Idiomatic way to prevent slicing? Announcing the arrival of Valued Associate #679: Cesar Manara Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern) Data science time! April 2019 and salary with experience Should we burninate the [wrap] tag? The Ask Question Wizard is Live!Force function to be called only with specific typesWhat is object slicing?How to initialize a const field in constructor?What's the point of g++ -Wreorder?Java: Can Vector<derived> be called as Vector<base>?Easiest way to convert int to string in C++capture variables inside of subclass?Detecting if a type can be derived from in C++Implementing polymorphic operator==() in C++ idiomatic wayIs using inline classes inside a function permitted to be used as template types?Short-circuit evaluation and assignment in C++

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Idiomatic way to prevent slicing?



Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)
Data science time! April 2019 and salary with experience
Should we burninate the [wrap] tag?
The Ask Question Wizard is Live!Force function to be called only with specific typesWhat is object slicing?How to initialize a const field in constructor?What's the point of g++ -Wreorder?Java: Can Vector<derived> be called as Vector<base>?Easiest way to convert int to string in C++capture variables inside of subclass?Detecting if a type can be derived from in C++Implementing polymorphic operator==() in C++ idiomatic wayIs using inline classes inside a function permitted to be used as template types?Short-circuit evaluation and assignment in C++



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15















Sometimes it can be an annoyance that c++ defaults to allow slicing. For example



struct foo int a; ;
struct bar : foo int b; ;

int main()
bar x1,2;
foo y = x; // <- I dont want this to compile!



This compiles and runs as expected! Though, what if I dont want to enable slicing?



What is the idiomatic way to write foo such that one cannot slice instances of any derived class?










share|improve this question
























  • The easiest way is to not use inheritance. You have have foo be a member variable of bar.

    – KevinZ
    Apr 12 at 13:50











  • @KevinZ stackoverflow.com/a/55600208/4117728

    – user463035818
    Apr 12 at 13:51

















15















Sometimes it can be an annoyance that c++ defaults to allow slicing. For example



struct foo int a; ;
struct bar : foo int b; ;

int main()
bar x1,2;
foo y = x; // <- I dont want this to compile!



This compiles and runs as expected! Though, what if I dont want to enable slicing?



What is the idiomatic way to write foo such that one cannot slice instances of any derived class?










share|improve this question
























  • The easiest way is to not use inheritance. You have have foo be a member variable of bar.

    – KevinZ
    Apr 12 at 13:50











  • @KevinZ stackoverflow.com/a/55600208/4117728

    – user463035818
    Apr 12 at 13:51













15












15








15


4






Sometimes it can be an annoyance that c++ defaults to allow slicing. For example



struct foo int a; ;
struct bar : foo int b; ;

int main()
bar x1,2;
foo y = x; // <- I dont want this to compile!



This compiles and runs as expected! Though, what if I dont want to enable slicing?



What is the idiomatic way to write foo such that one cannot slice instances of any derived class?










share|improve this question
















Sometimes it can be an annoyance that c++ defaults to allow slicing. For example



struct foo int a; ;
struct bar : foo int b; ;

int main()
bar x1,2;
foo y = x; // <- I dont want this to compile!



This compiles and runs as expected! Though, what if I dont want to enable slicing?



What is the idiomatic way to write foo such that one cannot slice instances of any derived class?







c++ inheritance object-slicing






share|improve this question















share|improve this question













share|improve this question




share|improve this question








edited Apr 12 at 21:48







user463035818

















asked Apr 9 at 19:32









user463035818user463035818

19.3k42971




19.3k42971












  • The easiest way is to not use inheritance. You have have foo be a member variable of bar.

    – KevinZ
    Apr 12 at 13:50











  • @KevinZ stackoverflow.com/a/55600208/4117728

    – user463035818
    Apr 12 at 13:51

















  • The easiest way is to not use inheritance. You have have foo be a member variable of bar.

