Overload pure virtual function with non pure virtual version

up vote
7
down vote

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With Base and Derived defined like this:

class Base 

 public:
 virtual int f1(int a) const = 0;
 virtual int f2(int a, int b) const return a+b;
;

class Derived : public Base 

 public:
 int f1(int a) const return a; 
; 

int main() 
 Derived obj;
 cout << obj.f1(1) << endl;
 cout << obj.f2(1, 2) << endl;

The result is

1
3

obj.f1(1) uses the f1 implementation from Derived and obj.f2(1, 2) uses the implementation inherited from Base, which is what I want.

Now, I would like these two functions to have the same name, f, so the base class provides an implementation when there are two parameters and the derived class must implement the single parameter version (that's why it's pure virtual).

However, if I do this (just rename f1 and f2 to f):

class Base 

 public:
 virtual int f(int a) const = 0;
 virtual int f(int a, int b) const return a + b;
;

class Derived : public Base 

 public:
 int f(int a) const return a; 
;

int main() 
 Derived obj;
 cout << obj.f(1) << endl;
 cout << obj.f(1, 2) << endl;

I get the following error:

20:23: error: no matching function for call to 'Derived::f(int, int)'
20:23: note: candidate is:
14:13: note: virtual int Derived::f(int) const
14:13: note: candidate expects 1 argument, 2 provided

Why is this? Is it not possible to do this kind of overloading?

asked 46 mins ago

Milo

1,031926

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up vote
7
down vote

favorite

With Base and Derived defined like this:

class Base 

 public:
 virtual int f1(int a) const = 0;
 virtual int f2(int a, int b) const return a+b;
;

class Derived : public Base 

 public:
 int f1(int a) const return a; 
; 

int main() 
 Derived obj;
 cout << obj.f1(1) << endl;
 cout << obj.f2(1, 2) << endl;

The result is

1
3

obj.f1(1) uses the f1 implementation from Derived and obj.f2(1, 2) uses the implementation inherited from Base, which is what I want.

However, if I do this (just rename f1 and f2 to f):

class Base 

 public:
 virtual int f(int a) const = 0;
 virtual int f(int a, int b) const return a + b;
;

class Derived : public Base 

 public:
 int f(int a) const return a; 
;

int main() 
 Derived obj;
 cout << obj.f(1) << endl;
 cout << obj.f(1, 2) << endl;

I get the following error:

20:23: error: no matching function for call to 'Derived::f(int, int)'
20:23: note: candidate is:
14:13: note: virtual int Derived::f(int) const
14:13: note: candidate expects 1 argument, 2 provided

Why is this? Is it not possible to do this kind of overloading?

asked 46 mins ago

Milo

1,031926

add a commentÂ |Â

up vote
7
down vote

favorite

With Base and Derived defined like this:

class Base 

 public:
 virtual int f1(int a) const = 0;
 virtual int f2(int a, int b) const return a+b;
;

class Derived : public Base 

 public:
 int f1(int a) const return a; 
; 

int main() 
 Derived obj;
 cout << obj.f1(1) << endl;
 cout << obj.f2(1, 2) << endl;

The result is

1
3

obj.f1(1) uses the f1 implementation from Derived and obj.f2(1, 2) uses the implementation inherited from Base, which is what I want.

However, if I do this (just rename f1 and f2 to f):

class Base 

 public:
 virtual int f(int a) const = 0;
 virtual int f(int a, int b) const return a + b;
;

class Derived : public Base 

 public:
 int f(int a) const return a; 
;

int main() 
 Derived obj;
 cout << obj.f(1) << endl;
 cout << obj.f(1, 2) << endl;

I get the following error:

20:23: error: no matching function for call to 'Derived::f(int, int)'
20:23: note: candidate is:
14:13: note: virtual int Derived::f(int) const
14:13: note: candidate expects 1 argument, 2 provided

Why is this? Is it not possible to do this kind of overloading?

asked 46 mins ago

Milo

1,031926

With Base and Derived defined like this:

class Base 

 public:
 virtual int f1(int a) const = 0;
 virtual int f2(int a, int b) const return a+b;
;

class Derived : public Base 

 public:
 int f1(int a) const return a; 
; 

int main() 
 Derived obj;
 cout << obj.f1(1) << endl;
 cout << obj.f2(1, 2) << endl;

The result is

1
3

obj.f1(1) uses the f1 implementation from Derived and obj.f2(1, 2) uses the implementation inherited from Base, which is what I want.

