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Why doesn't a class having private constructor prevent inheriting from this class? How to control which classes can inherit from a certain base?

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}

24

1

class B {

private:

    friend class C;

    B() = default;

};



class C : public B {};

class D : public B {};



int main() {

    C {};

    D {};

    return 0;

}

I assumed that since only class C is a friend of B, and B's constructor is private, then only class C is valid and D is not allowed to instantiate B. But that's not how it works. Where am I wrong with my reasoning, and how to achieve this kind of control over which classes are allowed to subclass a certain base?

Update: as pointed out by others in the comments, the snippet above works as I initially expected under C++14, but not C++17. Changing the instantiation to C c; D d; in main() does work as expected in C++17 mode as well.

c++ c++11 inheritance c++17

share|improve this question

edited 13 hours ago

Violet Giraffe

asked 13 hours ago

Violet GiraffeViolet Giraffe

15k29139256

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  See this: stackoverflow.com/questions/32235294/…
  
  – Diodacus
  13 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  @Diodacus: so what, declaring a private constructor as default makes it public, despite being declared in the private: section?
  
  – Violet Giraffe
  13 hours ago
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  
  I got the error you expect: "'D::D(void)': attempting to reference a deleted function" (msvs 2017)
  
  – vahancho
  13 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  how to achieve this kind of control over which classes are allowed to sublcass a certain base; I understand your wish, but can you explain why you would want this? Because it means that whenever you want to make an extra class, through inheritance, you need to alter the base class and this might be considered as anti-pattern
  
  – Stefan
  13 hours ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  @Stefan: I hear you, but there are exactly two classes for which it is semantically meaningful to subclass B, and I'm trying to express/enforce this logical constraint in C++.
  
  – Violet Giraffe
  13 hours ago
  
  
  

  
  
  
  

 | 
show 16 more comments

24

1

class B {

private:

    friend class C;

    B() = default;

};



class C : public B {};

class D : public B {};



int main() {

    C {};

    D {};

    return 0;

}

I assumed that since only class C is a friend of B, and B's constructor is private, then only class C is valid and D is not allowed to instantiate B. But that's not how it works. Where am I wrong with my reasoning, and how to achieve this kind of control over which classes are allowed to subclass a certain base?

Update: as pointed out by others in the comments, the snippet above works as I initially expected under C++14, but not C++17. Changing the instantiation to C c; D d; in main() does work as expected in C++17 mode as well.

c++ c++11 inheritance c++17

share|improve this question

edited 13 hours ago

Violet Giraffe

asked 13 hours ago

Violet GiraffeViolet Giraffe

15k29139256

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  See this: stackoverflow.com/questions/32235294/…
  
  – Diodacus
  13 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  @Diodacus: so what, declaring a private constructor as default makes it public, despite being declared in the private: section?
  
  – Violet Giraffe
  13 hours ago
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  
  I got the error you expect: "'D::D(void)': attempting to reference a deleted function" (msvs 2017)
  
  – vahancho
  13 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  how to achieve this kind of control over which classes are allowed to sublcass a certain base; I understand your wish, but can you explain why you would want this? Because it means that whenever you want to make an extra class, through inheritance, you need to alter the base class and this might be considered as anti-pattern
  
  – Stefan
  13 hours ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  @Stefan: I hear you, but there are exactly two classes for which it is semantically meaningful to subclass B, and I'm trying to express/enforce this logical constraint in C++.
  
  – Violet Giraffe
  13 hours ago
  
  
  

  
  
  
  

 | 
show 16 more comments

class B {

private:

    friend class C;

    B() = default;

};



class C : public B {};

class D : public B {};



int main() {

    C {};

    D {};

    return 0;

}

I assumed that since only class C is a friend of B, and B's constructor is private, then only class C is valid and D is not allowed to instantiate B. But that's not how it works. Where am I wrong with my reasoning, and how to achieve this kind of control over which classes are allowed to subclass a certain base?