    – KevinZ
    Apr 12 at 13:50











  • @KevinZ stackoverflow.com/a/55600208/4117728

    – user463035818
    Apr 12 at 13:51
















The easiest way is to not use inheritance. You have have foo be a member variable of bar.

– KevinZ
Apr 12 at 13:50





The easiest way is to not use inheritance. You have have foo be a member variable of bar.

– KevinZ
Apr 12 at 13:50













@KevinZ stackoverflow.com/a/55600208/4117728

– user463035818
Apr 12 at 13:51





@KevinZ stackoverflow.com/a/55600208/4117728

– user463035818
Apr 12 at 13:51












3 Answers
3






active

oldest

votes


















16














I'm not sure if there is a named idiom for it but you can add a deleted function to the overload set that is a better match then the base classes slicing operations. If you change foo to



struct foo 

int a;
foo() = default; // you have to add this because of the template constructor

template<typename T>
foo(const T&) = delete; // error trying to copy anything but a foo

template<typename T>
foo& operator=(const T&) = delete; // error assigning anything else but a foo
;


then you can only ever copy construct or copy assign a foo to foo. Any other type will pick the function template and you'll get an error about using a deleted function. This does mean that your class, and the classes that use it can no longer be an aggregate though. Since the members that are added are templates, they are not considered copy constructors or copy assignment operators so you'll get the default copy and move constructors and assignment operators.






share|improve this answer

























  • Note that this doesn't prevent explicit slicing like this: foo y = static_cast<foo&>(x);. That said, perhaps it's not a problem to OP.

    – eerorika
    Apr 9 at 20:02











  • if I understand correctly this is a nice way to prevent implicit conversions for function parameters in general

    – user463035818
    Apr 9 at 20:05






  • 1





    @user463035818 Yep. I've been using it since I've asked that Q.

    – NathanOliver
    Apr 9 at 20:06






  • 4





    I look at it as reverse SFINAE. You make the overloads you want to compile, and then add a deleted template stopping everything else.

    – NathanOliver
    Apr 9 at 20:08












  • actually I was a bit hestitant to accept this answer. The technique is great, but in fact it opens the door to specializing all kinds of unwanted assignments, though if I have to choose between the javaish "protect against every possible stupidity at any cost" vs a pythonic "we are all adults" then I know what to pick ;)

    – user463035818
    Apr 10 at 8:23


















5














Since 2011, the idiomatic way has been to use auto:



#include <iostream>
struct foo int a; ;
struct bar : foo int b; ;

int main()
bar x1,2;
auto y = x; // <- y is a bar



If you wish to actively prevent slicing, there are a number of ways:



Usually the most preferable way, unless you specifically need inheritance (you often don't) is to use encapsulation:



#include <iostream>

struct foo int a; ;
struct bar

bar(int a, int b)
: foo_(a)
, b(b)


int b;

int get_a() const return foo_.a;

private:
foo foo_;
;

int main()
bar x1,2;
// foo y = x; // <- does not compile




Another more specialised way might be to alter the permissions around copy operators:



#include <iostream>

struct foo
int a;
protected:
foo(foo const&) = default;
foo(foo&&) = default;
foo& operator=(foo const&) = default;
foo& operator=(foo&&) = default;

;

struct bar : foo

bar(int a, int b)
: fooa, bb


int b;
;

int main()
auto x = bar (1,2);
// foo y = x; // <- does not compile






share|improve this answer






























    3














    You can prevent the base from being copied outside of member functions of derived classes and the base itself by declaring the copy constructor protected:



    struct foo 
    // ...
    protected:
    foo(foo&) = default;
    ;





    share|improve this answer


















    • 6





      but then I cannot copy foos anymore :( I'd like to prevent only copying a bar to a foo if possible

