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Why am I allowed to create multiple unique pointers from a single object?



The Next CEO of Stack OverflowWhy use pointers?Pretty-print C++ STL containersC++ pointers local in functionsHow C++ reference worksWhy is reading lines from stdin much slower in C++ than Python?Why should I use a pointer rather than the object itself?Pointers as Function Parameters ExampleReturn class object pointer from another classwhy should we use std::move semantic with unique pointers?Why does my object appear to be on the heap without using `new`?










13















Why am I allowed to create multiple unique pointers from a single object?



#include <iostream>
#include <memory>

using namespace std;

class Class

public:
Class(int a): int_(a)std::cout << "constr" << std::endl;
~Class()std::cout << "destr" << std::endl;
int int_;

;

int main()

Class a(4);
std::unique_ptr<Class> ptr = std::make_unique<Class>(a);
std::unique_ptr<Class> ptr2 = std::make_unique<Class>(a);
std::unique_ptr<Class> ptr3 = std::make_unique<Class>(a);
std::cout << ptr->int_ << std::endl;
std::cout << ptr2->int_ << std::endl;
std::cout << ptr3->int_ << std::endl;

return 0;



Output:



constr
4
4
4
destr
destr
destr
destr









share|improve this question









New contributor




Lokas Beard is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.















  • 21





    Copy construction.

    – user4581301
    2 days ago






  • 1





    Try this: a.int_ = 5; after you create each unique_ptr and before you print the contents of each pointer

    – alter igel
    2 days ago






  • 1





    Creating several unique_ptr to the same instance will almost certainly lead to undefined behavior, but that's not what you are doing here. Something like auto ptr2 = std::unique_ptr<Class>(ptr.get()); would create a second std::unique_ptr that points to the same instance as ptr does, and that would be problematic.

    – François Andrieux
    2 days ago











  • Demo with user-defined copy constructor (called 3 times). Exemplifies the underlying copy construction.

    – TrebledJ
    yesterday















13















Why am I allowed to create multiple unique pointers from a single object?



#include <iostream>
#include <memory>

using namespace std;

class Class

public:
Class(int a): int_(a)std::cout << "constr" << std::endl;
~Class()std::cout << "destr" << std::endl;
int int_;

;

int main()

Class a(4);
std::unique_ptr<Class> ptr = std::make_unique<Class>(a);
std::unique_ptr<Class> ptr2 = std::make_unique<Class>(a);
std::unique_ptr<Class> ptr3 = std::make_unique<Class>(a);
std::cout << ptr->int_ << std::endl;
std::cout << ptr2->int_ << std::endl;
std::cout << ptr3->int_ << std::endl;

return 0;



Output:



constr
4
4
4
destr
destr
destr
destr









share|improve this question









New contributor




Lokas Beard is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.















  • 21





    Copy construction.

    – user4581301
    2 days ago






  • 1





    Try this: a.int_ = 5; after you create each unique_ptr and before you print the contents of each pointer

    – alter igel
    2 days ago






  • 1





    Creating several unique_ptr to the same instance will almost certainly lead to undefined behavior, but that's not what you are doing here. Something like auto ptr2 = std::unique_ptr<Class>(ptr.get()); would create a second std::unique_ptr that points to the same instance as ptr does, and that would be problematic.

    – François Andrieux
    2 days ago











  • Demo with user-defined copy constructor (called 3 times). Exemplifies the underlying copy construction.

    – TrebledJ
    yesterday













13












13








13


2






Why am I allowed to create multiple unique pointers from a single object?



#include <iostream>
#include <memory>

using namespace std;

class Class

public:
Class(int a): int_(a)std::cout << "constr" << std::endl;
~Class()std::cout << "destr" << std::endl;
int int_;

;

int main()

Class a(4);
std::unique_ptr<Class> ptr = std::make_unique<Class>(a);
std::unique_ptr<Class> ptr2 = std::make_unique<Class>(a);
std::unique_ptr<Class> ptr3 = std::make_unique<Class>(a);
std::cout << ptr->int_ << std::endl;
std::cout << ptr2->int_ << std::endl;
std::cout << ptr3->int_ << std::endl;

return 0;



Output:



constr
4
4
4
destr
destr
destr
destr









share|improve this question









New contributor




Lokas Beard is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.












