherited pure virtual function resolves to B::foo() though the shared A vtable. This is a somewhat surprising side-effect -- that you can wind up invoking methods implemented on a cousin type.
The answer by #cdhowie gives you a solution.
To understand the problem the compiler is complaining about, take a set of simpler classes:
struct A
{
virtual void foo() = 0;
};
struct B : A
{
virtual void foo() {}
}
struct C : A
{
void bar() {}
}
struct D : B, C
{
};
The class hierarchy of D is:
A A
| |
B C
\ /
D
With this inheritance structure, D has two virtual tables, one corresponding to the B inheritance hierarchy and one corresponding to C inheritance hierarchy. The difference being that in the B hierarchy, there is an implementation of A::foo() while there isn't one in the C hierarchy.
Let's say you were allowed to construct an object of type D.
D d;
C* cp = &d;
Now cp points to the C hierarchy of D, and uses a virtual table in which foo is not implemented. That will be a run time error that the compiler is helping you avoid at compile time.
I know this is a late answer but since you are deriving from a pure virtual function for class C, you have to implement it, then in those functions you call the base class:
virtual void foo() const { // for class C
B::foo();
}