Non-Movable C++17 Unique Pointer
I came across this answer Prevent moving of a unique_ptr C++11. However while trying it out on a compiler online, this works with C++11(std::move compiler error) but with C++17, I'm seeing that the std::move below is successful. Shouldn't the compiler throw an error on that line? Also if some semantics have changed in C++17, what is the correct way to create a non movable unique_ptr in C++17 and onward.
template <typename T>
using scoped_ptr = const std::unique_ptr<T>;
int main()
{
auto p = scoped_ptr<int>(new int(5));
auto p2 = std::move(p); // should be error?
std::cout << *p2 << std::endl; // 5
return 0;
}
You can try it online here.
c++ c++11 c++17 unique-ptr
asked Nov 24 '18 at 6:08
tangytangy
1,010720
add a comment |
I came across this answer Prevent moving of a unique_ptr C++11. However while trying it out on a compiler online, this works with C++11(std::move compiler error) but with C++17, I'm seeing that the std::move below is successful. Shouldn't the compiler throw an error on that line? Also if some semantics have changed in C++17, what is the correct way to create a non movable unique_ptr in C++17 and onward.
template <typename T>
using scoped_ptr = const std::unique_ptr<T>;
int main()
{
auto p = scoped_ptr<int>(new int(5));
auto p2 = std::move(p); // should be error?
std::cout << *p2 << std::endl; // 5
return 0;
}
You can try it online here.
c++ c++11 c++17 unique-ptr
asked Nov 24 '18 at 6:08
tangytangy
1,010720
add a comment |
I came across this answer Prevent moving of a unique_ptr C++11. However while trying it out on a compiler online, this works with C++11(std::move compiler error) but with C++17, I'm seeing that the std::move below is successful. Shouldn't the compiler throw an error on that line? Also if some semantics have changed in C++17, what is the correct way to create a non movable unique_ptr in C++17 and onward.
template <typename T>
using scoped_ptr = const std::unique_ptr<T>;
int main()
{
auto p = scoped_ptr<int>(new int(5));
auto p2 = std::move(p); // should be error?
std::cout << *p2 << std::endl; // 5
return 0;
}
You can try it online here.
c++ c++11 c++17 unique-ptr
asked Nov 24 '18 at 6:08
tangytangy
1,010720
I came across this answer Prevent moving of a unique_ptr C++11. However while trying it out on a compiler online, this works with C++11(std::move compiler error) but with C++17, I'm seeing that the std::move below is successful. Shouldn't the compiler throw an error on that line? Also if some semantics have changed in C++17, what is the correct way to create a non movable unique_ptr in C++17 and onward.
template <typename T>
using scoped_ptr = const std::unique_ptr<T>;
int main()
{
auto p = scoped_ptr<int>(new int(5));
auto p2 = std::move(p); // should be error?
std::cout << *p2 << std::endl; // 5
return 0;
}
You can try it online here.
c++ c++11 c++17 unique-ptr
c++ c++11 c++17 unique-ptr
asked Nov 24 '18 at 6:08
tangytangy
1,010720
asked Nov 24 '18 at 6:08
tangytangy
1,010720
edited Nov 24 '18 at 6:14
tangy
asked Nov 24 '18 at 6:08
tangytangy
1,010720
asked Nov 24 '18 at 6:08
tangytangy
1,010720
asked Nov 24 '18 at 6:08
tangytangy
1,010720
1,010720
add a comment |
add a comment |
3 Answers
3
active
oldest
votes
p is not const. See here for it to fail the way you expect.
auto deduces like a template<class T>void foo(T) does. T is never deduced as const, and neither is auto p=.
Meanwhile, the auto p = line works because you compiled it in c++17 mode. In c++11 it does not compile. This is because how prvalues differ in 17; some call the difference guaranteed elision.