Why am I allowed to create multiple unique pointers from a single object?



#include <iostream>
#include <memory>

using namespace std;

class Class

public:
Class(int a): int_(a)std::cout << "constr" << std::endl;
~Class()std::cout << "destr" << std::endl;
int int_;

;

int main()

Class a(4);
std::unique_ptr<Class> ptr = std::make_unique<Class>(a);
std::unique_ptr<Class> ptr2 = std::make_unique<Class>(a);
std::unique_ptr<Class> ptr3 = std::make_unique<Class>(a);
std::cout << ptr->int_ << std::endl;
std::cout << ptr2->int_ << std::endl;
std::cout << ptr3->int_ << std::endl;

return 0;



Output:



constr
4
4
4
destr
destr
destr
destr






c++ pointers unique-ptr






share|improve this question









New contributor




Lokas Beard is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.











share|improve this question









New contributor




Lokas Beard is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.









share|improve this question




share|improve this question








edited yesterday









TrebledJ

3,46911228




3,46911228






New contributor




Lokas Beard is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.









asked 2 days ago









Lokas BeardLokas Beard

715




715




New contributor




Lokas Beard is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.





New contributor





Lokas Beard is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.






Lokas Beard is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.







  • 21





    Copy construction.

    – user4581301
    2 days ago






  • 1





    Try this: a.int_ = 5; after you create each unique_ptr and before you print the contents of each pointer

    – alter igel
    2 days ago






  • 1





    Creating several unique_ptr to the same instance will almost certainly lead to undefined behavior, but that's not what you are doing here. Something like auto ptr2 = std::unique_ptr<Class>(ptr.get()); would create a second std::unique_ptr that points to the same instance as ptr does, and that would be problematic.

    – François Andrieux
    2 days ago











  • Demo with user-defined copy constructor (called 3 times). Exemplifies the underlying copy construction.

    – TrebledJ
    yesterday












  • 21





    Copy construction.

    – user4581301
    2 days ago






  • 1





    Try this: a.int_ = 5; after you create each unique_ptr and before you print the contents of each pointer

    – alter igel
    2 days ago






  • 1





    Creating several unique_ptr to the same instance will almost certainly lead to undefined behavior, but that's not what you are doing here. Something like auto ptr2 = std::unique_ptr<Class>(ptr.get()); would create a second std::unique_ptr that points to the same instance as ptr does, and that would be problematic.

    – François Andrieux
    2 days ago











  • Demo with user-defined copy constructor (called 3 times). Exemplifies the underlying copy construction.

    – TrebledJ
    yesterday







21




21





Copy construction.

– user4581301
2 days ago





Copy construction.

– user4581301
2 days ago




1




1





Try this: a.int_ = 5; after you create each unique_ptr and before you print the contents of each pointer

– alter igel
2 days ago





Try this: a.int_ = 5; after you create each unique_ptr and before you print the contents of each pointer

– alter igel
2 days ago




1




1





Creating several unique_ptr to the same instance will almost certainly lead to undefined behavior, but that's not what you are doing here. Something like auto ptr2 = std::unique_ptr<Class>(ptr.get()); would create a second std::unique_ptr that points to the same instance as ptr does, and that would be problematic.

– François Andrieux
2 days ago





Creating several unique_ptr to the same instance will almost certainly lead to undefined behavior, but that's not what you are doing here. Something like auto ptr2 = std::unique_ptr<Class>(ptr.get()); would create a second std::unique_ptr that points to the same instance as ptr does, and that would be problematic.

– François Andrieux
2 days ago













Demo with user-defined copy constructor (called 3 times). Exemplifies the underlying copy construction.

– TrebledJ
yesterday





Demo with user-defined copy constructor (called 3 times). Exemplifies the underlying copy construction.

– TrebledJ
yesterday












3 Answers
3






active

oldest

votes


















36














Why not?