      – user463035818
      Apr 9 at 19:42











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    3 Answers
    3






    active

    oldest

    votes








    3 Answers
    3






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    16














    I'm not sure if there is a named idiom for it but you can add a deleted function to the overload set that is a better match then the base classes slicing operations. If you change foo to



    struct foo 

    int a;
    foo() = default; // you have to add this because of the template constructor

    template<typename T>
    foo(const T&) = delete; // error trying to copy anything but a foo

    template<typename T>
    foo& operator=(const T&) = delete; // error assigning anything else but a foo
    ;


    then you can only ever copy construct or copy assign a foo to foo. Any other type will pick the function template and you'll get an error about using a deleted function. This does mean that your class, and the classes that use it can no longer be an aggregate though. Since the members that are added are templates, they are not considered copy constructors or copy assignment operators so you'll get the default copy and move constructors and assignment operators.






    share|improve this answer

























    • Note that this doesn't prevent explicit slicing like this: foo y = static_cast<foo&>(x);. That said, perhaps it's not a problem to OP.

      – eerorika
      Apr 9 at 20:02











    • if I understand correctly this is a nice way to prevent implicit conversions for function parameters in general

      – user463035818
      Apr 9 at 20:05






    • 1





      @user463035818 Yep. I've been using it since I've asked that Q.

      – NathanOliver
      Apr 9 at 20:06






    • 4





      I look at it as reverse SFINAE. You make the overloads you want to compile, and then add a deleted template stopping everything else.

      – NathanOliver
      Apr 9 at 20:08












    • actually I was a bit hestitant to accept this answer. The technique is great, but in fact it opens the door to specializing all kinds of unwanted assignments, though if I have to choose between the javaish "protect against every possible stupidity at any cost" vs a pythonic "we are all adults" then I know what to pick ;)

      – user463035818
      Apr 10 at 8:23















    16














    I'm not sure if there is a named idiom for it but you can add a deleted function to the overload set that is a better match then the base classes slicing operations. If you change foo to



    struct foo 

    int a;
    foo() = default; // you have to add this because of the template constructor

    template<typename T>
    foo(const T&) = delete; // error trying to copy anything but a foo

    template<typename T>
    foo& operator=(const T&) = delete; // error assigning anything else but a foo
    ;


    then you can only ever copy construct or copy assign a foo to foo. Any other type will pick the function template and you'll get an error about using a deleted function. This does mean that your class, and the classes that use it can no longer be an aggregate though. Since the members that are added are templates, they are not considered copy constructors or copy assignment operators so you'll get the default copy and move constructors and assignment operators.






    share|improve this answer

























    • Note that this doesn't prevent explicit slicing like this: foo y = static_cast<foo&>(x);. That said, perhaps it's not a problem to OP.

      – eerorika
      Apr 9 at 20:02











    • if I understand correctly this is a nice way to prevent implicit conversions for function parameters in general

      – user463035818
      Apr 9 at 20:05






    • 1





      @user463035818 Yep. I've been using it since I've asked that Q.

      – NathanOliver
      Apr 9 at 20:06






    • 4





      I look at it as reverse SFINAE. You make the overloads you want to compile, and then add a deleted template stopping everything else.

      – NathanOliver
      Apr 9 at 20:08












    • actually I was a bit hestitant to accept this answer. The technique is great, but in fact it opens the door to specializing all kinds of unwanted assignments, though if I have to choose between the javaish "protect against every possible stupidity at any cost" vs a pythonic "we are all adults" then I know what to pick ;)

      – user463035818
      Apr 10 at 8:23













    16












    16








    16







    I'm not sure if there is a named idiom for it but you can add a deleted function to the overload set that is a better match then the base classes slicing operations. If you change foo to



    struct foo 

    int a;
    foo() = default; // you have to add this because of the template constructor

    template<typename T>
    foo(const T&) = delete; // error trying to copy anything but a foo

    template<typename T>
    foo& operator=(const T&) = delete; // error assigning anything else but a foo
    ;


    then you can only ever copy construct or copy assign a foo to foo. Any other type will pick the function template and you'll get an error about using a deleted function. This does mean that your class, and the classes that use it can no longer be an aggregate though. Since the members that are added are templates, they are not considered copy constructors or copy assignment operators so you'll get the default copy and move constructors and assignment operators.






