Inheriting private CRTP constructor using friend declaration
I have problems understanding the correct behavior of inheriting constructors from a base class.
In my particular case, I have a crtp-base class with private constructors in order to prevent an instantiation of the base class (as abstract class).
Now, the crtp-base befriends the derived class and the derived class inherits the base class constructors with a using statement. This works well for the default, copy and move constructor but fails for custom constructors. Is there a way to achieve this without reimplenting all constructors in the derived class?
#include <iostream>
template <typename d_t>
class base
{
friend d_t;
base()
{
std::cout << "base: ctor()n";
}
base(base const & other) = default;
base(base && other) = default;
base(int)
{
std::cout << "base: ctor()n";
}
};
class derived: public base<derived>
{
public:
using base<derived>::base;
};
int main()
{
derived d{};
derived d1{d};
derived d2{std::move(d1)};
// derived d3{1}; //does not compile!
}
EDIT
AFAIK cppreference says that accessibility of constructors is not changed by the access specifier of the using declaration in the derived class, which is different to other member functions. But I have seen this kind of code compiling and running and I am not sure if I understood the using-declaration correctly.
Of course I'll investigate the other code further to see what is going on but I wanted to know if there is something hidden that I might miss.
c++ using friend crtp
asked Nov 23 '18 at 0:27
guest3450132guest3450132
11
|
show 3 more comments
I have problems understanding the correct behavior of inheriting constructors from a base class.
In my particular case, I have a crtp-base class with private constructors in order to prevent an instantiation of the base class (as abstract class).
Now, the crtp-base befriends the derived class and the derived class inherits the base class constructors with a using statement. This works well for the default, copy and move constructor but fails for custom constructors. Is there a way to achieve this without reimplenting all constructors in the derived class?
#include <iostream>
template <typename d_t>
class base
{
friend d_t;
base()
{
std::cout << "base: ctor()n";
}
base(base const & other) = default;
base(base && other) = default;
base(int)
{
std::cout << "base: ctor()n";
}
};
class derived: public base<derived>
{
public:
using base<derived>::base;
};
int main()
{
derived d{};
derived d1{d};
derived d2{std::move(d1)};
// derived d3{1}; //does not compile!
}
EDIT
AFAIK cppreference says that accessibility of constructors is not changed by the access specifier of the using declaration in the derived class, which is different to other member functions. But I have seen this kind of code compiling and running and I am not sure if I understood the using-declaration correctly.
Of course I'll investigate the other code further to see what is going on but I wanted to know if there is something hidden that I might miss.
c++ using friend crtp
asked Nov 23 '18 at 0:27
guest3450132guest3450132
11
friend d_t;that looks wrong?
– Yakk - Adam Nevraumont
Nov 23 '18 at 0:39
@Yakk-AdamNevraumont Why?
– songyuanyao
Nov 23 '18 at 0:50
_tis reserved in POSIX?
– user4581301
Nov 23 '18 at 1:07
1
Making the derived class a friend does not make much sense as you would get almost the same effect by make everything protected instead of private. The main difference would be if you derive other classes from your derived classes but given that you are using the CRTP pattern, it is probably not the case. Thus in the end, I think you are using the wrong tool for the job.
– Phil1970
Nov 23 '18 at 1:12
@Phil1970 true. I coluld make it protected. And if I understand en.cppreference.com/w/cpp/language/using_declaration correct then it should not even work. It is just that I've seen this kind of code compiling and I am completely out of explanation why? So I was wondering if there is a bit that I am not aware of which makes this work.
– guest3450132
Nov 23 '18 at 9:03
|
show 3 more comments
I have problems understanding the correct behavior of inheriting constructors from a base class.
In my particular case, I have a crtp-base class with private constructors in order to prevent an instantiation of the base class (as abstract class).
Now, the crtp-base befriends the derived class and the derived class inherits the base class constructors with a using statement. This works well for the default, copy and move constructor but fails for custom constructors. Is there a way to achieve this without reimplenting all constructors in the derived class?