Update: as pointed out by others in the comments, the snippet above works as I initially expected under C++14, but not C++17. Changing the instantiation to C c; D d; in main() does work as expected in C++17 mode as well.

c++ c++11 inheritance c++17

share|improve this question

edited 13 hours ago

Violet Giraffe

asked 13 hours ago

Violet GiraffeViolet Giraffe

15k29139256

class B {

private:

    friend class C;

    B() = default;

};



class C : public B {};

class D : public B {};



int main() {

    C {};

    D {};

    return 0;

}

share|improve this question

edited 13 hours ago

Violet Giraffe

asked 13 hours ago

Violet GiraffeViolet Giraffe

15k29139256

share|improve this question

edited 13 hours ago

Violet Giraffe

asked 13 hours ago

Violet GiraffeViolet Giraffe

15k29139256

asked 13 hours ago

Violet GiraffeViolet Giraffe

15k29139256

asked 13 hours ago

Violet GiraffeViolet Giraffe

15k29139256

2 Answers
2

active

oldest

votes

21

This is a new feature added to C++17. What is going on is C is now considered an aggregate. Since it is an aggregate, it doesn't need a constructor. If we look at [dcl.init.aggr]/1 we get that an aggregate is

An aggregate is an array or a class with

• no user-provided, explicit, or inherited constructors ([class.ctor]),




• no private or protected non-static data members (Clause [class.access]),




• no virtual functions, and




• no virtual, private, or protected base classes ([class.mi]).

[ Note: Aggregate initialization does not allow accessing protected and private base class' members or constructors.  — end note ]

And we check of all those bullet points. You don't have any constructors declared in C or D so there is bullet 1. You don't have any data members so the second bullet doesn't matter, and your base class is public so the third bullet is satisfied.

The change that happened between C++11/14 and C++17 that allows this is that aggregates can now have base classes. You can see the old wording here where it expressly stated that bases classes are not allowed.

We can confirm this by checking the trait std::is_aggregate_v like

int main()

{

    std::cout << std::is_aggregate_v<C>;

}

which will print 1.

---

Do note that since C is a friend of B you can use

C c{};

C c1;

C c2 = C();

As valid ways to initialize a C. Since D is not a friend of B the only one that works is D d{}; as that is aggregate initialization. All of the other forms try to default initialize and that can't be done since D has a deleted default constructor.

share|improve this answer

edited 10 hours ago

answered 12 hours ago

NathanOliverNathanOliver

98.2k16138216

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  I guess writing defaulted constructor definition out of class like B::B() = default; will be considered a user-provided constructor, while defaulting it in class is considered a not-user-provided constructor?
  
  – VTT
  12 hours ago
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  
  @VTT That makes no difference. B() = default inside the class is still a user declared constructor.
  
  – NathanOliver
  12 hours ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  @NathanOliver you sure? I'm quite certain that during one of the many talks in cppcon one of the lecturers explained the difference, although it may be applied elsewhere, not in this very example. Can't find the link tho.
  
  – Fureeish
  12 hours ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  @Fureeish 100% sure. What moving the constructor out of the class does change is how the object is initialized: stackoverflow.com/questions/54350114/…
  
  – NathanOliver
  12 hours ago
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  
  @VioletGiraffe D d2(); is the most vexing parse so you have a function, not an object.
  
  – NathanOliver
  12 hours ago
  

  
  
  
  

 | 
show 14 more comments

-4

From What is the default access of constructor in c++:

If there is no user-declared constructor for class X, a constructor having no parameters is implicitly declared as defaulted. An implicitly-declared default constructor is an inline public member of its class.

If the class definition does not explicitly declare a copy constructor, one is declared implicitly. [...] An implicitly-declared copy/move constructor is an inline public member of its class.

Constructors for classes C and D are generated internally by compiler.

BTW.: If you want to play with inheritance, please make sure you have virtual destructor defined.

share|improve this answer

edited 13 hours ago

TrebledJ

3,60521228

answered 13 hours ago

DiodacusDiodacus

1966

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  I think you misunderstood the point of my confusion, although your link is still relevant. I know each C and D has a default public constructor, but D is not supposed to be able to instantiate the instance of its base class B because of the latter's constructor being private.
  
  – Violet Giraffe
  13 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  does this answer the question?
  
  – sp2danny
  13 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  So why B() = default; is threated as "no user-declared constructor"?
  
  – VTT
  13 hours ago
  

  
  
  
  

add a comment | 

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21

This is a new feature added to C++17. What is going on is C is now considered an aggregate. Since it is an aggregate, it doesn't need a constructor. If we look at [dcl.init.aggr]/1 we get that an aggregate is

An aggregate is an array or a class with

• no user-provided, explicit, or inherited constructors ([class.ctor]),




• no private or protected non-static data members (Clause [class.access]),




• no virtual functions, and




• no virtual, private, or protected base classes ([class.mi]).

[ Note: Aggregate initialization does not allow accessing protected and private base class' members or constructors.  — end note ]

And we check of all those bullet points. You don't have any constructors declared in C or D so there is bullet 1. You don't have any data members so the second bullet doesn't matter, and your base class is public so the third bullet is satisfied.