You are not creating multiple unique_ptr instances pointing to the same Class instance, but you are instead allocating three new Class instances on the heap, copy-constructed from a. Every unique_ptr points to a different instance.




std::unique_ptr<Class> ptr = std::make_unique<Class>(a);


The above means: create a new instance of Class on the heap, copy-constructed from a, and give ownership of it to a new std::unique_ptr instance with name ptr.






share|improve this answer























  • What i don't understand is he didn't provide a copy constructor. I know that if you don't create a custom constructor, the compiler will generate for you default, copy, move, etc. But, in this case, he provided a custom constructor.

    – Jordan Motta
    2 days ago






  • 11





    @JordanMotta: adding a non-default constructor does not inhibit implicit generation of the copy constructor. See i.stack.imgur.com/b2VBV.png

    – Vittorio Romeo
    2 days ago



















20















Why am I allowed to create multiple unique pointers from a single object?




You're not allowed to do that*, so it's a good thing you're not doing that!



Don't forget, this:



std::unique_ptr<Class> ptr = std::make_unique<Class>(a);


is this**:



std::unique_ptr<Class> ptr(new Class(a));


not this:



std::unique_ptr<Class> ptr(&a);


std::make_unique creates a thing and gives you a unique_ptr to that thing. It does so by forwarding its arguments to the thing's constructor. Admittedly this can be confusing when you pass in the name of an existing object, leading to the copy constructor being used.



tl;dr: You're creating copies of a.



* Well, with a no-op deleter you could do it safely, but let's save that conversation for another day…
** More or less…






share|improve this answer




















  • 1





    you can prove this to yourself by ptr2->_int = 6; and then 7 and so on. You will see that changing the Class one pointer points to has no effect on the others. They are just a bunch of copies.

    – Kate Gregory
    2 days ago











  • Cunning of you to place the TD;DR version at the bottom of the answer, guaranteeing that most people will read the the full version first. Another simple proof is to remove the copy constructor (Class(const Class &) =delete;) the copy constructor and watch the compiler choke.

    – user4581301
    2 days ago






  • 1





    @user4581301 Conclusions and summaries have been used for centuries. Feel free to write your own answer.

    – Lightness Races in Orbit
    yesterday


















10















Why am I allowed to create multiple unique pointers from a single object?




You're allowed to make multiple copies of the object, because the class is copyable.




"Unique" in unique_ptr doesn't mean that the pointed object is the unique instance of its class. It means that no other pointer should have ownership of the pointed object. In your example, each unique pointer points to a separate instance; each of the uniquely owned by the respective pointer.



You could violate the uniqueness like this:



std::unique_ptr<Class> ptr(&a);


Could: The program would be well-formed and compiler would be required to compile it. But the behaviour of the program would then be undefined, so you very much should not do that.






share|improve this answer

























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    3 Answers
    3






    active

    oldest

    votes








    3 Answers
    3






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    36














    Why not?



    You are not creating multiple unique_ptr instances pointing to the same Class instance, but you are instead allocating three new Class instances on the heap, copy-constructed from a. Every unique_ptr points to a different instance.




    std::unique_ptr<Class> ptr = std::make_unique<Class>(a);


    The above means: create a new instance of Class on the heap, copy-constructed from a, and give ownership of it to a new std::unique_ptr instance with name ptr.






    share|improve this answer























    • What i don't understand is he didn't provide a copy constructor. I know that if you don't create a custom constructor, the compiler will generate for you default, copy, move, etc. But, in this case, he provided a custom constructor.

      – Jordan Motta
      2 days ago






    • 11





      @JordanMotta: adding a non-default constructor does not inhibit implicit generation of the copy constructor. See i.stack.imgur.com/b2VBV.png

      – Vittorio Romeo
      2 days ago
















    36














    Why not?



    You are not creating multiple unique_ptr instances pointing to the same Class instance, but you are instead allocating three new Class instances on the heap, copy-constructed from a. Every unique_ptr points to a different instance.




    std::unique_ptr<Class> ptr = std::make_unique<Class>(a);


    The above means: create a new instance of Class on the heap, copy-constructed from a, and give ownership of it to a new std::unique_ptr instance with name ptr.






    share|improve this answer























    • What i don't understand is he didn't provide a copy constructor. I know that if you don't create a custom constructor, the compiler will generate for you default, copy, move, etc. But, in this case, he provided a custom constructor.

