deleting or throwing compiler error when a virtual base function is called from a derived class in c++
-1
I have the following code:
class A
{
public:
virtual void f(int a) = 0;
virtual void f(int a, int b) = 0;
};
class B : public A
{
public:
// do not want f(int a,int b) accessible
void f(int a);
};
class C : public A
{
public:
// do not want f(int a) accessible
void f(int a, int b);
};
I am aware that purely virtual functions cannot be deleted. Is there any way to disable these functions such that a compile time error occurs if an instance of B tries to call f(int,int) or when an instance of C tries to call f(int)
c++ oop c++11 polymorphism virtual-functions
asked Nov 23 '18 at 20:02
basilbasil
3091722
add a comment |
-1
I have the following code:
class A
{
public:
virtual void f(int a) = 0;
virtual void f(int a, int b) = 0;
};
class B : public A
{
public:
// do not want f(int a,int b) accessible
void f(int a);
};
class C : public A
{
public:
// do not want f(int a) accessible
void f(int a, int b);
};
I am aware that purely virtual functions cannot be deleted. Is there any way to disable these functions such that a compile time error occurs if an instance of B tries to call f(int,int) or when an instance of C tries to call f(int)
c++ oop c++11 polymorphism virtual-functions
asked Nov 23 '18 at 20:02
basilbasil
3091722
Did you mean detecting instead of deleting (in the question title)?
– François Andrieux
Nov 23 '18 at 20:03
They are both pure virtual, so they would need to be implemented. that's the compilation error if you try to instantiate B or C.
– Matthieu Brucher
Nov 23 '18 at 20:04
1
What would happen if it is called fromAinterface (B b; A& a = b; a.f(4, 2);)?
– Jarod42
Nov 23 '18 at 20:04
The nature ofvirtualis that you can't necessarily know at compile time which function will actually be called at a given call site. Edit : Maybe what you want is to implement bothA::f(int)andA::f(int, int)so that they throw, terminate or whatever. That way trying to call it with an instance that doesn't provide it can be detected.
– François Andrieux
Nov 23 '18 at 20:05
3
This just seems wrong. Neither B nor C is really an A here. How about just splitting A into two distinct interfaces?
– StoryTeller
Nov 23 '18 at 20:07
add a comment |
-1
-1
-1
I have the following code:
class A
{
public:
virtual void f(int a) = 0;
virtual void f(int a, int b) = 0;
};
class B : public A
{
public:
// do not want f(int a,int b) accessible
void f(int a);
};
class C : public A
{
public:
// do not want f(int a) accessible
void f(int a, int b);
};
I am aware that purely virtual functions cannot be deleted. Is there any way to disable these functions such that a compile time error occurs if an instance of B tries to call f(int,int) or when an instance of C tries to call f(int)
c++ oop c++11 polymorphism virtual-functions
asked Nov 23 '18 at 20:02
basilbasil
3091722
I have the following code:
class A
{
public:
virtual void f(int a) = 0;
virtual void f(int a, int b) = 0;
};
class B : public A
{
public:
// do not want f(int a,int b) accessible
void f(int a);
};
class C : public A
{
public:
// do not want f(int a) accessible
void f(int a, int b);
};
I am aware that purely virtual functions cannot be deleted. Is there any way to disable these functions such that a compile time error occurs if an instance of B tries to call f(int,int) or when an instance of C tries to call f(int)
c++ oop c++11 polymorphism virtual-functions
c++ oop c++11 polymorphism virtual-functions
asked Nov 23 '18 at 20:02
basilbasil
3091722
asked Nov 23 '18 at 20:02
basilbasil
3091722
asked Nov 23 '18 at 20:02
basilbasil
3091722
asked Nov 23 '18 at 20:02
basilbasil
3091722
asked Nov 23 '18 at 20:02
basilbasil
3091722
3091722
Did you mean detecting instead of deleting (in the question title)?
– François Andrieux
Nov 23 '18 at 20:03
They are both pure virtual, so they would need to be implemented. that's the compilation error if you try to instantiate B or C.
– Matthieu Brucher
Nov 23 '18 at 20:04
1
What would happen if it is called fromAinterface (B b; A& a = b; a.f(4, 2);)?
– Jarod42
Nov 23 '18 at 20:04
The nature ofvirtualis that you can't necessarily know at compile time which function will actually be called at a given call site. Edit : Maybe what you want is to implement bothA::f(int)andA::f(int, int)so that they throw, terminate or whatever. That way trying to call it with an instance that doesn't provide it can be detected.
– François Andrieux
Nov 23 '18 at 20:05
3
This just seems wrong. Neither B nor C is really an A here. How about just splitting A into two distinct interfaces?
– StoryTeller
Nov 23 '18 at 20:07
add a comment |
Did you mean detecting instead of deleting (in the question title)?