The change that happened between C++11/14 and C++17 that allows this is that aggregates can now have base classes. You can see the old wording here where it expressly stated that bases classes are not allowed.

We can confirm this by checking the trait std::is_aggregate_v like

int main()

{

    std::cout << std::is_aggregate_v<C>;

}

which will print 1.

---

Do note that since C is a friend of B you can use

C c{};

C c1;

C c2 = C();

As valid ways to initialize a C. Since D is not a friend of B the only one that works is D d{}; as that is aggregate initialization. All of the other forms try to default initialize and that can't be done since D has a deleted default constructor.

share|improve this answer

edited 10 hours ago

answered 12 hours ago

NathanOliverNathanOliver

98.2k16138216

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  I guess writing defaulted constructor definition out of class like B::B() = default; will be considered a user-provided constructor, while defaulting it in class is considered a not-user-provided constructor?
  
  – VTT
  12 hours ago
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  
  @VTT That makes no difference. B() = default inside the class is still a user declared constructor.
  
  – NathanOliver
  12 hours ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  @NathanOliver you sure? I'm quite certain that during one of the many talks in cppcon one of the lecturers explained the difference, although it may be applied elsewhere, not in this very example. Can't find the link tho.
  
  – Fureeish
  12 hours ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  @Fureeish 100% sure. What moving the constructor out of the class does change is how the object is initialized: stackoverflow.com/questions/54350114/…
  
  – NathanOliver
  12 hours ago
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  
  @VioletGiraffe D d2(); is the most vexing parse so you have a function, not an object.
  
  – NathanOliver
  12 hours ago
  

  
  
  
  

 | 
show 14 more comments

21

This is a new feature added to C++17. What is going on is C is now considered an aggregate. Since it is an aggregate, it doesn't need a constructor. If we look at [dcl.init.aggr]/1 we get that an aggregate is

An aggregate is an array or a class with

• no user-provided, explicit, or inherited constructors ([class.ctor]),




• no private or protected non-static data members (Clause [class.access]),




• no virtual functions, and




• no virtual, private, or protected base classes ([class.mi]).

[ Note: Aggregate initialization does not allow accessing protected and private base class' members or constructors.  — end note ]

And we check of all those bullet points. You don't have any constructors declared in C or D so there is bullet 1. You don't have any data members so the second bullet doesn't matter, and your base class is public so the third bullet is satisfied.

The change that happened between C++11/14 and C++17 that allows this is that aggregates can now have base classes. You can see the old wording here where it expressly stated that bases classes are not allowed.

We can confirm this by checking the trait std::is_aggregate_v like

int main()

{

    std::cout << std::is_aggregate_v<C>;

}

which will print 1.

---

Do note that since C is a friend of B you can use

C c{};

C c1;

C c2 = C();

As valid ways to initialize a C. Since D is not a friend of B the only one that works is D d{}; as that is aggregate initialization. All of the other forms try to default initialize and that can't be done since D has a deleted default constructor.

share|improve this answer

edited 10 hours ago

answered 12 hours ago

NathanOliverNathanOliver

98.2k16138216

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  I guess writing defaulted constructor definition out of class like B::B() = default; will be considered a user-provided constructor, while defaulting it in class is considered a not-user-provided constructor?
  
  – VTT
  12 hours ago
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  
  @VTT That makes no difference. B() = default inside the class is still a user declared constructor.
  
  – NathanOliver
  12 hours ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  @NathanOliver you sure? I'm quite certain that during one of the many talks in cppcon one of the lecturers explained the difference, although it may be applied elsewhere, not in this very example. Can't find the link tho.
  
  – Fureeish
  12 hours ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  @Fureeish 100% sure. What moving the constructor out of the class does change is how the object is initialized: stackoverflow.com/questions/54350114/…
  
  – NathanOliver
  12 hours ago
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  
  @VioletGiraffe D d2(); is the most vexing parse so you have a function, not an object.
  
  – NathanOliver
  12 hours ago
  

  
  
  
  

 | 
show 14 more comments

This is a new feature added to C++17. What is going on is C is now considered an aggregate. Since it is an aggregate, it doesn't need a constructor. If we look at [dcl.init.aggr]/1 we get that an aggregate is

An aggregate is an array or a class with

• no user-provided, explicit, or inherited constructors ([class.ctor]),




• no private or protected non-static data members (Clause [class.access]),




• no virtual functions, and




• no virtual, private, or protected base classes ([class.mi]).