If you want an immobile unique ptr:
template<class T, class D>
struct immobile_ptr:private std::unique_ptr<T, D>{
using unique_ptr<T>::operator*;
using unique_ptr<T>::operator->;
using unique_ptr<T>::get;
using unique_ptr<T>::operator bool;
// etc
// manually forward some ctors, as using grabs some move ctors in this case
};
template<class T, class...Args>
immobile_ptr<T> make_immobile_ptr(Args&&...args); // todo
an alternative might be to take a unique ptr with an immobile destroyer.
template<class X>
struct nomove_destroy:std::destroy<T>{
nomove_destroy(nomove_destroy&&)=delete;
nomove_destroy()=default;
nomove_destroy& operator=(nomove_destroy&&)=delete;
};
template<class T>
using nomove_ptr=std::unique_ptr<T,nomove_destroy<T>>;
But I am uncertain if that will work.
answered Nov 24 '18 at 6:18
Yakk - Adam NevraumontYakk - Adam Nevraumont
186k19198379
1
If I read overload #6 correctly, the deleter trick should work. Neat!
– Quentin
Nov 24 '18 at 13:08
add a comment |
Note that p is declared as non-reference type, the const part of the argument scoped_ptr<int>(new int(5)) is ignored in type deduction. Then the type deduction result for p is std::unique_ptr<int>, not const std::unique_ptr<int> (i.e. scoped_ptr<int> as you expected).
What you want might be
auto& p = scoped_ptr<int>(new int(5)); // p is of type const std::unique_ptr<int>& now
answered Nov 24 '18 at 6:16
songyuanyaosongyuanyao
92k11175240
auto&doesn't make it of non-reference typeconst std::unique_ptr<int>, does it?
– ShadowRanger
Nov 24 '18 at 6:22
@ShadowRanger You meanconst std::unique_ptr<int>&?
– songyuanyao
Nov 24 '18 at 6:23
Works across versions this way coliru.stacked-crooked.com/a/aef0f48df5604e38
– tangy
Nov 24 '18 at 6:31
add a comment |
Welcome to the world of type deduction in C++. Try
auto & p = scoped_ptr<int>(new int(5));
or
auto && p = scoped_ptr<int>(new int(5));
instead. This lecture may be helpful: https://www.youtube.com/watch?v=wQxj20X-tIU
edited Nov 25 '18 at 9:44
Jon Harper
2,9482929
answered Nov 24 '18 at 6:23
AmosAmos
1,67531029
Thanks for the suggested lecture - I think I need to watch this and read the relevant material from Modern Effective C++ again :)
– tangy
Nov 24 '18 at 6:26
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
p is not const. See here for it to fail the way you expect.
auto deduces like a template<class T>void foo(T) does. T is never deduced as const, and neither is auto p=.
Meanwhile, the auto p = line works because you compiled it in c++17 mode. In c++11 it does not compile. This is because how prvalues differ in 17; some call the difference guaranteed elision.
If you want an immobile unique ptr:
template<class T, class D>
struct immobile_ptr:private std::unique_ptr<T, D>{
using unique_ptr<T>::operator*;
using unique_ptr<T>::operator->;
using unique_ptr<T>::get;
using unique_ptr<T>::operator bool;
// etc
// manually forward some ctors, as using grabs some move ctors in this case
};
template<class T, class...Args>
immobile_ptr<T> make_immobile_ptr(Args&&...args); // todo
an alternative might be to take a unique ptr with an immobile destroyer.
template<class X>
struct nomove_destroy:std::destroy<T>{
nomove_destroy(nomove_destroy&&)=delete;
nomove_destroy()=default;
nomove_destroy& operator=(nomove_destroy&&)=delete;
};
template<class T>
using nomove_ptr=std::unique_ptr<T,nomove_destroy<T>>;
But I am uncertain if that will work.
answered Nov 24 '18 at 6:18
Yakk - Adam NevraumontYakk - Adam Nevraumont
186k19198379
1
If I read overload #6 correctly, the deleter trick should work. Neat!
– Quentin
Nov 24 '18 at 13:08
add a comment |
p is not const. See here for it to fail the way you expect.
auto deduces like a template<class T>void foo(T) does. T is never deduced as const, and neither is auto p=.
Meanwhile, the auto p = line works because you compiled it in c++17 mode. In c++11 it does not compile. This is because how prvalues differ in 17; some call the difference guaranteed elision.
