Why doesn't a class having private constructor prevent inheriting from this class? How to control which...
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}
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
|
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
See this: stackoverflow.com/questions/32235294/…
– Diodacus
2 days ago
@Diodacus: so what, declaring a private constructor as default makes it public, despite being declared in theprivate:
section?
– Violet Giraffe
2 days ago
2
I got the error you expect: "'D::D(void)': attempting to reference a deleted function" (msvs 2017)
– vahancho
2 days 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
2 days ago
1
@Stefan: I hear you, but there are exactly two classes for which it is semantically meaningful to subclassB
, and I'm trying to express/enforce this logical constraint in C++.
– Violet Giraffe
2 days 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
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
c++ c++11 inheritance c++17
edited 2 days ago
Violet Giraffe
asked 2 days ago
Violet GiraffeViolet Giraffe
15k29139256
15k29139256
See this: stackoverflow.com/questions/32235294/…
– Diodacus
2 days ago
@Diodacus: so what, declaring a private constructor as default makes it public, despite being declared in theprivate:
section?
– Violet Giraffe
2 days ago
2
I got the error you expect: "'D::D(void)': attempting to reference a deleted function" (msvs 2017)
– vahancho
2 days 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
2 days ago
1
@Stefan: I hear you, but there are exactly two classes for which it is semantically meaningful to subclassB
, and I'm trying to express/enforce this logical constraint in C++.
– Violet Giraffe
2 days ago
|
show 16 more comments
See this: stackoverflow.com/questions/32235294/…
– Diodacus
2 days ago
@Diodacus: so what, declaring a private constructor as default makes it public, despite being declared in theprivate:
section?
– Violet Giraffe
2 days ago
2
I got the error you expect: "'D::D(void)': attempting to reference a deleted function" (msvs 2017)
– vahancho
2 days 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
2 days ago
1
@Stefan: I hear you, but there are exactly two classes for which it is semantically meaningful to subclassB
, and I'm trying to express/enforce this logical constraint in C++.
– Violet Giraffe
2 days ago
See this: stackoverflow.com/questions/32235294/…
– Diodacus
2 days ago
See this: stackoverflow.com/questions/32235294/…
– Diodacus
2 days ago
@Diodacus: so what, declaring a private constructor as default makes it public, despite being declared in the
private:
section?– Violet Giraffe
2 days ago
@Diodacus: so what, declaring a private constructor as default makes it public, despite being declared in the
private:
section?– Violet Giraffe
2 days ago
2
2
I got the error you expect: "'D::D(void)': attempting to reference a deleted function" (msvs 2017)
– vahancho
2 days ago
I got the error you expect: "'D::D(void)': attempting to reference a deleted function" (msvs 2017)
– vahancho
2 days 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
2 days 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
2 days 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
2 days 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
2 days ago
|
show 16 more comments
2 Answers
2
active
oldest
votes
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.
1
I guess writing defaulted constructor definition out of class likeB::B() = default;
will be considered a user-provided constructor, while defaulting it in class is considered a not-user-provided constructor?
– VTT
2 days ago
2
@VTT That makes no difference.B() = default
inside the class is still a user declared constructor.
– NathanOliver
2 days 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
2 days 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
2 days ago
2
@VioletGiraffeD d2();
is the most vexing parse so you have a function, not an object.
– NathanOliver
2 days ago
|
show 14 more comments
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.
1
I think you misunderstood the point of my confusion, although your link is still relevant. I know eachC
andD
has a default public constructor, butD
is not supposed to be able to instantiate the instance of its base classB
because of the latter's constructor being private.
– Violet Giraffe
2 days ago
does this answer the question?
– sp2danny
2 days ago
So whyB() = default;
is threated as "no user-declared constructor"?
– VTT
2 days ago
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
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.
1
I guess writing defaulted constructor definition out of class likeB::B() = default;
will be considered a user-provided constructor, while defaulting it in class is considered a not-user-provided constructor?
– VTT
2 days ago
2
@VTT That makes no difference.B() = default
inside the class is still a user declared constructor.
– NathanOliver
2 days 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
2 days 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
2 days ago
2
@VioletGiraffeD d2();
is the most vexing parse so you have a function, not an object.
– NathanOliver
2 days 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.
1
I guess writing defaulted constructor definition out of class likeB::B() = default;
will be considered a user-provided constructor, while defaulting it in class is considered a not-user-provided constructor?