– François Andrieux
Nov 23 '18 at 20:03
They are both pure virtual, so they would need to be implemented. that's the compilation error if you try to instantiate B or C.
– Matthieu Brucher
Nov 23 '18 at 20:04
1
What would happen if it is called fromAinterface (B b; A& a = b; a.f(4, 2);)?
– Jarod42
Nov 23 '18 at 20:04
The nature ofvirtualis that you can't necessarily know at compile time which function will actually be called at a given call site. Edit : Maybe what you want is to implement bothA::f(int)andA::f(int, int)so that they throw, terminate or whatever. That way trying to call it with an instance that doesn't provide it can be detected.
– François Andrieux
Nov 23 '18 at 20:05
3
This just seems wrong. Neither B nor C is really an A here. How about just splitting A into two distinct interfaces?
– StoryTeller
Nov 23 '18 at 20:07
Did you mean detecting instead of deleting (in the question title)?
– François Andrieux
Nov 23 '18 at 20:03
Did you mean detecting instead of deleting (in the question title)?
– François Andrieux
Nov 23 '18 at 20:03
They are both pure virtual, so they would need to be implemented. that's the compilation error if you try to instantiate B or C.
– Matthieu Brucher
Nov 23 '18 at 20:04
They are both pure virtual, so they would need to be implemented. that's the compilation error if you try to instantiate B or C.
– Matthieu Brucher
Nov 23 '18 at 20:04
1
1
What would happen if it is called from
A interface (B b; A& a = b; a.f(4, 2);)?– Jarod42
Nov 23 '18 at 20:04
What would happen if it is called from
A interface (B b; A& a = b; a.f(4, 2);)?– Jarod42
Nov 23 '18 at 20:04
The nature of
virtual is that you can't necessarily know at compile time which function will actually be called at a given call site. Edit : Maybe what you want is to implement both A::f(int) and A::f(int, int) so that they throw, terminate or whatever. That way trying to call it with an instance that doesn't provide it can be detected.– François Andrieux
Nov 23 '18 at 20:05
The nature of
virtual is that you can't necessarily know at compile time which function will actually be called at a given call site. Edit : Maybe what you want is to implement both A::f(int) and A::f(int, int) so that they throw, terminate or whatever. That way trying to call it with an instance that doesn't provide it can be detected.– François Andrieux
Nov 23 '18 at 20:05
3
3
This just seems wrong. Neither B nor C is really an A here. How about just splitting A into two distinct interfaces?
– StoryTeller
Nov 23 '18 at 20:07
This just seems wrong. Neither B nor C is really an A here. How about just splitting A into two distinct interfaces?
– StoryTeller
Nov 23 '18 at 20:07
add a comment |
2 Answers
2
active
oldest
votes
1
There's no way to do that. You'd need a more complex class hierarchy. Something like this:
class A
{
public:
virtual ~A() {}
};
class BaseForB : public A
{
public:
virtual void f(int a) = 0;
};
class BaseForC : public A
{
public:
virtual void f(int a, int b) = 0;
};
class B : public BaseForB
{
public:
void f(int a) override
{
// details...
}
};
class C : public BaseForC
{
public:
void f(int a, int b) override
{
// details...
}
};
answered Nov 23 '18 at 20:11
Peter RudermanPeter Ruderman
10.2k2352
add a comment |
0
One option is to put a class between A B and A which implements a private version of the function that is final and not callable by instances of B like this:
class A
{
public:
virtual void f(int a) = 0;
virtual void f(int a, int b) = 0;
};
class A_B : public A {
using A::f;
private:
void f(int a, int b) final override {}
};
class B : public A_B
{
public:
// do not want f(int a,int b) accessible
void f(int a);
};
answered Nov 23 '18 at 20:41
Ian4264 Ian4264
13514
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
1
There's no way to do that. You'd need a more complex class hierarchy. Something like this:
class A
{
public:
virtual ~A() {}
};
class BaseForB : public A
{
public:
virtual void f(int a) = 0;
};
class BaseForC : public A
{
public:
virtual void f(int a, int b) = 0;
};
class B : public BaseForB
{
public:
void f(int a) override
{
// details...
}
};
class C : public BaseForC
{
public:
void f(int a, int b) override
{
// details...
}
};
answered Nov 23 '18 at 20:11
Peter RudermanPeter Ruderman
10.2k2352
add a comment |
1
There's no way to do that. You'd need a more complex class hierarchy. Something like this:
class A
{
public:
virtual ~A() {}
};
class BaseForB : public A
{
public:
virtual void f(int a) = 0;
};
class BaseForC : public A
{
public:
virtual void f(int a, int b) = 0;
};
class B : public BaseForB
{
public:
void f(int a) override
{
// details...