    share|improve this answer















    I'm not sure if there is a named idiom for it but you can add a deleted function to the overload set that is a better match then the base classes slicing operations. If you change foo to



    struct foo 

    int a;
    foo() = default; // you have to add this because of the template constructor

    template<typename T>
    foo(const T&) = delete; // error trying to copy anything but a foo

    template<typename T>
    foo& operator=(const T&) = delete; // error assigning anything else but a foo
    ;


    then you can only ever copy construct or copy assign a foo to foo. Any other type will pick the function template and you'll get an error about using a deleted function. This does mean that your class, and the classes that use it can no longer be an aggregate though. Since the members that are added are templates, they are not considered copy constructors or copy assignment operators so you'll get the default copy and move constructors and assignment operators.







    share|improve this answer














    share|improve this answer



    share|improve this answer








    edited Apr 9 at 19:56

























    answered Apr 9 at 19:43









    NathanOliverNathanOliver

    99.1k16138219




    99.1k16138219












    • Note that this doesn't prevent explicit slicing like this: foo y = static_cast<foo&>(x);. That said, perhaps it's not a problem to OP.

      – eerorika
      Apr 9 at 20:02











    • if I understand correctly this is a nice way to prevent implicit conversions for function parameters in general

      – user463035818
      Apr 9 at 20:05






    • 1





      @user463035818 Yep. I've been using it since I've asked that Q.

      – NathanOliver
      Apr 9 at 20:06






    • 4





      I look at it as reverse SFINAE. You make the overloads you want to compile, and then add a deleted template stopping everything else.

      – NathanOliver
      Apr 9 at 20:08












    • actually I was a bit hestitant to accept this answer. The technique is great, but in fact it opens the door to specializing all kinds of unwanted assignments, though if I have to choose between the javaish "protect against every possible stupidity at any cost" vs a pythonic "we are all adults" then I know what to pick ;)

      – user463035818
      Apr 10 at 8:23

















    • Note that this doesn't prevent explicit slicing like this: foo y = static_cast<foo&>(x);. That said, perhaps it's not a problem to OP.

      – eerorika
      Apr 9 at 20:02











    • if I understand correctly this is a nice way to prevent implicit conversions for function parameters in general

      – user463035818
      Apr 9 at 20:05






    • 1





      @user463035818 Yep. I've been using it since I've asked that Q.

      – NathanOliver
      Apr 9 at 20:06






    • 4





      I look at it as reverse SFINAE. You make the overloads you want to compile, and then add a deleted template stopping everything else.

      – NathanOliver
      Apr 9 at 20:08












    • actually I was a bit hestitant to accept this answer. The technique is great, but in fact it opens the door to specializing all kinds of unwanted assignments, though if I have to choose between the javaish "protect against every possible stupidity at any cost" vs a pythonic "we are all adults" then I know what to pick ;)

      – user463035818
      Apr 10 at 8:23
















    Note that this doesn't prevent explicit slicing like this: foo y = static_cast<foo&>(x);. That said, perhaps it's not a problem to OP.

    – eerorika
    Apr 9 at 20:02





    Note that this doesn't prevent explicit slicing like this: foo y = static_cast<foo&>(x);. That said, perhaps it's not a problem to OP.

    – eerorika
    Apr 9 at 20:02













    if I understand correctly this is a nice way to prevent implicit conversions for function parameters in general

    – user463035818
    Apr 9 at 20:05





    if I understand correctly this is a nice way to prevent implicit conversions for function parameters in general

    – user463035818
    Apr 9 at 20:05




    1




    1





    @user463035818 Yep. I've been using it since I've asked that Q.

    – NathanOliver
    Apr 9 at 20:06





    @user463035818 Yep. I've been using it since I've asked that Q.