However, if I do this (just rename f1 and f2 to f):

class Base 

 public:
 virtual int f(int a) const = 0;
 virtual int f(int a, int b) const return a + b;
;

class Derived : public Base 

 public:
 int f(int a) const return a; 
;

int main() 
 Derived obj;
 cout << obj.f(1) << endl;
 cout << obj.f(1, 2) << endl;

I get the following error:

20:23: error: no matching function for call to 'Derived::f(int, int)'
20:23: note: candidate is:
14:13: note: virtual int Derived::f(int) const
14:13: note: candidate expects 1 argument, 2 provided

Why is this? Is it not possible to do this kind of overloading?

c++

asked 46 mins ago

Milo

1,031926

asked 46 mins ago

Milo

1,031926

asked 46 mins ago

Milo

1,031926

asked 46 mins ago

Milo

1,031926

asked 46 mins ago

Milo

1,031926

add a commentÂ |Â

5 Answers
5

active

oldest

votes

up vote
7
down vote

You need to write

class Derived : public Base 

 public:
 using Base::f;
 int f(int a) const return a; 
;

Note the using statement. That brings the base class version back into scope.

answered 45 mins ago

Bathsheba

167k26237352

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up vote
3
down vote

Go to the concise answer by @Bathsheba to see the proper way at solving the issue -- I'm just putting this here to answer the question:

Why is this? Is it not possible to do this kind of overloading?

No, not possible as written in the original question due to [basic.scope.hidingÃ‚Â¶3]:

In a member function definition, the declaration of a name at block
scope hides the declaration of a member of the class with the same
name; see [basic.scope.class]. The declaration of a member in a
derived class hides the declaration of a member of a base class of the
same name; see [class.member.lookup].

This clause is about names, not overloads. As such, it doesn't matter if there are other overloads in the base class, they all share the same name, which is hidden according to quote above.

edited 12 mins ago

StoryTeller

82.9k12166230

answered 18 mins ago

SkepticalEmpiricist

3,473720

add a commentÂ |Â

up vote
1
down vote

You can either pull all definitions of Base into scope with by writing using Base::f; or you can explcitily write for some versions of f something like this: int f(int a, int b) const override return Base::f(a,b);

class Derived : public Base 

public:
 int f(int a) const return a; 
 int f(int a, int b) const override return Base::f(a,b); 
;

The version with using is already mentioned in this answer:

class Derived : public Base 

 public:
 using Base::f;
 int f(int a) const return a; 
;

Note: I know that the second version does not work on MSVC 2010. I am also aware this compiler is ancient, but just for people who care ;)

edited 31 mins ago

answered 38 mins ago

P i

2,340832

add a commentÂ |Â

up vote
0
down vote

In addition to the other solutions, if you don't want to redefine f you could explicitly invoke f of the base class using

int main() 
 Derived obj;
 cout << obj.f1(1) << endl;
 cout << obj.Base::f2(1, 2) << endl;

answered 2 mins ago

dan

1462

add a commentÂ |Â

up vote
0
down vote

Why is this?

Calling one of multiple overloads of a member function is resolved by the compiler during "overload resolution". But before available overloads can be evaluated with respect to how well they match the invocation, they must be collected. This step is the "name lookup". Here,

cout << obj.f(1, 2) << endl;

the static type of the instance obj, which is Derived, is searched for a function signature with the name f. There is exactly one, because once any f has been overridden, this is the only interface Derived offers its clients. Those from Base are not automatically brought in - but why this limitation?

Think of an overload set as a bundled entity of a class interface. Base offers an overload set Base::f, and Derived offers Derived::f. Again, once one single function of the former overrides one function of the latter, the whole overload set is overridden (unless enriched by using Base::f). This makes sense, because f (the overload set) is the means a class designer gives client code to use its class.

edited 1 min ago

answered 21 mins ago

lubgr

6,59021340

add a commentÂ |Â

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5 Answers
5

active

oldest

votes

5 Answers
5

active

oldest

votes

up vote
7
down vote

You need to write

class Derived : public Base 

 public:
 using Base::f;
 int f(int a) const return a; 
;

Note the using statement. That brings the base class version back into scope.

answered 45 mins ago

Bathsheba

167k26237352

add a commentÂ |Â

up vote
7
down vote

You need to write

class Derived : public Base 

 public:
 using Base::f;
 int f(int a) const return a; 
;

