Kdtyjb

Why does Kotter return in Welcome Back Kotter?

Is it legal for company to use my work email to pretend I still work there?

Are astronomers waiting to see something in an image from a gravitational lens that they've already seen in an adjacent image?

Do I have a twin with permutated remainders?

Why doesn't Newton's third law mean a person bounces back to where they started when they hit the ground?

Is it unprofessional to ask if a job posting on GlassDoor is real?

meaning of に in 本当に?

Was any UN Security Council vote triple-vetoed?

What would happen to a modern skyscraper if it rains micro blackholes?

What are the disadvantages of having a left skewed distribution?

Can I ask the recruiters in my resume to put the reason why I am rejected?

What's that red-plus icon near a text?

Why is 150k or 200k jobs considered good when there's 300k+ births a month?

Horror movie about a virus at the prom; beginning and end are stylized as a cartoon

Modeling an IP Address

How old can references or sources in a thesis be?

How do I deal with an unproductive colleague in a small company?

Did Shadowfax go to Valinor?

Codimension of non-flat locus

Two films in a tank, only one comes out with a development error – why?

How does quantile regression compare to logistic regression with the variable split at the quantile?

When a company launches a new product do they "come out" with a new product or do they "come up" with a new product?

Convert two switches to a dual stack, and add outlet - possible here?

Languages that we cannot (dis)prove to be Context-Free

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

C++ zero initialization - Why is `b` in this program uninitialized, but `a` is initialized?Deleted default constructor. Objects can still be created… sometimesWhat is the default access of constructor in c++Extending a singleton class which has private constructor and destructor gives compile time warningHow to Restrict creation of objects from certain classCan a friend class invoke a private constructor in c++? (and what's Singleton)how can a base class disable the derived class's constructorCan you force classes inheriting from abstract base class to only have the public methods defined in base case?Inherit from class with private initializer?How to enforce private constructors in children of a base classConditional access to base class from inherited class in c++How to unit test a class with private constructor?Why does C++ forbid private inheritance of a final class?

.everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty{ height:90px;width:728px;box-sizing:border-box;
}

24

1

class B {

private:

    friend class C;

    B() = default;

};



class C : public B {};

class D : public B {};



int main() {

    C {};

    D {};

    return 0;

}

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

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

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

share|improve this question

edited 10 hours ago

Violet Giraffe

asked 11 hours ago

Violet GiraffeViolet Giraffe

15k29139256

• 
  
  
  

  
  

  
  
  
  
  
  

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

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

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

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  
  I got the error you expect: "'D::D(void)': attempting to reference a deleted function" (msvs 2017)
  
  – vahancho
  11 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
  11 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
  10 hours ago
  
  
  

  
  
  
  

 | 
show 16 more comments

24

1

class B {

private:

    friend class C;

    B() = default;

};



class C : public B {};

class D : public B {};



int main() {

    C {};

    D {};

    return 0;

}

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

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

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

share|improve this question

edited 10 hours ago

Violet Giraffe

asked 11 hours ago

Violet GiraffeViolet Giraffe

15k29139256

• 
  
  
  

  
  

  
  
  
  
  
  

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

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

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

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  
  I got the error you expect: "'D::D(void)': attempting to reference a deleted function" (msvs 2017)
  
  – vahancho
  11 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
  11 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
  10 hours ago
  
  
  

  
  
  
  

 | 
show 16 more comments

class B {

private:

    friend class C;

    B() = default;

};



class C : public B {};

class D : public B {};



int main() {

    C {};

    D {};

    return 0;

}

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

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

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

share|improve this question

edited 10 hours ago

Violet Giraffe

asked 11 hours ago

Violet GiraffeViolet Giraffe

15k29139256

class B {

private:

    friend class C;

    B() = default;

};



class C : public B {};

class D : public B {};



int main() {

    C {};

    D {};

    return 0;

}

share|improve this question

edited 10 hours ago

Violet Giraffe

asked 11 hours ago

Violet GiraffeViolet Giraffe

15k29139256

share|improve this question

edited 10 hours ago

Violet Giraffe

asked 11 hours ago

Violet GiraffeViolet Giraffe

15k29139256

asked 11 hours ago

Violet GiraffeViolet Giraffe

15k29139256

asked 11 hours ago

Violet GiraffeViolet Giraffe

15k29139256

2 Answers
2

active

oldest

votes

21

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

An aggregate is an array or a class with

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




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




• no virtual functions, and




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

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

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

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

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

int main()

{

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

}

which will print 1.