    – NathanOliver
    Apr 9 at 20:06




    4




    4





    I look at it as reverse SFINAE. You make the overloads you want to compile, and then add a deleted template stopping everything else.

    – NathanOliver
    Apr 9 at 20:08






    I look at it as reverse SFINAE. You make the overloads you want to compile, and then add a deleted template stopping everything else.

    – NathanOliver
    Apr 9 at 20:08














    actually I was a bit hestitant to accept this answer. The technique is great, but in fact it opens the door to specializing all kinds of unwanted assignments, though if I have to choose between the javaish "protect against every possible stupidity at any cost" vs a pythonic "we are all adults" then I know what to pick ;)

    – user463035818
    Apr 10 at 8:23





    actually I was a bit hestitant to accept this answer. The technique is great, but in fact it opens the door to specializing all kinds of unwanted assignments, though if I have to choose between the javaish "protect against every possible stupidity at any cost" vs a pythonic "we are all adults" then I know what to pick ;)

    – user463035818
    Apr 10 at 8:23













    5














    Since 2011, the idiomatic way has been to use auto:



    #include <iostream>
    struct foo int a; ;
    struct bar : foo int b; ;

    int main()
    bar x1,2;
    auto y = x; // <- y is a bar



    If you wish to actively prevent slicing, there are a number of ways:



    Usually the most preferable way, unless you specifically need inheritance (you often don't) is to use encapsulation:



    #include <iostream>

    struct foo int a; ;
    struct bar

    bar(int a, int b)
    : foo_(a)
    , b(b)


    int b;

    int get_a() const return foo_.a;

    private:
    foo foo_;
    ;

    int main()
    bar x1,2;
    // foo y = x; // <- does not compile




    Another more specialised way might be to alter the permissions around copy operators:



    #include <iostream>

    struct foo
    int a;
    protected:
    foo(foo const&) = default;
    foo(foo&&) = default;
    foo& operator=(foo const&) = default;
    foo& operator=(foo&&) = default;

    ;

    struct bar : foo

    bar(int a, int b)
    : fooa, bb


    int b;
    ;

    int main()
    auto x = bar (1,2);
    // foo y = x; // <- does not compile






    share|improve this answer



























      5














      Since 2011, the idiomatic way has been to use auto:



      #include <iostream>
      struct foo int a; ;
      struct bar : foo int b; ;

      int main()
      bar x1,2;
      auto y = x; // <- y is a bar



      If you wish to actively prevent slicing, there are a number of ways:



      Usually the most preferable way, unless you specifically need inheritance (you often don't) is to use encapsulation:



      #include <iostream>

      struct foo int a; ;
      struct bar

      bar(int a, int b)
      : foo_(a)
      , b(b)


      int b;

      int get_a() const return foo_.a;

      private:
      foo foo_;
      ;

      int main()
      bar x1,2;
      // foo y = x; // <- does not compile




      Another more specialised way might be to alter the permissions around copy operators:



      #include <iostream>

      struct foo
      int a;
      protected:
      foo(foo const&) = default;
      foo(foo&&) = default;
      foo& operator=(foo const&) = default;
      foo& operator=(foo&&) = default;

      ;

      struct bar : foo

      bar(int a, int b)
      : fooa, bb


      int b;
      ;

      int main()
      auto x = bar (1,2);
      // foo y = x; // <- does not compile






      share|improve this answer

























        5












        5








        5







        Since 2011, the idiomatic way has been to use auto:



        #include <iostream>
        struct foo int a; ;
        struct bar : foo int b; ;

        int main()
        bar x1,2;
        auto y = x; // <- y is a bar



        If you wish to actively prevent slicing, there are a number of ways:



        Usually the most preferable way, unless you specifically need inheritance (you often don't) is to use encapsulation:



        #include <iostream>

        struct foo int a; ;
        struct bar

        bar(int a, int b)
        : foo_(a)
        , b(b)


        int b;

        int get_a() const return foo_.a;

        private:
        foo foo_;
        ;

        int main()
        bar x1,2;
        // foo y = x; // <- does not compile




        Another more specialised way might be to alter the permissions around copy operators:



        #include <iostream>

        struct foo
        int a;
        protected:
        foo(foo const&) = default;
        foo(foo&&) = default;
        foo& operator=(foo const&) = default;
        foo& operator=(foo&&) = default;

        ;

        struct bar : foo

        bar(int a, int b)
        : fooa, bb


        int b;
        ;

        int main()
        auto x = bar (1,2);
        // foo y = x; // <- does not compile






        share|improve this answer













        Since 2011, the idiomatic way has been to use auto:



        #include <iostream>
        struct foo int a; ;
        struct bar : foo int b; ;

        int main()
        bar x1,2;
        auto y = x; // <- y is a bar



        If you wish to actively prevent slicing, there are a number of ways:



        Usually the most preferable way, unless you specifically need inheritance (you often don't) is to use encapsulation:



        #include <iostream>

        struct foo int a; ;
        struct bar

        bar(int a, int b)
        : foo_(a)
        , b(b)


        int b;

        int get_a() const return foo_.a;

        private:
        foo foo_;
        ;

        int main()
        bar x1,2;
        // foo y = x; // <- does not compile




        Another more specialised way might be to alter the permissions around copy operators:



        #include <iostream>

        struct foo
        int a;
        protected:
        foo(foo const&) = default;
        foo(foo&&) = default;
        foo& operator=(foo const&) = default;
        foo& operator=(foo&&) = default;

        ;

        struct bar : foo

        bar(int a, int b)
        : fooa, bb


        int b;
        ;

        int main()
        auto x = bar (1,2);
        // foo y = x; // <- does not compile







        share|improve this answer












        share|improve this answer



        share|improve this answer










        answered Apr 9 at 19:43









        Richard HodgesRichard Hodges

        57.1k658105




        57.1k658105





















            3














            You can prevent the base from being copied outside of member functions of derived classes and the base itself by declaring the copy constructor protected:



            struct foo 
            // ...
            protected:
            foo(foo&) = default;
            ;





            share|improve this answer


















            • 6





              but then I cannot copy foos anymore :( I'd like to prevent only copying a bar to a foo if possible

              – user463035818
              Apr 9 at 19:42















            3














            You can prevent the base from being copied outside of member functions of derived classes and the base itself by declaring the copy constructor protected:



            struct foo 
            // ...
            protected:
            foo(foo&) = default;
            ;





            share|improve this answer


















            • 6





              but then I cannot copy foos anymore :( I'd like to prevent only copying a bar to a foo if possible

              – user463035818
              Apr 9 at 19:42













            3












            3








            3







            You can prevent the base from being copied outside of member functions of derived classes and the base itself by declaring the copy constructor protected:



            struct foo 
            // ...
            protected:
            foo(foo&) = default;
            ;





            share|improve this answer













            You can prevent the base from being copied outside of member functions of derived classes and the base itself by declaring the copy constructor protected:



            struct foo 
            // ...
            protected:
            foo(foo&) = default;
            ;






            share|improve this answer












            share|improve this answer



            share|improve this answer










            answered Apr 9 at 19:41









            eerorikaeerorika

            90.2k665137




            90.2k665137







            • 6





              but then I cannot copy foos anymore :( I'd like to prevent only copying a bar to a foo if possible

              – user463035818
              Apr 9 at 19:42












            • 6





              but then I cannot copy foos anymore :( I'd like to prevent only copying a bar to a foo if possible

              – user463035818
              Apr 9 at 19:42







            6




            6





            but then I cannot copy foos anymore :( I'd like to prevent only copying a bar to a foo if possible

            – user463035818
            Apr 9 at 19:42





            but then I cannot copy foos anymore :( I'd like to prevent only copying a bar to a foo if possible

            – user463035818
            Apr 9 at 19:42

















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