Note the using statement. That brings the base class version back into scope.

answered 45 mins ago

Bathsheba

167k26237352

add a commentÂ |Â

up vote
7
down vote

You need to write

class Derived : public Base 

 public:
 using Base::f;
 int f(int a) const return a; 
;

Note the using statement. That brings the base class version back into scope.

answered 45 mins ago

Bathsheba

167k26237352

You need to write

class Derived : public Base 

 public:
 using Base::f;
 int f(int a) const return a; 
;

Note the using statement. That brings the base class version back into scope.

answered 45 mins ago

Bathsheba

167k26237352

answered 45 mins ago

Bathsheba

167k26237352

answered 45 mins ago

Bathsheba

167k26237352

answered 45 mins ago

Bathsheba

167k26237352

add a commentÂ |Â

up vote
3
down vote

Go to the concise answer by @Bathsheba to see the proper way at solving the issue -- I'm just putting this here to answer the question:

Why is this? Is it not possible to do this kind of overloading?

No, not possible as written in the original question due to [basic.scope.hidingÃ‚Â¶3]:

In a member function definition, the declaration of a name at block
scope hides the declaration of a member of the class with the same
name; see [basic.scope.class]. The declaration of a member in a
derived class hides the declaration of a member of a base class of the
same name; see [class.member.lookup].

This clause is about names, not overloads. As such, it doesn't matter if there are other overloads in the base class, they all share the same name, which is hidden according to quote above.

edited 12 mins ago

StoryTeller

82.9k12166230

answered 18 mins ago

SkepticalEmpiricist

3,473720

add a commentÂ |Â

up vote
3
down vote

Go to the concise answer by @Bathsheba to see the proper way at solving the issue -- I'm just putting this here to answer the question:

Why is this? Is it not possible to do this kind of overloading?

No, not possible as written in the original question due to [basic.scope.hidingÃ‚Â¶3]:

In a member function definition, the declaration of a name at block
scope hides the declaration of a member of the class with the same
name; see [basic.scope.class]. The declaration of a member in a
derived class hides the declaration of a member of a base class of the
same name; see [class.member.lookup].

This clause is about names, not overloads. As such, it doesn't matter if there are other overloads in the base class, they all share the same name, which is hidden according to quote above.

edited 12 mins ago

StoryTeller

82.9k12166230

answered 18 mins ago

SkepticalEmpiricist

3,473720

add a commentÂ |Â

up vote
3
down vote

Go to the concise answer by @Bathsheba to see the proper way at solving the issue -- I'm just putting this here to answer the question:

Why is this? Is it not possible to do this kind of overloading?

No, not possible as written in the original question due to [basic.scope.hidingÃ‚Â¶3]:

In a member function definition, the declaration of a name at block
scope hides the declaration of a member of the class with the same
name; see [basic.scope.class]. The declaration of a member in a
derived class hides the declaration of a member of a base class of the
same name; see [class.member.lookup].

This clause is about names, not overloads. As such, it doesn't matter if there are other overloads in the base class, they all share the same name, which is hidden according to quote above.

edited 12 mins ago

StoryTeller

82.9k12166230

answered 18 mins ago

SkepticalEmpiricist

3,473720

Go to the concise answer by @Bathsheba to see the proper way at solving the issue -- I'm just putting this here to answer the question:

Why is this? Is it not possible to do this kind of overloading?

No, not possible as written in the original question due to [basic.scope.hidingÃ‚Â¶3]:

In a member function definition, the declaration of a name at block
scope hides the declaration of a member of the class with the same
name; see [basic.scope.class]. The declaration of a member in a
derived class hides the declaration of a member of a base class of the
same name; see [class.member.lookup].