}
};
class C : public BaseForC
{
public:
void f(int a, int b) override
{
// details...
}
};
answered Nov 23 '18 at 20:11
Peter RudermanPeter Ruderman
10.2k2352
add a comment |
1
1
1
There's no way to do that. You'd need a more complex class hierarchy. Something like this:
class A
{
public:
virtual ~A() {}
};
class BaseForB : public A
{
public:
virtual void f(int a) = 0;
};
class BaseForC : public A
{
public:
virtual void f(int a, int b) = 0;
};
class B : public BaseForB
{
public:
void f(int a) override
{
// details...
}
};
class C : public BaseForC
{
public:
void f(int a, int b) override
{
// details...
}
};
answered Nov 23 '18 at 20:11
Peter RudermanPeter Ruderman
10.2k2352
There's no way to do that. You'd need a more complex class hierarchy. Something like this:
class A
{
public:
virtual ~A() {}
};
class BaseForB : public A
{
public:
virtual void f(int a) = 0;
};
class BaseForC : public A
{
public:
virtual void f(int a, int b) = 0;
};
class B : public BaseForB
{
public:
void f(int a) override
{
// details...
}
};
class C : public BaseForC
{
public:
void f(int a, int b) override
{
// details...
}
};
answered Nov 23 '18 at 20:11
Peter RudermanPeter Ruderman
10.2k2352
edited Nov 23 '18 at 20:22
answered Nov 23 '18 at 20:11
Peter RudermanPeter Ruderman
10.2k2352
answered Nov 23 '18 at 20:11
Peter RudermanPeter Ruderman
10.2k2352
answered Nov 23 '18 at 20:11
Peter RudermanPeter Ruderman
10.2k2352
10.2k2352
add a comment |
add a comment |
0
One option is to put a class between A B and A which implements a private version of the function that is final and not callable by instances of B like this:
class A
{
public:
virtual void f(int a) = 0;
virtual void f(int a, int b) = 0;
};
class A_B : public A {
using A::f;
private:
void f(int a, int b) final override {}
};
class B : public A_B
{
public:
// do not want f(int a,int b) accessible
void f(int a);
};
answered Nov 23 '18 at 20:41
Ian4264 Ian4264
13514
add a comment |
0
One option is to put a class between A B and A which implements a private version of the function that is final and not callable by instances of B like this:
class A
{
public:
virtual void f(int a) = 0;
virtual void f(int a, int b) = 0;
};
class A_B : public A {
using A::f;
private:
void f(int a, int b) final override {}
};
class B : public A_B
{
public:
// do not want f(int a,int b) accessible
void f(int a);
};
answered Nov 23 '18 at 20:41
Ian4264 Ian4264
13514
add a comment |
0
0
0
One option is to put a class between A B and A which implements a private version of the function that is final and not callable by instances of B like this:
class A
{
public:
virtual void f(int a) = 0;
virtual void f(int a, int b) = 0;
};
class A_B : public A {
using A::f;
private:
void f(int a, int b) final override {}
};
class B : public A_B
{
public:
// do not want f(int a,int b) accessible
void f(int a);
};
answered Nov 23 '18 at 20:41
Ian4264 Ian4264
13514
One option is to put a class between A B and A which implements a private version of the function that is final and not callable by instances of B like this:
class A
{
public:
virtual void f(int a) = 0;
virtual void f(int a, int b) = 0;
};
class A_B : public A {
using A::f;
private:
void f(int a, int b) final override {}
};
class B : public A_B
{
public:
// do not want f(int a,int b) accessible
void f(int a);
};
answered Nov 23 '18 at 20:41
Ian4264 Ian4264
13514
answered Nov 23 '18 at 20:41
Ian4264 Ian4264
13514
answered Nov 23 '18 at 20:41
Ian4264 Ian4264
13514
answered Nov 23 '18 at 20:41
Ian4264 Ian4264
13514
13514
add a comment |
add a comment |
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Did you mean detecting instead of deleting (in the question title)?
– François Andrieux
Nov 23 '18 at 20:03
They are both pure virtual, so they would need to be implemented. that's the compilation error if you try to instantiate B or C.
– Matthieu Brucher
Nov 23 '18 at 20:04
1
What would happen if it is called from
Ainterface (B b; A& a = b; a.f(4, 2);)?– Jarod42
Nov 23 '18 at 20:04
The nature of
virtualis that you can't necessarily know at compile time which function will actually be called at a given call site. Edit : Maybe what you want is to implement bothA::f(int)andA::f(int, int)so that they throw, terminate or whatever. That way trying to call it with an instance that doesn't provide it can be detected.– François Andrieux
Nov 23 '18 at 20:05
3
This just seems wrong. Neither B nor C is really an A here. How about just splitting A into two distinct interfaces?
– StoryTeller
Nov 23 '18 at 20:07