If you want an immobile unique ptr:
template<class T, class D>
struct immobile_ptr:private std::unique_ptr<T, D>{
using unique_ptr<T>::operator*;
using unique_ptr<T>::operator->;
using unique_ptr<T>::get;
using unique_ptr<T>::operator bool;
// etc
// manually forward some ctors, as using grabs some move ctors in this case
};
template<class T, class...Args>
immobile_ptr<T> make_immobile_ptr(Args&&...args); // todo
an alternative might be to take a unique ptr with an immobile destroyer.
template<class X>
struct nomove_destroy:std::destroy<T>{
nomove_destroy(nomove_destroy&&)=delete;
nomove_destroy()=default;
nomove_destroy& operator=(nomove_destroy&&)=delete;
};
template<class T>
using nomove_ptr=std::unique_ptr<T,nomove_destroy<T>>;
But I am uncertain if that will work.
answered Nov 24 '18 at 6:18
Yakk - Adam NevraumontYakk - Adam Nevraumont
186k19198379
1
If I read overload #6 correctly, the deleter trick should work. Neat!
– Quentin
Nov 24 '18 at 13:08
add a comment |
p is not const. See here for it to fail the way you expect.
auto deduces like a template<class T>void foo(T) does. T is never deduced as const, and neither is auto p=.
Meanwhile, the auto p = line works because you compiled it in c++17 mode. In c++11 it does not compile. This is because how prvalues differ in 17; some call the difference guaranteed elision.
If you want an immobile unique ptr:
template<class T, class D>
struct immobile_ptr:private std::unique_ptr<T, D>{
using unique_ptr<T>::operator*;
using unique_ptr<T>::operator->;
using unique_ptr<T>::get;
using unique_ptr<T>::operator bool;
// etc
// manually forward some ctors, as using grabs some move ctors in this case
};
template<class T, class...Args>
immobile_ptr<T> make_immobile_ptr(Args&&...args); // todo
an alternative might be to take a unique ptr with an immobile destroyer.
template<class X>
struct nomove_destroy:std::destroy<T>{
nomove_destroy(nomove_destroy&&)=delete;
nomove_destroy()=default;
nomove_destroy& operator=(nomove_destroy&&)=delete;
};
template<class T>
using nomove_ptr=std::unique_ptr<T,nomove_destroy<T>>;
But I am uncertain if that will work.
answered Nov 24 '18 at 6:18
Yakk - Adam NevraumontYakk - Adam Nevraumont
186k19198379
p is not const. See here for it to fail the way you expect.
auto deduces like a template<class T>void foo(T) does. T is never deduced as const, and neither is auto p=.
Meanwhile, the auto p = line works because you compiled it in c++17 mode. In c++11 it does not compile. This is because how prvalues differ in 17; some call the difference guaranteed elision.
If you want an immobile unique ptr:
template<class T, class D>
struct immobile_ptr:private std::unique_ptr<T, D>{
using unique_ptr<T>::operator*;
using unique_ptr<T>::operator->;
using unique_ptr<T>::get;
using unique_ptr<T>::operator bool;
// etc
// manually forward some ctors, as using grabs some move ctors in this case
};
template<class T, class...Args>
immobile_ptr<T> make_immobile_ptr(Args&&...args); // todo
an alternative might be to take a unique ptr with an immobile destroyer.
template<class X>
struct nomove_destroy:std::destroy<T>{
nomove_destroy(nomove_destroy&&)=delete;
nomove_destroy()=default;
nomove_destroy& operator=(nomove_destroy&&)=delete;
};
template<class T>
using nomove_ptr=std::unique_ptr<T,nomove_destroy<T>>;
But I am uncertain if that will work.
answered Nov 24 '18 at 6:18
Yakk - Adam NevraumontYakk - Adam Nevraumont
186k19198379
edited Nov 24 '18 at 6:24
answered Nov 24 '18 at 6:18
Yakk - Adam NevraumontYakk - Adam Nevraumont
186k19198379
answered Nov 24 '18 at 6:18
Yakk - Adam NevraumontYakk - Adam Nevraumont
186k19198379
answered Nov 24 '18 at 6:18
Yakk - Adam NevraumontYakk - Adam Nevraumont
186k19198379
186k19198379
1
If I read overload #6 correctly, the deleter trick should work. Neat!