      – Jordan Motta
      2 days ago






    • 11





      @JordanMotta: adding a non-default constructor does not inhibit implicit generation of the copy constructor. See i.stack.imgur.com/b2VBV.png

      – Vittorio Romeo
      2 days ago














    36












    36








    36







    Why not?



    You are not creating multiple unique_ptr instances pointing to the same Class instance, but you are instead allocating three new Class instances on the heap, copy-constructed from a. Every unique_ptr points to a different instance.




    std::unique_ptr<Class> ptr = std::make_unique<Class>(a);


    The above means: create a new instance of Class on the heap, copy-constructed from a, and give ownership of it to a new std::unique_ptr instance with name ptr.






    share|improve this answer













    Why not?



    You are not creating multiple unique_ptr instances pointing to the same Class instance, but you are instead allocating three new Class instances on the heap, copy-constructed from a. Every unique_ptr points to a different instance.




    std::unique_ptr<Class> ptr = std::make_unique<Class>(a);


    The above means: create a new instance of Class on the heap, copy-constructed from a, and give ownership of it to a new std::unique_ptr instance with name ptr.







    share|improve this answer












    share|improve this answer



    share|improve this answer










    answered 2 days ago









    Vittorio RomeoVittorio Romeo

    59.4k17163307




    59.4k17163307












    • What i don't understand is he didn't provide a copy constructor. I know that if you don't create a custom constructor, the compiler will generate for you default, copy, move, etc. But, in this case, he provided a custom constructor.

      – Jordan Motta
      2 days ago






    • 11





      @JordanMotta: adding a non-default constructor does not inhibit implicit generation of the copy constructor. See i.stack.imgur.com/b2VBV.png

      – Vittorio Romeo
      2 days ago


















    • What i don't understand is he didn't provide a copy constructor. I know that if you don't create a custom constructor, the compiler will generate for you default, copy, move, etc. But, in this case, he provided a custom constructor.

      – Jordan Motta
      2 days ago






    • 11





      @JordanMotta: adding a non-default constructor does not inhibit implicit generation of the copy constructor. See i.stack.imgur.com/b2VBV.png

      – Vittorio Romeo
      2 days ago

















    What i don't understand is he didn't provide a copy constructor. I know that if you don't create a custom constructor, the compiler will generate for you default, copy, move, etc. But, in this case, he provided a custom constructor.

    – Jordan Motta
    2 days ago





    What i don't understand is he didn't provide a copy constructor. I know that if you don't create a custom constructor, the compiler will generate for you default, copy, move, etc. But, in this case, he provided a custom constructor.

    – Jordan Motta
    2 days ago




    11




    11





    @JordanMotta: adding a non-default constructor does not inhibit implicit generation of the copy constructor. See i.stack.imgur.com/b2VBV.png

    – Vittorio Romeo
    2 days ago






    @JordanMotta: adding a non-default constructor does not inhibit implicit generation of the copy constructor. See i.stack.imgur.com/b2VBV.png

    – Vittorio Romeo
    2 days ago














    20















    Why am I allowed to create multiple unique pointers from a single object?




    You're not allowed to do that*, so it's a good thing you're not doing that!



    Don't forget, this:



    std::unique_ptr<Class> ptr = std::make_unique<Class>(a);


    is this**:



    std::unique_ptr<Class> ptr(new Class(a));


    not this:



    std::unique_ptr<Class> ptr(&a);


    std::make_unique creates a thing and gives you a unique_ptr to that thing. It does so by forwarding its arguments to the thing's constructor. Admittedly this can be confusing when you pass in the name of an existing object, leading to the copy constructor being used.



    tl;dr: You're creating copies of a.



    * Well, with a no-op deleter you could do it safely, but let's save that conversation for another day…
    ** More or less…






    share|improve this answer




















    • 1





      you can prove this to yourself by ptr2->_int = 6; and then 7 and so on. You will see that changing the Class one pointer points to has no effect on the others. They are just a bunch of copies.

      – Kate Gregory
      2 days ago











    • Cunning of you to place the TD;DR version at the bottom of the answer, guaranteeing that most people will read the the full version first. Another simple proof is to remove the copy constructor (Class(const Class &) =delete;) the copy constructor and watch the compiler choke.

