Why doesn't a class having private constructor prevent inheriting from this class? How to control which...





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24















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.










share|improve this question

























  • See this: stackoverflow.com/questions/32235294/…

    – Diodacus
    18 hours ago











  • @Diodacus: so what, declaring a private constructor as default makes it public, despite being declared in the private: section?

    – Violet Giraffe
    18 hours ago






  • 2





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

    – vahancho
    18 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
    18 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
    18 hours ago




















24















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.










share|improve this question

























  • See this: stackoverflow.com/questions/32235294/…

    – Diodacus
    18 hours ago











  • @Diodacus: so what, declaring a private constructor as default makes it public, despite being declared in the private: section?

    – Violet Giraffe
    18 hours ago






  • 2





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

    – vahancho
    18 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
    18 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
    18 hours ago
















24












24








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.










share|improve this question
















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















share|improve this question













share|improve this question




share|improve this question








edited 17 hours ago







Violet Giraffe

















asked 18 hours ago









Violet GiraffeViolet Giraffe

15k29139256




15k29139256













  • See this: stackoverflow.com/questions/32235294/…

    – Diodacus
    18 hours ago











  • @Diodacus: so what, declaring a private constructor as default makes it public, despite being declared in the private: section?

    – Violet Giraffe
    18 hours ago






  • 2





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

    – vahancho
    18 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
    18 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
    18 hours ago





















  • See this: stackoverflow.com/questions/32235294/…

    – Diodacus
    18 hours ago











  • @Diodacus: so what, declaring a private constructor as default makes it public, despite being declared in the private: section?

    – Violet Giraffe
    18 hours ago






  • 2





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

    – vahancho
    18 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
    18 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
    18 hours ago



















See this: stackoverflow.com/questions/32235294/…

– Diodacus
18 hours ago





See this: stackoverflow.com/questions/32235294/…

– Diodacus
18 hours ago













@Diodacus: so what, declaring a private constructor as default makes it public, despite being declared in the private: section?

– Violet Giraffe
18 hours ago





@Diodacus: so what, declaring a private constructor as default makes it public, despite being declared in the private: section?

– Violet Giraffe
18 hours ago




2




2





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

– vahancho
18 hours ago





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

– vahancho
18 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
18 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
18 hours ago




1




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
18 hours ago







@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
18 hours ago














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





















  • 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
    17 hours ago






  • 2





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

    – NathanOliver
    17 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
    17 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
    17 hours ago






  • 2





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

    – NathanOliver
    17 hours ago



















-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





















  • 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
    18 hours ago











  • does this answer the question?

    – sp2danny
    18 hours ago











  • So why B() = default; is threated as "no user-declared constructor"?

    – VTT
    18 hours ago












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2 Answers
2






active

oldest

votes








2 Answers
2






active

oldest

votes









active

oldest

votes






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





















  • 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
    17 hours ago






  • 2





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

    – NathanOliver
    17 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
    17 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
    17 hours ago






  • 2





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

    – NathanOliver
    17 hours ago
















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





















  • 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
    17 hours ago






  • 2





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

    – NathanOliver
    17 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
    17 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
    17 hours ago






  • 2





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

    – NathanOliver
    17 hours ago














21












21








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















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














share|improve this answer



share|improve this answer








edited 15 hours ago

























answered 17 hours ago









NathanOliverNathanOliver

98.2k16138217




98.2k16138217








  • 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
    17 hours ago






  • 2





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

    – NathanOliver
    17 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
    17 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
    17 hours ago






  • 2





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

    – NathanOliver
    17 hours ago














  • 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
    17 hours ago






  • 2





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

    – NathanOliver
    17 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
    17 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
    17 hours ago






  • 2





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

    – NathanOliver
    17 hours ago








1




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
17 hours ago





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
17 hours ago




2




2





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

– NathanOliver
17 hours ago





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

– NathanOliver
17 hours ago




1




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
17 hours ago





@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
17 hours ago




1




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
17 hours ago





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

– NathanOliver
17 hours ago




2




2





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

– NathanOliver
17 hours ago





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

– NathanOliver
17 hours ago













-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





















  • 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
    18 hours ago











  • does this answer the question?

    – sp2danny
    18 hours ago











  • So why B() = default; is threated as "no user-declared constructor"?

    – VTT
    18 hours ago
















-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





















  • 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
    18 hours ago











  • does this answer the question?

    – sp2danny
    18 hours ago











  • So why B() = default; is threated as "no user-declared constructor"?

    – VTT
    18 hours ago














-4












-4








-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















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














share|improve this answer



share|improve this answer








edited 17 hours ago









TrebledJ

3,62021228




3,62021228










answered 18 hours ago









DiodacusDiodacus

1966




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
    18 hours ago











  • does this answer the question?

    – sp2danny
    18 hours ago











  • So why B() = default; is threated as "no user-declared constructor"?

    – VTT
    18 hours ago














  • 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
    18 hours ago











  • does this answer the question?

    – sp2danny
    18 hours ago











  • So why B() = default; is threated as "no user-declared constructor"?

    – VTT
    18 hours ago








1




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
18 hours ago





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
18 hours ago













does this answer the question?

– sp2danny
18 hours ago





does this answer the question?

– sp2danny
18 hours ago













So why B() = default; is threated as "no user-declared constructor"?

– VTT
18 hours ago





So why B() = default; is threated as "no user-declared constructor"?

– VTT
18 hours ago


















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