#include <iostream>
template <typename d_t>
class base
{
friend d_t;
base()
{
std::cout << "base: ctor()n";
}
base(base const & other) = default;
base(base && other) = default;
base(int)
{
std::cout << "base: ctor()n";
}
};
class derived: public base<derived>
{
public:
using base<derived>::base;
};
int main()
{
derived d{};
derived d1{d};
derived d2{std::move(d1)};
// derived d3{1}; //does not compile!
}
EDIT
AFAIK cppreference says that accessibility of constructors is not changed by the access specifier of the using declaration in the derived class, which is different to other member functions. But I have seen this kind of code compiling and running and I am not sure if I understood the using-declaration correctly.
Of course I'll investigate the other code further to see what is going on but I wanted to know if there is something hidden that I might miss.
c++ using friend crtp
asked Nov 23 '18 at 0:27
guest3450132guest3450132
11
I have problems understanding the correct behavior of inheriting constructors from a base class.
In my particular case, I have a crtp-base class with private constructors in order to prevent an instantiation of the base class (as abstract class).
Now, the crtp-base befriends the derived class and the derived class inherits the base class constructors with a using statement. This works well for the default, copy and move constructor but fails for custom constructors. Is there a way to achieve this without reimplenting all constructors in the derived class?
#include <iostream>
template <typename d_t>
class base
{
friend d_t;
base()
{
std::cout << "base: ctor()n";
}
base(base const & other) = default;
base(base && other) = default;
base(int)
{
std::cout << "base: ctor()n";
}
};
class derived: public base<derived>
{
public:
using base<derived>::base;
};
int main()
{
derived d{};
derived d1{d};
derived d2{std::move(d1)};
// derived d3{1}; //does not compile!
}
EDIT
AFAIK cppreference says that accessibility of constructors is not changed by the access specifier of the using declaration in the derived class, which is different to other member functions. But I have seen this kind of code compiling and running and I am not sure if I understood the using-declaration correctly.
Of course I'll investigate the other code further to see what is going on but I wanted to know if there is something hidden that I might miss.
c++ using friend crtp
c++ using friend crtp
asked Nov 23 '18 at 0:27
guest3450132guest3450132
11
asked Nov 23 '18 at 0:27
guest3450132guest3450132
11
edited Nov 23 '18 at 9:16
guest3450132
asked Nov 23 '18 at 0:27
guest3450132guest3450132
11
asked Nov 23 '18 at 0:27
guest3450132guest3450132
11
asked Nov 23 '18 at 0:27
guest3450132guest3450132
11
11
friend d_t;that looks wrong?
– Yakk - Adam Nevraumont
Nov 23 '18 at 0:39
@Yakk-AdamNevraumont Why?
– songyuanyao
Nov 23 '18 at 0:50
_tis reserved in POSIX?
– user4581301
Nov 23 '18 at 1:07
1
Making the derived class a friend does not make much sense as you would get almost the same effect by make everything protected instead of private. The main difference would be if you derive other classes from your derived classes but given that you are using the CRTP pattern, it is probably not the case. Thus in the end, I think you are using the wrong tool for the job.
– Phil1970
Nov 23 '18 at 1:12
@Phil1970 true. I coluld make it protected. And if I understand en.cppreference.com/w/cpp/language/using_declaration correct then it should not even work. It is just that I've seen this kind of code compiling and I am completely out of explanation why? So I was wondering if there is a bit that I am not aware of which makes this work.
– guest3450132
Nov 23 '18 at 9:03
|
show 3 more comments
friend d_t;that looks wrong?
– Yakk - Adam Nevraumont
Nov 23 '18 at 0:39
@Yakk-AdamNevraumont Why?
– songyuanyao
Nov 23 '18 at 0:50
_tis reserved in POSIX?
– user4581301
Nov 23 '18 at 1:07
1
Making the derived class a friend does not make much sense as you would get almost the same effect by make everything protected instead of private. The main difference would be if you derive other classes from your derived classes but given that you are using the CRTP pattern, it is probably not the case. Thus in the end, I think you are using the wrong tool for the job.