– Quentin
Nov 24 '18 at 13:08
add a comment |
1
If I read overload #6 correctly, the deleter trick should work. Neat!
– Quentin
Nov 24 '18 at 13:08
1
1
If I read overload #6 correctly, the deleter trick should work. Neat!
– Quentin
Nov 24 '18 at 13:08
If I read overload #6 correctly, the deleter trick should work. Neat!
– Quentin
Nov 24 '18 at 13:08
add a comment |
Note that p is declared as non-reference type, the const part of the argument scoped_ptr<int>(new int(5)) is ignored in type deduction. Then the type deduction result for p is std::unique_ptr<int>, not const std::unique_ptr<int> (i.e. scoped_ptr<int> as you expected).
What you want might be
auto& p = scoped_ptr<int>(new int(5)); // p is of type const std::unique_ptr<int>& now
answered Nov 24 '18 at 6:16
songyuanyaosongyuanyao
92k11175240
auto&doesn't make it of non-reference typeconst std::unique_ptr<int>, does it?
– ShadowRanger
Nov 24 '18 at 6:22
@ShadowRanger You meanconst std::unique_ptr<int>&?
– songyuanyao
Nov 24 '18 at 6:23
Works across versions this way coliru.stacked-crooked.com/a/aef0f48df5604e38
– tangy
Nov 24 '18 at 6:31
add a comment |
Note that p is declared as non-reference type, the const part of the argument scoped_ptr<int>(new int(5)) is ignored in type deduction. Then the type deduction result for p is std::unique_ptr<int>, not const std::unique_ptr<int> (i.e. scoped_ptr<int> as you expected).
What you want might be
auto& p = scoped_ptr<int>(new int(5)); // p is of type const std::unique_ptr<int>& now
answered Nov 24 '18 at 6:16
songyuanyaosongyuanyao
92k11175240
auto&doesn't make it of non-reference typeconst std::unique_ptr<int>, does it?
– ShadowRanger
Nov 24 '18 at 6:22
@ShadowRanger You meanconst std::unique_ptr<int>&?
– songyuanyao
Nov 24 '18 at 6:23
Works across versions this way coliru.stacked-crooked.com/a/aef0f48df5604e38
– tangy
Nov 24 '18 at 6:31
add a comment |
Note that p is declared as non-reference type, the const part of the argument scoped_ptr<int>(new int(5)) is ignored in type deduction. Then the type deduction result for p is std::unique_ptr<int>, not const std::unique_ptr<int> (i.e. scoped_ptr<int> as you expected).
What you want might be
auto& p = scoped_ptr<int>(new int(5)); // p is of type const std::unique_ptr<int>& now
answered Nov 24 '18 at 6:16
songyuanyaosongyuanyao
92k11175240
Note that p is declared as non-reference type, the const part of the argument scoped_ptr<int>(new int(5)) is ignored in type deduction. Then the type deduction result for p is std::unique_ptr<int>, not const std::unique_ptr<int> (i.e. scoped_ptr<int> as you expected).
What you want might be
auto& p = scoped_ptr<int>(new int(5)); // p is of type const std::unique_ptr<int>& now
answered Nov 24 '18 at 6:16
songyuanyaosongyuanyao
92k11175240
edited Nov 24 '18 at 6:23
answered Nov 24 '18 at 6:16
songyuanyaosongyuanyao
92k11175240
answered Nov 24 '18 at 6:16
songyuanyaosongyuanyao
92k11175240
answered Nov 24 '18 at 6:16
songyuanyaosongyuanyao
92k11175240
92k11175240
auto&doesn't make it of non-reference typeconst std::unique_ptr<int>, does it?
– ShadowRanger
Nov 24 '18 at 6:22
@ShadowRanger You meanconst std::unique_ptr<int>&?
– songyuanyao
Nov 24 '18 at 6:23
Works across versions this way coliru.stacked-crooked.com/a/aef0f48df5604e38
– tangy
Nov 24 '18 at 6:31
add a comment |
auto&doesn't make it of non-reference typeconst std::unique_ptr<int>, does it?
– ShadowRanger
Nov 24 '18 at 6:22
@ShadowRanger You meanconst std::unique_ptr<int>&?