      – user4581301
      2 days ago






    • 1





      @user4581301 Conclusions and summaries have been used for centuries. Feel free to write your own answer.

      – Lightness Races in Orbit
      yesterday















    20















    Why am I allowed to create multiple unique pointers from a single object?




    You're not allowed to do that*, so it's a good thing you're not doing that!



    Don't forget, this:



    std::unique_ptr<Class> ptr = std::make_unique<Class>(a);


    is this**:



    std::unique_ptr<Class> ptr(new Class(a));


    not this:



    std::unique_ptr<Class> ptr(&a);


    std::make_unique creates a thing and gives you a unique_ptr to that thing. It does so by forwarding its arguments to the thing's constructor. Admittedly this can be confusing when you pass in the name of an existing object, leading to the copy constructor being used.



    tl;dr: You're creating copies of a.



    * Well, with a no-op deleter you could do it safely, but let's save that conversation for another day…
    ** More or less…






    share|improve this answer




















    • 1





      you can prove this to yourself by ptr2->_int = 6; and then 7 and so on. You will see that changing the Class one pointer points to has no effect on the others. They are just a bunch of copies.

      – Kate Gregory
      2 days ago











    • Cunning of you to place the TD;DR version at the bottom of the answer, guaranteeing that most people will read the the full version first. Another simple proof is to remove the copy constructor (Class(const Class &) =delete;) the copy constructor and watch the compiler choke.

      – user4581301
      2 days ago






    • 1





      @user4581301 Conclusions and summaries have been used for centuries. Feel free to write your own answer.

      – Lightness Races in Orbit
      yesterday













    20












    20








    20








    Why am I allowed to create multiple unique pointers from a single object?




    You're not allowed to do that*, so it's a good thing you're not doing that!



    Don't forget, this:



    std::unique_ptr<Class> ptr = std::make_unique<Class>(a);


    is this**:



    std::unique_ptr<Class> ptr(new Class(a));


    not this:



    std::unique_ptr<Class> ptr(&a);


    std::make_unique creates a thing and gives you a unique_ptr to that thing. It does so by forwarding its arguments to the thing's constructor. Admittedly this can be confusing when you pass in the name of an existing object, leading to the copy constructor being used.



    tl;dr: You're creating copies of a.



    * Well, with a no-op deleter you could do it safely, but let's save that conversation for another day…
    ** More or less…






    share|improve this answer
















    Why am I allowed to create multiple unique pointers from a single object?




    You're not allowed to do that*, so it's a good thing you're not doing that!



    Don't forget, this:



    std::unique_ptr<Class> ptr = std::make_unique<Class>(a);


    is this**:



    std::unique_ptr<Class> ptr(new Class(a));


    not this:



    std::unique_ptr<Class> ptr(&a);


    std::make_unique creates a thing and gives you a unique_ptr to that thing. It does so by forwarding its arguments to the thing's constructor. Admittedly this can be confusing when you pass in the name of an existing object, leading to the copy constructor being used.



    tl;dr: You're creating copies of a.



    * Well, with a no-op deleter you could do it safely, but let's save that conversation for another day…
    ** More or less…







    share|improve this answer














    share|improve this answer



    share|improve this answer








    edited 2 days ago

























    answered 2 days ago









    Lightness Races in OrbitLightness Races in Orbit

    294k54477811




    294k54477811







    • 1





      you can prove this to yourself by ptr2->_int = 6; and then 7 and so on. You will see that changing the Class one pointer points to has no effect on the others. They are just a bunch of copies.

      – Kate Gregory
      2 days ago











    • Cunning of you to place the TD;DR version at the bottom of the answer, guaranteeing that most people will read the the full version first. Another simple proof is to remove the copy constructor (Class(const Class &) =delete;) the copy constructor and watch the compiler choke.

      – user4581301
      2 days ago






    • 1





      @user4581301 Conclusions and summaries have been used for centuries. Feel free to write your own answer.

      – Lightness Races in Orbit
      yesterday












    • 1





      you can prove this to yourself by ptr2->_int = 6; and then 7 and so on. You will see that changing the Class one pointer points to has no effect on the others. They are just a bunch of copies.