– VTT
2 days ago
2
@VTT That makes no difference.B() = default
inside the class is still a user declared constructor.
– NathanOliver
2 days 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
2 days 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
2 days ago
2
@VioletGiraffeD d2();
is the most vexing parse so you have a function, not an object.
– NathanOliver
2 days 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.
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.
edited 2 days ago
answered 2 days ago
NathanOliverNathanOliver
98.2k16138217
98.2k16138217
1
I guess writing defaulted constructor definition out of class likeB::B() = default;
will be considered a user-provided constructor, while defaulting it in class is considered a not-user-provided constructor?
– VTT
2 days ago
2
@VTT That makes no difference.B() = default
inside the class is still a user declared constructor.
– NathanOliver
2 days 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
2 days 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
2 days ago
2
@VioletGiraffeD d2();
is the most vexing parse so you have a function, not an object.
– NathanOliver
2 days ago
|
show 14 more comments
1
I guess writing defaulted constructor definition out of class likeB::B() = default;
will be considered a user-provided constructor, while defaulting it in class is considered a not-user-provided constructor?
– VTT
2 days ago
2
@VTT That makes no difference.B() = default
inside the class is still a user declared constructor.
– NathanOliver
2 days 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
2 days 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
2 days ago
2
@VioletGiraffeD d2();
is the most vexing parse so you have a function, not an object.
– NathanOliver
2 days 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
2 days 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
2 days ago
2
2
@VTT That makes no difference.
B() = default
inside the class is still a user declared constructor.– NathanOliver
2 days ago
@VTT That makes no difference.
B() = default
inside the class is still a user declared constructor.– NathanOliver
2 days 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
2 days 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
2 days 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
2 days 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
2 days ago
2
2
@VioletGiraffe
D d2();
is the most vexing parse so you have a function, not an object.– NathanOliver
2 days ago
@VioletGiraffe
D d2();
is the most vexing parse so you have a function, not an object.– NathanOliver
2 days ago
|
show 14 more comments
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.
1
I think you misunderstood the point of my confusion, although your link is still relevant. I know eachC
andD
has a default public constructor, butD
is not supposed to be able to instantiate the instance of its base classB
because of the latter's constructor being private.
– Violet Giraffe
2 days ago
does this answer the question?
– sp2danny
2 days ago
So whyB() = default;
is threated as "no user-declared constructor"?
– VTT
2 days ago
add a comment |
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.
1
I think you misunderstood the point of my confusion, although your link is still relevant. I know eachC
andD
has a default public constructor, butD
is not supposed to be able to instantiate the instance of its base classB
because of the latter's constructor being private.
– Violet Giraffe
2 days ago
does this answer the question?
– sp2danny
2 days ago
So whyB() = default;
is threated as "no user-declared constructor"?
– VTT
2 days ago
add a comment |
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.
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.
edited 2 days ago
TrebledJ
3,62921328
3,62921328
answered 2 days ago
DiodacusDiodacus
1946
1946
1
I think you misunderstood the point of my confusion, although your link is still relevant. I know eachC
andD
has a default public constructor, butD
is not supposed to be able to instantiate the instance of its base classB
because of the latter's constructor being private.
– Violet Giraffe
2 days ago
does this answer the question?
– sp2danny
2 days ago
So whyB() = default;
is threated as "no user-declared constructor"?
– VTT
2 days ago
add a comment |
1
I think you misunderstood the point of my confusion, although your link is still relevant. I know eachC
andD
has a default public constructor, butD
is not supposed to be able to instantiate the instance of its base classB
because of the latter's constructor being private.
– Violet Giraffe
2 days ago
does this answer the question?
– sp2danny
2 days ago
So whyB() = default;
is threated as "no user-declared constructor"?
– VTT
2 days 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
2 days 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
2 days ago
does this answer the question?
– sp2danny
2 days ago
does this answer the question?
– sp2danny
2 days ago
So why
B() = default;
is threated as "no user-declared constructor"?– VTT
2 days ago
So why
B() = default;
is threated as "no user-declared constructor"?– VTT
2 days ago
add a comment |
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See this: stackoverflow.com/questions/32235294/…
– Diodacus
2 days ago
@Diodacus: so what, declaring a private constructor as default makes it public, despite being declared in the
private:
section?– Violet Giraffe
2 days ago
2
I got the error you expect: "'D::D(void)': attempting to reference a deleted function" (msvs 2017)
– vahancho
2 days 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
2 days 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
2 days ago