– songyuanyao
Nov 24 '18 at 6:23
Works across versions this way coliru.stacked-crooked.com/a/aef0f48df5604e38
– tangy
Nov 24 '18 at 6:31
auto& doesn't make it of non-reference type const std::unique_ptr<int>, does it?– ShadowRanger
Nov 24 '18 at 6:22
auto& doesn't make it of non-reference type const std::unique_ptr<int>, does it?– ShadowRanger
Nov 24 '18 at 6:22
@ShadowRanger You mean
const std::unique_ptr<int>& ?– songyuanyao
Nov 24 '18 at 6:23
@ShadowRanger You mean
const std::unique_ptr<int>& ?– songyuanyao
Nov 24 '18 at 6:23
Works across versions this way coliru.stacked-crooked.com/a/aef0f48df5604e38
– tangy
Nov 24 '18 at 6:31
Works across versions this way coliru.stacked-crooked.com/a/aef0f48df5604e38
– tangy
Nov 24 '18 at 6:31
add a comment |
Welcome to the world of type deduction in C++. Try
auto & p = scoped_ptr<int>(new int(5));
or
auto && p = scoped_ptr<int>(new int(5));
instead. This lecture may be helpful: https://www.youtube.com/watch?v=wQxj20X-tIU
edited Nov 25 '18 at 9:44
Jon Harper
2,9482929
answered Nov 24 '18 at 6:23
AmosAmos
1,67531029
Thanks for the suggested lecture - I think I need to watch this and read the relevant material from Modern Effective C++ again :)
– tangy
Nov 24 '18 at 6:26
add a comment |
Welcome to the world of type deduction in C++. Try
auto & p = scoped_ptr<int>(new int(5));
or
auto && p = scoped_ptr<int>(new int(5));
instead. This lecture may be helpful: https://www.youtube.com/watch?v=wQxj20X-tIU
edited Nov 25 '18 at 9:44
Jon Harper
2,9482929
answered Nov 24 '18 at 6:23
AmosAmos
1,67531029
Thanks for the suggested lecture - I think I need to watch this and read the relevant material from Modern Effective C++ again :)
– tangy
Nov 24 '18 at 6:26
add a comment |
Welcome to the world of type deduction in C++. Try
auto & p = scoped_ptr<int>(new int(5));
or
auto && p = scoped_ptr<int>(new int(5));
instead. This lecture may be helpful: https://www.youtube.com/watch?v=wQxj20X-tIU
edited Nov 25 '18 at 9:44
Jon Harper
2,9482929
answered Nov 24 '18 at 6:23
AmosAmos
1,67531029
Welcome to the world of type deduction in C++. Try
auto & p = scoped_ptr<int>(new int(5));
or
auto && p = scoped_ptr<int>(new int(5));
instead. This lecture may be helpful: https://www.youtube.com/watch?v=wQxj20X-tIU
edited Nov 25 '18 at 9:44
Jon Harper
2,9482929
answered Nov 24 '18 at 6:23
AmosAmos
1,67531029
edited Nov 25 '18 at 9:44
Jon Harper
2,9482929
edited Nov 25 '18 at 9:44
Jon Harper
2,9482929
edited Nov 25 '18 at 9:44
Jon Harper
2,9482929
2,9482929
answered Nov 24 '18 at 6:23
AmosAmos
1,67531029
answered Nov 24 '18 at 6:23
AmosAmos
1,67531029
answered Nov 24 '18 at 6:23
AmosAmos
1,67531029
1,67531029
Thanks for the suggested lecture - I think I need to watch this and read the relevant material from Modern Effective C++ again :)
– tangy
Nov 24 '18 at 6:26
add a comment |
Thanks for the suggested lecture - I think I need to watch this and read the relevant material from Modern Effective C++ again :)
– tangy
Nov 24 '18 at 6:26
Thanks for the suggested lecture - I think I need to watch this and read the relevant material from Modern Effective C++ again :)
– tangy
Nov 24 '18 at 6:26
Thanks for the suggested lecture - I think I need to watch this and read the relevant material from Modern Effective C++ again :)
– tangy
Nov 24 '18 at 6:26
add a comment |
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