[ Note: Aggregate initialization does not allow accessing protected and private base class' members or constructors.  — end note ]

And we check of all those bullet points. You don't have any constructors declared in C or D so there is bullet 1. You don't have any data members so the second bullet doesn't matter, and your base class is public so the third bullet is satisfied.

The change that happened between C++11/14 and C++17 that allows this is that aggregates can now have base classes. You can see the old wording here where it expressly stated that bases classes are not allowed.

We can confirm this by checking the trait std::is_aggregate_v like

int main()

{

    std::cout << std::is_aggregate_v<C>;

}

which will print 1.

---

Do note that since C is a friend of B you can use

C c{};

C c1;

C c2 = C();

As valid ways to initialize a C. Since D is not a friend of B the only one that works is D d{}; as that is aggregate initialization. All of the other forms try to default initialize and that can't be done since D has a deleted default constructor.

share|improve this answer

edited 10 hours ago

answered 12 hours ago

NathanOliverNathanOliver

98.2k16138216

int main()

{

    std::cout << std::is_aggregate_v<C>;

}

C c{};

C c1;

C c2 = C();

share|improve this answer

edited 10 hours ago

answered 12 hours ago

NathanOliverNathanOliver

98.2k16138216

answered 12 hours ago

NathanOliverNathanOliver

98.2k16138216

answered 12 hours ago

NathanOliverNathanOliver

98.2k16138216

-4

From What is the default access of constructor in c++:

If there is no user-declared constructor for class X, a constructor having no parameters is implicitly declared as defaulted. An implicitly-declared default constructor is an inline public member of its class.

If the class definition does not explicitly declare a copy constructor, one is declared implicitly. [...] An implicitly-declared copy/move constructor is an inline public member of its class.

Constructors for classes C and D are generated internally by compiler.

BTW.: If you want to play with inheritance, please make sure you have virtual destructor defined.

share|improve this answer

edited 13 hours ago

TrebledJ

3,60521228

answered 13 hours ago

DiodacusDiodacus

1966

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  I think you misunderstood the point of my confusion, although your link is still relevant. I know each C and D has a default public constructor, but D is not supposed to be able to instantiate the instance of its base class B because of the latter's constructor being private.
  
  – Violet Giraffe
  13 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  does this answer the question?
  
  – sp2danny
  13 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  So why B() = default; is threated as "no user-declared constructor"?
  
  – VTT
  13 hours ago
  

  
  
  
  

add a comment | 

-4

From What is the default access of constructor in c++:

If there is no user-declared constructor for class X, a constructor having no parameters is implicitly declared as defaulted. An implicitly-declared default constructor is an inline public member of its class.

If the class definition does not explicitly declare a copy constructor, one is declared implicitly. [...] An implicitly-declared copy/move constructor is an inline public member of its class.

Constructors for classes C and D are generated internally by compiler.

BTW.: If you want to play with inheritance, please make sure you have virtual destructor defined.

share|improve this answer

edited 13 hours ago

TrebledJ

3,60521228

answered 13 hours ago

DiodacusDiodacus

1966

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  I think you misunderstood the point of my confusion, although your link is still relevant. I know each C and D has a default public constructor, but D is not supposed to be able to instantiate the instance of its base class B because of the latter's constructor being private.
  
  – Violet Giraffe
  13 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  does this answer the question?
  
  – sp2danny
  13 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  So why B() = default; is threated as "no user-declared constructor"?
  
  – VTT
  13 hours ago
  

  
  
  
  

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From What is the default access of constructor in c++:

If there is no user-declared constructor for class X, a constructor having no parameters is implicitly declared as defaulted. An implicitly-declared default constructor is an inline public member of its class.

If the class definition does not explicitly declare a copy constructor, one is declared implicitly. [...] An implicitly-declared copy/move constructor is an inline public member of its class.

Constructors for classes C and D are generated internally by compiler.

BTW.: If you want to play with inheritance, please make sure you have virtual destructor defined.

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edited 13 hours ago

TrebledJ

3,60521228

answered 13 hours ago

DiodacusDiodacus

1966

share|improve this answer

edited 13 hours ago

TrebledJ

3,60521228

answered 13 hours ago

DiodacusDiodacus

1966

edited 13 hours ago

TrebledJ

3,60521228

edited 13 hours ago

TrebledJ

3,60521228

answered 13 hours ago

DiodacusDiodacus

1966

answered 13 hours ago

DiodacusDiodacus

1966