      – Kate Gregory
      2 days ago











    • Cunning of you to place the TD;DR version at the bottom of the answer, guaranteeing that most people will read the the full version first. Another simple proof is to remove the copy constructor (Class(const Class &) =delete;) the copy constructor and watch the compiler choke.

      – user4581301
      2 days ago






    • 1





      @user4581301 Conclusions and summaries have been used for centuries. Feel free to write your own answer.

      – Lightness Races in Orbit
      yesterday







    1




    1





    you can prove this to yourself by ptr2->_int = 6; and then 7 and so on. You will see that changing the Class one pointer points to has no effect on the others. They are just a bunch of copies.

    – Kate Gregory
    2 days ago





    you can prove this to yourself by ptr2->_int = 6; and then 7 and so on. You will see that changing the Class one pointer points to has no effect on the others. They are just a bunch of copies.

    – Kate Gregory
    2 days ago













    Cunning of you to place the TD;DR version at the bottom of the answer, guaranteeing that most people will read the the full version first. Another simple proof is to remove the copy constructor (Class(const Class &) =delete;) the copy constructor and watch the compiler choke.

    – user4581301
    2 days ago





    Cunning of you to place the TD;DR version at the bottom of the answer, guaranteeing that most people will read the the full version first. Another simple proof is to remove the copy constructor (Class(const Class &) =delete;) the copy constructor and watch the compiler choke.

    – user4581301
    2 days ago




    1




    1





    @user4581301 Conclusions and summaries have been used for centuries. Feel free to write your own answer.

    – Lightness Races in Orbit
    yesterday





    @user4581301 Conclusions and summaries have been used for centuries. Feel free to write your own answer.

    – Lightness Races in Orbit
    yesterday











    10















    Why am I allowed to create multiple unique pointers from a single object?




    You're allowed to make multiple copies of the object, because the class is copyable.




    "Unique" in unique_ptr doesn't mean that the pointed object is the unique instance of its class. It means that no other pointer should have ownership of the pointed object. In your example, each unique pointer points to a separate instance; each of the uniquely owned by the respective pointer.



    You could violate the uniqueness like this:



    std::unique_ptr<Class> ptr(&a);


    Could: The program would be well-formed and compiler would be required to compile it. But the behaviour of the program would then be undefined, so you very much should not do that.






    share|improve this answer





























      10















      Why am I allowed to create multiple unique pointers from a single object?




      You're allowed to make multiple copies of the object, because the class is copyable.




      "Unique" in unique_ptr doesn't mean that the pointed object is the unique instance of its class. It means that no other pointer should have ownership of the pointed object. In your example, each unique pointer points to a separate instance; each of the uniquely owned by the respective pointer.



      You could violate the uniqueness like this:



      std::unique_ptr<Class> ptr(&a);


      Could: The program would be well-formed and compiler would be required to compile it. But the behaviour of the program would then be undefined, so you very much should not do that.






      share|improve this answer



























        10












        10








        10








        Why am I allowed to create multiple unique pointers from a single object?




        You're allowed to make multiple copies of the object, because the class is copyable.




        "Unique" in unique_ptr doesn't mean that the pointed object is the unique instance of its class. It means that no other pointer should have ownership of the pointed object. In your example, each unique pointer points to a separate instance; each of the uniquely owned by the respective pointer.



        You could violate the uniqueness like this:



        std::unique_ptr<Class> ptr(&a);


        Could: The program would be well-formed and compiler would be required to compile it. But the behaviour of the program would then be undefined, so you very much should not do that.






        share|improve this answer
















        Why am I allowed to create multiple unique pointers from a single object?




        You're allowed to make multiple copies of the object, because the class is copyable.




        "Unique" in unique_ptr doesn't mean that the pointed object is the unique instance of its class. It means that no other pointer should have ownership of the pointed object. In your example, each unique pointer points to a separate instance; each of the uniquely owned by the respective pointer.



        You could violate the uniqueness like this:



        std::unique_ptr<Class> ptr(&a);


        Could: The program would be well-formed and compiler would be required to compile it. But the behaviour of the program would then be undefined, so you very much should not do that.







        share|improve this answer














        share|improve this answer



        share|improve this answer








        edited 2 days ago

























        answered 2 days ago









        eerorikaeerorika

        88.4k663135




        88.4k663135




















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