– Phil1970
Nov 23 '18 at 1:12
@Phil1970 true. I coluld make it protected. And if I understand en.cppreference.com/w/cpp/language/using_declaration correct then it should not even work. It is just that I've seen this kind of code compiling and I am completely out of explanation why? So I was wondering if there is a bit that I am not aware of which makes this work.
– guest3450132
Nov 23 '18 at 9:03
friend d_t; that looks wrong?– Yakk - Adam Nevraumont
Nov 23 '18 at 0:39
friend d_t; that looks wrong?– Yakk - Adam Nevraumont
Nov 23 '18 at 0:39
@Yakk-AdamNevraumont Why?
– songyuanyao
Nov 23 '18 at 0:50
@Yakk-AdamNevraumont Why?
– songyuanyao
Nov 23 '18 at 0:50
_t is reserved in POSIX?– user4581301
Nov 23 '18 at 1:07
_t is reserved in POSIX?– user4581301
Nov 23 '18 at 1:07
1
1
Making the derived class a friend does not make much sense as you would get almost the same effect by make everything protected instead of private. The main difference would be if you derive other classes from your derived classes but given that you are using the CRTP pattern, it is probably not the case. Thus in the end, I think you are using the wrong tool for the job.
– Phil1970
Nov 23 '18 at 1:12
Making the derived class a friend does not make much sense as you would get almost the same effect by make everything protected instead of private. The main difference would be if you derive other classes from your derived classes but given that you are using the CRTP pattern, it is probably not the case. Thus in the end, I think you are using the wrong tool for the job.
– Phil1970
Nov 23 '18 at 1:12
@Phil1970 true. I coluld make it protected. And if I understand en.cppreference.com/w/cpp/language/using_declaration correct then it should not even work. It is just that I've seen this kind of code compiling and I am completely out of explanation why? So I was wondering if there is a bit that I am not aware of which makes this work.
– guest3450132
Nov 23 '18 at 9:03
@Phil1970 true. I coluld make it protected. And if I understand en.cppreference.com/w/cpp/language/using_declaration correct then it should not even work. It is just that I've seen this kind of code compiling and I am completely out of explanation why? So I was wondering if there is a bit that I am not aware of which makes this work.
– guest3450132
Nov 23 '18 at 9:03
|
show 3 more comments
1 Answer
1
active
oldest
votes
The short answer is yes, just make them public in the base class. Inherited constructors have the same accessibility in the derived class as they do in the base. There's no way to make a private base class constructor public in the derived class.
But abstract base classes can't be instantiated on their own anyway. There's no need to make their constructors private.
CRTP has no bearing on any of the above. If you only needed it for the friend declaration, then you can do without the pattern.
class base
{
public:
base()
{
}
base(int)
{
std::cout << "base: ctor()n";
}
virtual void foo() = 0;
};
class derived: public base
{
public:
using base::base;
void foo() override
{
// details...
}
};
int main()
{
derived d{};
derived d1{d};
derived d2{std::move(d1)};
derived d3{1}; //Works!!
}
Further reading on cppreference: Using-declaration
answered Nov 23 '18 at 3:42
Peter RudermanPeter Ruderman
10.2k2352
Thanks for the reply. I know that this solves the problem, but going to virtual polymorphism is not an option. The code is completely based on static polymorphism.
– guest3450132
Nov 23 '18 at 9:09
What did you mean by "as abstract class" in the original question?
– Peter Ruderman
Nov 23 '18 at 12:36
Abstract means, that in order to avoid misuse of the CRTP base class, it should not be instantiable without a dreived class. To prevent a wrong inheritance (a derived class that gives a wrong derived type to the CRTP) you need to make the constructors private and befriend the derived class. This way only the derived class can call the base class constructors and thus can be instantiated.
– guest3450132
Nov 30 '18 at 9:21
add a comment |
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The short answer is yes, just make them public in the base class. Inherited constructors have the same accessibility in the derived class as they do in the base. There's no way to make a private base class constructor public in the derived class.
But abstract base classes can't be instantiated on their own anyway. There's no need to make their constructors private.