This clause is about names, not overloads. As such, it doesn't matter if there are other overloads in the base class, they all share the same name, which is hidden according to quote above.

edited 12 mins ago

StoryTeller

82.9k12166230

answered 18 mins ago

SkepticalEmpiricist

3,473720

edited 12 mins ago

StoryTeller

82.9k12166230

edited 12 mins ago

StoryTeller

82.9k12166230

edited 12 mins ago

StoryTeller

82.9k12166230

answered 18 mins ago

SkepticalEmpiricist

3,473720

answered 18 mins ago

SkepticalEmpiricist

3,473720

answered 18 mins ago

SkepticalEmpiricist

3,473720

add a commentÂ |Â

up vote
1
down vote

class Derived : public Base 

public:
 int f(int a) const return a; 
 int f(int a, int b) const override return Base::f(a,b); 
;

The version with using is already mentioned in this answer:

class Derived : public Base 

 public:
 using Base::f;
 int f(int a) const return a; 
;

Note: I know that the second version does not work on MSVC 2010. I am also aware this compiler is ancient, but just for people who care ;)

edited 31 mins ago

answered 38 mins ago

P i

2,340832

add a commentÂ |Â

up vote
1
down vote

class Derived : public Base 

public:
 int f(int a) const return a; 
 int f(int a, int b) const override return Base::f(a,b); 
;

The version with using is already mentioned in this answer:

class Derived : public Base 

 public:
 using Base::f;
 int f(int a) const return a; 
;

Note: I know that the second version does not work on MSVC 2010. I am also aware this compiler is ancient, but just for people who care ;)

edited 31 mins ago

answered 38 mins ago

P i

2,340832

add a commentÂ |Â

up vote
1
down vote

class Derived : public Base 

public:
 int f(int a) const return a; 
 int f(int a, int b) const override return Base::f(a,b); 
;

The version with using is already mentioned in this answer:

class Derived : public Base 

 public:
 using Base::f;
 int f(int a) const return a; 
;

Note: I know that the second version does not work on MSVC 2010. I am also aware this compiler is ancient, but just for people who care ;)

edited 31 mins ago

answered 38 mins ago

P i

2,340832

class Derived : public Base 

public:
 int f(int a) const return a; 
 int f(int a, int b) const override return Base::f(a,b); 
;

The version with using is already mentioned in this answer:

class Derived : public Base 

 public:
 using Base::f;
 int f(int a) const return a; 
;

Note: I know that the second version does not work on MSVC 2010. I am also aware this compiler is ancient, but just for people who care ;)

edited 31 mins ago

answered 38 mins ago

P i

2,340832

edited 31 mins ago

answered 38 mins ago

P i

2,340832

answered 38 mins ago

P i

2,340832

answered 38 mins ago

P i

2,340832

add a commentÂ |Â

up vote
0
down vote

In addition to the other solutions, if you don't want to redefine f you could explicitly invoke f of the base class using

int main() 
 Derived obj;
 cout << obj.f1(1) << endl;
 cout << obj.Base::f2(1, 2) << endl;

answered 2 mins ago

dan

1462

add a commentÂ |Â

up vote
0
down vote

In addition to the other solutions, if you don't want to redefine f you could explicitly invoke f of the base class using

int main() 
 Derived obj;
 cout << obj.f1(1) << endl;
 cout << obj.Base::f2(1, 2) << endl;

answered 2 mins ago

dan

1462

add a commentÂ |Â

up vote
0
down vote

In addition to the other solutions, if you don't want to redefine f you could explicitly invoke f of the base class using

int main() 
 Derived obj;
 cout << obj.f1(1) << endl;
 cout << obj.Base::f2(1, 2) << endl;

answered 2 mins ago

dan

1462

In addition to the other solutions, if you don't want to redefine f you could explicitly invoke f of the base class using

int main() 
 Derived obj;
 cout << obj.f1(1) << endl;
 cout << obj.Base::f2(1, 2) << endl;

answered 2 mins ago

dan

1462

answered 2 mins ago

dan

1462

answered 2 mins ago

dan

1462

answered 2 mins ago

dan

1462

add a commentÂ |Â

up vote
0
down vote

Why is this?

cout << obj.f(1, 2) << endl;

edited 1 min ago

answered 21 mins ago

lubgr

6,59021340

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up vote
0
down vote

Why is this?

cout << obj.f(1, 2) << endl;

edited 1 min ago

answered 21 mins ago

lubgr

6,59021340

add a commentÂ |Â

up vote
0
down vote

Why is this?

cout << obj.f(1, 2) << endl;

edited 1 min ago

answered 21 mins ago

lubgr

6,59021340

Why is this?

cout << obj.f(1, 2) << endl;

edited 1 min ago

answered 21 mins ago

lubgr

6,59021340

edited 1 min ago

answered 21 mins ago

lubgr

6,59021340

answered 21 mins ago

lubgr

6,59021340

answered 21 mins ago

lubgr

6,59021340

add a commentÂ |Â

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