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



The Next CEO of Stack OverflowWhy is 'this' a pointer and not a reference?Cannot assign pointer in a self-referential object in Visual Studio 2010Pretty-print C++ STL containersHow C++ reference worksWhy is reading lines from stdin much slower in C++ than Python?Namespaces and the Pre-ProcessorWhy should I use a pointer rather than the object itself?Returning a STL container of unique pointersWhy does my object appear to be on the heap without using `new`?Print Object that is pointed to










11















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;










share|improve this question









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















  • 15





    Copy construction.

    – user4581301
    14 hours ago






  • 1





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

    – alter igel
    14 hours 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
    14 hours ago















11















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;










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.















  • 15





    Copy construction.

    – user4581301
    14 hours ago






  • 1





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

    – alter igel
    14 hours 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
    14 hours ago













11












11








11


1






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;










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;







c++






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 14 hours ago









Shridhar R Kulkarni

1,55821328




1,55821328






New contributor




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asked 14 hours ago









Lokas BeardLokas Beard

585




585




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.







  • 15





    Copy construction.

    – user4581301
    14 hours ago






  • 1





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

    – alter igel
    14 hours 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
    14 hours ago












  • 15





    Copy construction.

    – user4581301
    14 hours ago






  • 1





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

    – alter igel
    14 hours 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
    14 hours ago







15




15





Copy construction.

– user4581301
14 hours ago





Copy construction.

– user4581301
14 hours 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
14 hours ago





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

– alter igel
14 hours 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
14 hours 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
14 hours ago












3 Answers
3






active

oldest

votes


















28














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
    13 hours ago






  • 9





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

    – Vittorio Romeo
    13 hours ago



















14















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
    12 hours 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
    12 hours ago


















7















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









    28














    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
      13 hours ago






    • 9





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

      – Vittorio Romeo
      13 hours ago
















    28














    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
      13 hours ago






    • 9





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

      – Vittorio Romeo
      13 hours ago














    28












    28








    28







    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 14 hours ago









    Vittorio RomeoVittorio Romeo

    59.3k17163307




    59.3k17163307












    • 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
      13 hours ago






    • 9





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

      – Vittorio Romeo
      13 hours 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
      13 hours ago






    • 9





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

      – Vittorio Romeo
      13 hours 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
    13 hours 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
    13 hours ago




    9




    9





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

    – Vittorio Romeo
    13 hours 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
    13 hours ago














    14















    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
      12 hours 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
      12 hours ago















    14















    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
      12 hours 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
      12 hours ago













    14












    14








    14








    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 13 hours ago

























    answered 13 hours 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
      12 hours 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
      12 hours ago












    • 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
      12 hours 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
      12 hours ago







    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
    12 hours 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
    12 hours 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
    12 hours 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
    12 hours ago











    7















    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





























      7















      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



























        7












        7








        7








        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 13 hours ago

























        answered 13 hours ago









        eerorikaeerorika

        88.2k663134




        88.2k663134




















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