CRTP has no bearing on any of the above. If you only needed it for the friend declaration, then you can do without the pattern.
class base
{
public:
base()
{
}
base(int)
{
std::cout << "base: ctor()n";
}
virtual void foo() = 0;
};
class derived: public base
{
public:
using base::base;
void foo() override
{
// details...
}
};
int main()
{
derived d{};
derived d1{d};
derived d2{std::move(d1)};
derived d3{1}; //Works!!
}
Further reading on cppreference: Using-declaration
answered Nov 23 '18 at 3:42
Peter RudermanPeter Ruderman
10.2k2352
Thanks for the reply. I know that this solves the problem, but going to virtual polymorphism is not an option. The code is completely based on static polymorphism.
– guest3450132
Nov 23 '18 at 9:09
What did you mean by "as abstract class" in the original question?
– Peter Ruderman
Nov 23 '18 at 12:36
Abstract means, that in order to avoid misuse of the CRTP base class, it should not be instantiable without a dreived class. To prevent a wrong inheritance (a derived class that gives a wrong derived type to the CRTP) you need to make the constructors private and befriend the derived class. This way only the derived class can call the base class constructors and thus can be instantiated.
– guest3450132
Nov 30 '18 at 9:21
add a comment |
The short answer is yes, just make them public in the base class. Inherited constructors have the same accessibility in the derived class as they do in the base. There's no way to make a private base class constructor public in the derived class.
But abstract base classes can't be instantiated on their own anyway. There's no need to make their constructors private.
CRTP has no bearing on any of the above. If you only needed it for the friend declaration, then you can do without the pattern.
class base
{
public:
base()
{
}
base(int)
{
std::cout << "base: ctor()n";
}
virtual void foo() = 0;
};
class derived: public base
{
public:
using base::base;
void foo() override
{
// details...
}
};
int main()
{
derived d{};
derived d1{d};
derived d2{std::move(d1)};
derived d3{1}; //Works!!
}
Further reading on cppreference: Using-declaration
answered Nov 23 '18 at 3:42
Peter RudermanPeter Ruderman
10.2k2352
Thanks for the reply. I know that this solves the problem, but going to virtual polymorphism is not an option. The code is completely based on static polymorphism.
– guest3450132
Nov 23 '18 at 9:09
What did you mean by "as abstract class" in the original question?
– Peter Ruderman
Nov 23 '18 at 12:36
Abstract means, that in order to avoid misuse of the CRTP base class, it should not be instantiable without a dreived class. To prevent a wrong inheritance (a derived class that gives a wrong derived type to the CRTP) you need to make the constructors private and befriend the derived class. This way only the derived class can call the base class constructors and thus can be instantiated.
– guest3450132
Nov 30 '18 at 9:21
add a comment |
The short answer is yes, just make them public in the base class. Inherited constructors have the same accessibility in the derived class as they do in the base. There's no way to make a private base class constructor public in the derived class.
But abstract base classes can't be instantiated on their own anyway. There's no need to make their constructors private.
CRTP has no bearing on any of the above. If you only needed it for the friend declaration, then you can do without the pattern.
class base
{
public:
base()
{
}
base(int)
{
std::cout << "base: ctor()n";
}
virtual void foo() = 0;
};
class derived: public base
{
public:
using base::base;
void foo() override
{
// details...
}
};
int main()
{
derived d{};
derived d1{d};
derived d2{std::move(d1)};
derived d3{1}; //Works!!
}
Further reading on cppreference: Using-declaration
answered Nov 23 '18 at 3:42
Peter RudermanPeter Ruderman
10.2k2352
The short answer is yes, just make them public in the base class. Inherited constructors have the same accessibility in the derived class as they do in the base. There's no way to make a private base class constructor public in the derived class.
But abstract base classes can't be instantiated on their own anyway. There's no need to make their constructors private.
CRTP has no bearing on any of the above. If you only needed it for the friend declaration, then you can do without the pattern.
class base
{
public:
base()
{
}
base(int)
{
std::cout << "base: ctor()n";
}
virtual void foo() = 0;
};
class derived: public base
{
public:
using base::base;
void foo() override
{
// details...