---

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

C c{};

C c1;

C c2 = C();

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

share|improve this answer

edited 8 hours ago

answered 10 hours ago

NathanOliverNathanOliver

98.2k16138216

• 
  
  
  

  
  1
  

  
  
  
  
  
  

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

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  
  @VTT That makes no difference. B() = default inside the class is still a user declared constructor.
  
  – NathanOliver
  10 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
  10 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
  10 hours ago
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

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

  
  
  
  

 | 
show 14 more comments

-4

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

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

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

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

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

share|improve this answer

edited 10 hours ago

TrebledJ

3,60521228

answered 11 hours ago

DiodacusDiodacus

1966

• 
  
  
  

  
  1
  

  
  
  
  
  
  

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

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  does this answer the question?
  
  – sp2danny
  10 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

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

  
  
  
  

add a comment | 

Your Answer

StackExchange.ifUsing("editor", function () {
StackExchange.using("externalEditor", function () {
StackExchange.using("snippets", function () {
StackExchange.snippets.init();
});
});
}, "code-snippets");

StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "1"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);

StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});

function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});


}
});

draft saved

draft discarded

Sign up or log in

StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});

Sign up using Google

Sign up using Facebook

Sign up using Email and Password

Post as a guest

Name

Email

Required, but never shown

StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fstackoverflow.com%2fquestions%2f55535107%2fwhy-doesnt-a-class-having-private-constructor-prevent-inheriting-from-this-clas%23new-answer', 'question_page');
}
);

Post as a guest

Name

Email

Required, but never shown

21

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

An aggregate is an array or a class with

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




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




• no virtual functions, and




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

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

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

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

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

int main()

{

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

}

which will print 1.

---

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

C c{};

C c1;

C c2 = C();

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

share|improve this answer

edited 8 hours ago

answered 10 hours ago

NathanOliverNathanOliver

98.2k16138216

• 
  
  
  

  
  1
  

  
  
  
  
  
  

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

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  
  @VTT That makes no difference. B() = default inside the class is still a user declared constructor.
  
  – NathanOliver
  10 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
  10 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
  10 hours ago
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

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

  
  
  
  

 | 
show 14 more comments

21

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

An aggregate is an array or a class with

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




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




• no virtual functions, and




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

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

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

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

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

int main()

{

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

}

which will print 1.

---

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

C c{};

C c1;

C c2 = C();

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

share|improve this answer

edited 8 hours ago

answered 10 hours ago

NathanOliverNathanOliver

98.2k16138216

• 
  
  
  

  
  1
  

  
  
  
  
  
  

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

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  
  @VTT That makes no difference. B() = default inside the class is still a user declared constructor.
  
  – NathanOliver
  10 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
  10 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
  10 hours ago
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

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

  
  
  
  

 | 
show 14 more comments

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

An aggregate is an array or a class with

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




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




• no virtual functions, and




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

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

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

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

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

int main()

{

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

}

which will print 1.

---

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

C c{};

C c1;

C c2 = C();

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

share|improve this answer

edited 8 hours ago

answered 10 hours ago

NathanOliverNathanOliver

98.2k16138216

int main()

{

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

}

C c{};

C c1;

C c2 = C();

share|improve this answer

edited 8 hours ago

answered 10 hours ago

NathanOliverNathanOliver

98.2k16138216

answered 10 hours ago

NathanOliverNathanOliver

98.2k16138216

answered 10 hours ago

NathanOliverNathanOliver

98.2k16138216

-4

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

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

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

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

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

share|improve this answer

edited 10 hours ago

TrebledJ

3,60521228

answered 11 hours ago

DiodacusDiodacus

1966

• 
  
  
  

  
  1
  

  
  
  
  
  
  

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

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  does this answer the question?
  
  – sp2danny
  10 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

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

  
  
  
  

add a comment | 

-4

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

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

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

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

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

share|improve this answer

edited 10 hours ago

TrebledJ

3,60521228

answered 11 hours ago

DiodacusDiodacus

1966

• 
  
  
  

  
  1
  

  
  
  
  
  
  

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

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  does this answer the question?
  
  – sp2danny
  10 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

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

share|improve this answer

edited 10 hours ago

TrebledJ

3,60521228

answered 11 hours ago

DiodacusDiodacus

1966

share|improve this answer

edited 10 hours ago

TrebledJ

3,60521228

answered 11 hours ago

DiodacusDiodacus

1966

edited 10 hours ago

TrebledJ

3,60521228

edited 10 hours ago

TrebledJ

3,60521228

answered 11 hours ago

DiodacusDiodacus

1966

answered 11 hours ago

DiodacusDiodacus

1966