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
    Apr 5 at 12:14











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

    – Violet Giraffe
    Apr 5 at 12:17






  • 2





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

    – vahancho
    Apr 5 at 12:17






  • 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
    Apr 5 at 12:24








  • 1





    I have never seen "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++." actually survive prolonged contact with the future. Some future programmer, perhaps even you, will curse you for failing to see their patently obvious need for another subclass.

    – Eric Towers
    Apr 5 at 17:26


















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
    Apr 5 at 12:14











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

    – Violet Giraffe
    Apr 5 at 12:17






  • 2





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

    – vahancho
    Apr 5 at 12:17






  • 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
    Apr 5 at 12:24








  • 1





    I have never seen "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++." actually survive prolonged contact with the future. Some future programmer, perhaps even you, will curse you for failing to see their patently obvious need for another subclass.

    – Eric Towers
    Apr 5 at 17:26














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 Apr 5 at 12:29







Violet Giraffe

















asked Apr 5 at 12:12









Violet GiraffeViolet Giraffe

15.1k29139256




15.1k29139256













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

    – Diodacus
    Apr 5 at 12:14











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

    – Violet Giraffe
    Apr 5 at 12:17






  • 2





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

    – vahancho
    Apr 5 at 12:17






  • 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
    Apr 5 at 12:24








  • 1





    I have never seen "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++." actually survive prolonged contact with the future. Some future programmer, perhaps even you, will curse you for failing to see their patently obvious need for another subclass.

    – Eric Towers
    Apr 5 at 17:26



















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

    – Diodacus
    Apr 5 at 12:14











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

    – Violet Giraffe
    Apr 5 at 12:17






  • 2





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

    – vahancho
    Apr 5 at 12:17






  • 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
    Apr 5 at 12:24








  • 1





    I have never seen "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++." actually survive prolonged contact with the future. Some future programmer, perhaps even you, will curse you for failing to see their patently obvious need for another subclass.

    – Eric Towers
    Apr 5 at 17:26

















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

– Diodacus
Apr 5 at 12:14





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

– Diodacus
Apr 5 at 12:14













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

– Violet Giraffe
Apr 5 at 12:17





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

– Violet Giraffe
Apr 5 at 12:17




2




2





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

– vahancho
Apr 5 at 12:17





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

– vahancho
Apr 5 at 12:17




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
Apr 5 at 12:24







@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
Apr 5 at 12:24






1




1





I have never seen "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++." actually survive prolonged contact with the future. Some future programmer, perhaps even you, will curse you for failing to see their patently obvious need for another subclass.

– Eric Towers
Apr 5 at 17:26





I have never seen "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++." actually survive prolonged contact with the future. Some future programmer, perhaps even you, will curse you for failing to see their patently obvious need for another subclass.

– Eric Towers
Apr 5 at 17:26












2 Answers
2






active

oldest

votes


















24














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
    Apr 5 at 12:44






  • 2





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

    – NathanOliver
    Apr 5 at 12:45






  • 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
    Apr 5 at 12:46






  • 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
    Apr 5 at 12:49






  • 2





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

    – NathanOliver
    Apr 5 at 13:15



















-5














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
    Apr 5 at 12:19











  • does this answer the question?

    – sp2danny
    Apr 5 at 12:21











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

    – VTT
    Apr 5 at 12:23












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2 Answers
2






active

oldest

votes








2 Answers
2






active

oldest

votes









active

oldest

votes






active

oldest

votes









24














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
    Apr 5 at 12:44






  • 2





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

    – NathanOliver
    Apr 5 at 12:45






  • 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
    Apr 5 at 12:46






  • 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
    Apr 5 at 12:49






  • 2





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

    – NathanOliver
    Apr 5 at 13:15
















24














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
    Apr 5 at 12:44






  • 2





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

    – NathanOliver
    Apr 5 at 12:45






  • 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
    Apr 5 at 12:46






  • 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
    Apr 5 at 12:49






  • 2





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

    – NathanOliver
    Apr 5 at 13:15














24












24








24







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 Apr 5 at 14:50

























answered Apr 5 at 12:40









NathanOliverNathanOliver

99.3k16138219




99.3k16138219








  • 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
    Apr 5 at 12:44






  • 2





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

    – NathanOliver
    Apr 5 at 12:45






  • 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
    Apr 5 at 12:46






  • 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
    Apr 5 at 12:49






  • 2





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

    – NathanOliver
    Apr 5 at 13:15














  • 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
    Apr 5 at 12:44






  • 2





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

    – NathanOliver
    Apr 5 at 12:45






  • 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
    Apr 5 at 12:46






  • 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
    Apr 5 at 12:49






  • 2





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

    – NathanOliver
    Apr 5 at 13:15








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
Apr 5 at 12:44





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
Apr 5 at 12:44




2




2





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

– NathanOliver
Apr 5 at 12:45





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

– NathanOliver
Apr 5 at 12:45




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
Apr 5 at 12:46





@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
Apr 5 at 12:46




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
Apr 5 at 12:49





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

– NathanOliver
Apr 5 at 12:49




2




2





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

– NathanOliver
Apr 5 at 13:15





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

– NathanOliver
Apr 5 at 13:15













-5














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
    Apr 5 at 12:19











  • does this answer the question?

    – sp2danny
    Apr 5 at 12:21











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

    – VTT
    Apr 5 at 12:23
















-5














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
    Apr 5 at 12:19











  • does this answer the question?

    – sp2danny
    Apr 5 at 12:21











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

    – VTT
    Apr 5 at 12:23














-5












-5








-5







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 Apr 5 at 12:27









TrebledJ

4,00921431




4,00921431










answered Apr 5 at 12:16









DiodacusDiodacus

2687




2687








  • 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
    Apr 5 at 12:19











  • does this answer the question?

    – sp2danny
    Apr 5 at 12:21











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

    – VTT
    Apr 5 at 12:23














  • 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
    Apr 5 at 12:19











  • does this answer the question?

    – sp2danny
    Apr 5 at 12:21











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

    – VTT
    Apr 5 at 12:23








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
Apr 5 at 12:19





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
Apr 5 at 12:19













does this answer the question?

– sp2danny
Apr 5 at 12:21





does this answer the question?

– sp2danny
Apr 5 at 12:21













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

– VTT
Apr 5 at 12:23





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

– VTT
Apr 5 at 12:23


















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