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


Why 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`?






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24















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



















  • 21





    Copy construction.

    – user4581301
    Mar 29 at 15:01






  • 1





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

    – alter igel
    Mar 29 at 15:08






  • 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
    Mar 29 at 15:08











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

    – TrebledJ
    Mar 30 at 14:31

















24















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



















  • 21





    Copy construction.

    – user4581301
    Mar 29 at 15:01






  • 1





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

    – alter igel
    Mar 29 at 15:08






  • 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
    Mar 29 at 15:08











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

    – TrebledJ
    Mar 30 at 14:31













24












24








24


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
















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















share|improve this question













share|improve this question




share|improve this question








edited Mar 30 at 14:25









TrebledJ

4,10821432




4,10821432










asked Mar 29 at 15:00









Lokas BeardLokas Beard

1626




1626







  • 21





    Copy construction.

    – user4581301
    Mar 29 at 15:01






  • 1





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

    – alter igel
    Mar 29 at 15:08






  • 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
    Mar 29 at 15:08











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

    – TrebledJ
    Mar 30 at 14:31












  • 21





    Copy construction.

    – user4581301
    Mar 29 at 15:01






  • 1





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

    – alter igel
    Mar 29 at 15:08






  • 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
    Mar 29 at 15:08











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

    – TrebledJ
    Mar 30 at 14:31







21




21





Copy construction.

– user4581301
Mar 29 at 15:01





Copy construction.

– user4581301
Mar 29 at 15:01




1




1





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

– alter igel
Mar 29 at 15:08





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

– alter igel
Mar 29 at 15:08




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
Mar 29 at 15:08





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
Mar 29 at 15:08













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

– TrebledJ
Mar 30 at 14:31





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

– TrebledJ
Mar 30 at 14:31












3 Answers
3






active

oldest

votes


















40














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
    Mar 29 at 15:48






  • 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
    Mar 29 at 16:10



















21















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
    Mar 29 at 16:26











  • 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
    Mar 29 at 16:51






  • 1





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

    – Lightness Races in Orbit
    Mar 30 at 17:26


















11















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









    40














    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
      Mar 29 at 15:48






    • 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
      Mar 29 at 16:10
















    40














    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
      Mar 29 at 15:48






    • 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
      Mar 29 at 16:10














    40












    40








    40







    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 Mar 29 at 15:03









    Vittorio RomeoVittorio Romeo

    60.4k17168312




    60.4k17168312












    • 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
      Mar 29 at 15:48






    • 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
      Mar 29 at 16:10


















    • 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
      Mar 29 at 15:48






    • 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
      Mar 29 at 16:10

















    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
    Mar 29 at 15:48





    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
    Mar 29 at 15:48




    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
    Mar 29 at 16:10






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

    – Vittorio Romeo
    Mar 29 at 16:10














    21















    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
      Mar 29 at 16:26











    • 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
      Mar 29 at 16:51






    • 1





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

      – Lightness Races in Orbit
      Mar 30 at 17:26















    21















    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
      Mar 29 at 16:26











    • 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
      Mar 29 at 16:51






    • 1





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

      – Lightness Races in Orbit
      Mar 30 at 17:26













    21












    21








    21








    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 Mar 29 at 15:23

























    answered Mar 29 at 15:17









    Lightness Races in OrbitLightness Races in Orbit

    297k55480822




    297k55480822







    • 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
      Mar 29 at 16:26











    • 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
      Mar 29 at 16:51






    • 1





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

      – Lightness Races in Orbit
      Mar 30 at 17:26












    • 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
      Mar 29 at 16:26











    • 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
      Mar 29 at 16:51






    • 1





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

      – Lightness Races in Orbit
      Mar 30 at 17:26







    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
    Mar 29 at 16:26





    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
    Mar 29 at 16:26













    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
    Mar 29 at 16:51





    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
    Mar 29 at 16:51




    1




    1





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

    – Lightness Races in Orbit
    Mar 30 at 17:26





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

    – Lightness Races in Orbit
    Mar 30 at 17:26











    11















    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





























      11















      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



























        11












        11








        11








        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 Mar 29 at 15:25

























        answered Mar 29 at 15:12









        eerorikaeerorika

        90.8k666137




        90.8k666137



























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