}
};
int main()
{
derived d{};
derived d1{d};
derived d2{std::move(d1)};
derived d3{1}; //Works!!
}
Further reading on cppreference: Using-declaration
answered Nov 23 '18 at 3:42
Peter RudermanPeter Ruderman
10.2k2352
answered Nov 23 '18 at 3:42
Peter RudermanPeter Ruderman
10.2k2352
answered Nov 23 '18 at 3:42
Peter RudermanPeter Ruderman
10.2k2352
answered Nov 23 '18 at 3:42
Peter RudermanPeter Ruderman
10.2k2352
10.2k2352
Thanks for the reply. I know that this solves the problem, but going to virtual polymorphism is not an option. The code is completely based on static polymorphism.
– guest3450132
Nov 23 '18 at 9:09
What did you mean by "as abstract class" in the original question?
– Peter Ruderman
Nov 23 '18 at 12:36
Abstract means, that in order to avoid misuse of the CRTP base class, it should not be instantiable without a dreived class. To prevent a wrong inheritance (a derived class that gives a wrong derived type to the CRTP) you need to make the constructors private and befriend the derived class. This way only the derived class can call the base class constructors and thus can be instantiated.
– guest3450132
Nov 30 '18 at 9:21
add a comment |
Thanks for the reply. I know that this solves the problem, but going to virtual polymorphism is not an option. The code is completely based on static polymorphism.
– guest3450132
Nov 23 '18 at 9:09
What did you mean by "as abstract class" in the original question?
– Peter Ruderman
Nov 23 '18 at 12:36
Abstract means, that in order to avoid misuse of the CRTP base class, it should not be instantiable without a dreived class. To prevent a wrong inheritance (a derived class that gives a wrong derived type to the CRTP) you need to make the constructors private and befriend the derived class. This way only the derived class can call the base class constructors and thus can be instantiated.
– guest3450132
Nov 30 '18 at 9:21
Thanks for the reply. I know that this solves the problem, but going to virtual polymorphism is not an option. The code is completely based on static polymorphism.
– guest3450132
Nov 23 '18 at 9:09
Thanks for the reply. I know that this solves the problem, but going to virtual polymorphism is not an option. The code is completely based on static polymorphism.
– guest3450132
Nov 23 '18 at 9:09
What did you mean by "as abstract class" in the original question?
– Peter Ruderman
Nov 23 '18 at 12:36
What did you mean by "as abstract class" in the original question?
– Peter Ruderman
Nov 23 '18 at 12:36
Abstract means, that in order to avoid misuse of the CRTP base class, it should not be instantiable without a dreived class. To prevent a wrong inheritance (a derived class that gives a wrong derived type to the CRTP) you need to make the constructors private and befriend the derived class. This way only the derived class can call the base class constructors and thus can be instantiated.
– guest3450132
Nov 30 '18 at 9:21
Abstract means, that in order to avoid misuse of the CRTP base class, it should not be instantiable without a dreived class. To prevent a wrong inheritance (a derived class that gives a wrong derived type to the CRTP) you need to make the constructors private and befriend the derived class. This way only the derived class can call the base class constructors and thus can be instantiated.
– guest3450132
Nov 30 '18 at 9:21
add a comment |
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friend d_t;that looks wrong?– Yakk - Adam Nevraumont
Nov 23 '18 at 0:39
@Yakk-AdamNevraumont Why?
– songyuanyao
Nov 23 '18 at 0:50
_tis reserved in POSIX?– user4581301
Nov 23 '18 at 1:07
1
Making the derived class a friend does not make much sense as you would get almost the same effect by make everything protected instead of private. The main difference would be if you derive other classes from your derived classes but given that you are using the CRTP pattern, it is probably not the case. Thus in the end, I think you are using the wrong tool for the job.
– Phil1970
Nov 23 '18 at 1:12
@Phil1970 true. I coluld make it protected. And if I understand en.cppreference.com/w/cpp/language/using_declaration correct then it should not even work. It is just that I've seen this kind of code compiling and I am completely out of explanation why? So I was wondering if there is a bit that I am not aware of which makes this work.
– guest3450132
Nov 23 '18 at 9:03