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Is there a way to make member function NOT callable from constructor?



Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)
The Ask Question Wizard is Live!
Data science time! April 2019 and salary with experienceWhat are the differences between a pointer variable and a reference variable in C++?Virtual member call in a constructorHow do I call one constructor from another in Java?Can I call a constructor from another constructor (do constructor chaining) in C++?What is a clean, pythonic way to have multiple constructors in Python?Throwing exceptions from constructorsCall one constructor from anotherEquality-compare std::weak_ptrClass inheritance: Constructor and member functions of class not recognized by compilerIs it bad practice to have a constructor function return a Promise?



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19















I have member function (method) which uses



std::enable_shared_from_this::weak_from_this() 


In short: weak_from_this returns weak_ptr to this. One caveat is it can't be used from constructor.
If somebody would use my function from constructor of inherited class, weak_from_this inside it would return expired weak_ptr. I guard against that with assertion checking that it's not expired, but it's a run-time check.



Is there a way to check against it at compile time?










share|improve this question
























  • Note there is a difference in scope between a child class constructor body and the parent class constructor: the latter has been executed completely before you even start initializing the child class's members (if any), let alone enter the child class constructor body.

    – rubenvb
    Apr 8 at 14:58






  • 4





    Nice question. One way would be to make a dummy class with pure virtual function weak_from_this and inherit yours from it. This will make it a hard compile error.

    – SergeyA
    Apr 8 at 14:59






  • 5





    @SergeyA Why didn't you post that as an answer? All other people here seem to conclude that it's not possible so either your comment is wrong and misleading or they are wrong and you should show how it can be achieved.

    – Bakuriu
    Apr 8 at 21:17











  • @Bakuriu well, I did not have the energy to polish it to the full blown answer. It is possible that it is not a workable solution.

    – SergeyA
    Apr 9 at 14:07











  • Simply based on the fact the static type of an object doesn't (and cannot) depend on the fact its construction is completed. Note that if that were the case, you wouldn't be able in a ctor of T to store this in a global obj register, and use the obj later, w/o a cast, as the ptr stored would have "(in construction)T*" not plain T*.

    – curiousguy
    13 hours ago

















19















I have member function (method) which uses



std::enable_shared_from_this::weak_from_this() 


In short: weak_from_this returns weak_ptr to this. One caveat is it can't be used from constructor.
If somebody would use my function from constructor of inherited class, weak_from_this inside it would return expired weak_ptr. I guard against that with assertion checking that it's not expired, but it's a run-time check.



Is there a way to check against it at compile time?










share|improve this question
























  • Note there is a difference in scope between a child class constructor body and the parent class constructor: the latter has been executed completely before you even start initializing the child class's members (if any), let alone enter the child class constructor body.

    – rubenvb
    Apr 8 at 14:58






  • 4





    Nice question. One way would be to make a dummy class with pure virtual function weak_from_this and inherit yours from it. This will make it a hard compile error.

    – SergeyA
    Apr 8 at 14:59






  • 5





    @SergeyA Why didn't you post that as an answer? All other people here seem to conclude that it's not possible so either your comment is wrong and misleading or they are wrong and you should show how it can be achieved.

    – Bakuriu
    Apr 8 at 21:17











  • @Bakuriu well, I did not have the energy to polish it to the full blown answer. It is possible that it is not a workable solution.

    – SergeyA
    Apr 9 at 14:07











  • Simply based on the fact the static type of an object doesn't (and cannot) depend on the fact its construction is completed. Note that if that were the case, you wouldn't be able in a ctor of T to store this in a global obj register, and use the obj later, w/o a cast, as the ptr stored would have "(in construction)T*" not plain T*.

    – curiousguy
    13 hours ago













19












19








19


3






I have member function (method) which uses



std::enable_shared_from_this::weak_from_this() 


In short: weak_from_this returns weak_ptr to this. One caveat is it can't be used from constructor.
If somebody would use my function from constructor of inherited class, weak_from_this inside it would return expired weak_ptr. I guard against that with assertion checking that it's not expired, but it's a run-time check.



Is there a way to check against it at compile time?










share|improve this question
















I have member function (method) which uses



std::enable_shared_from_this::weak_from_this() 


In short: weak_from_this returns weak_ptr to this. One caveat is it can't be used from constructor.
If somebody would use my function from constructor of inherited class, weak_from_this inside it would return expired weak_ptr. I guard against that with assertion checking that it's not expired, but it's a run-time check.



Is there a way to check against it at compile time?







c++ constructor c++17 shared-ptr weak-ptr






share|improve this question















share|improve this question













share|improve this question




share|improve this question








edited 13 hours ago









curiousguy

4,68823046




4,68823046










asked Apr 8 at 14:53









KorriKorri

37629




37629












  • Note there is a difference in scope between a child class constructor body and the parent class constructor: the latter has been executed completely before you even start initializing the child class's members (if any), let alone enter the child class constructor body.

    – rubenvb
    Apr 8 at 14:58






  • 4





    Nice question. One way would be to make a dummy class with pure virtual function weak_from_this and inherit yours from it. This will make it a hard compile error.

    – SergeyA
    Apr 8 at 14:59






  • 5





    @SergeyA Why didn't you post that as an answer? All other people here seem to conclude that it's not possible so either your comment is wrong and misleading or they are wrong and you should show how it can be achieved.

    – Bakuriu
    Apr 8 at 21:17











  • @Bakuriu well, I did not have the energy to polish it to the full blown answer. It is possible that it is not a workable solution.

    – SergeyA
    Apr 9 at 14:07











  • Simply based on the fact the static type of an object doesn't (and cannot) depend on the fact its construction is completed. Note that if that were the case, you wouldn't be able in a ctor of T to store this in a global obj register, and use the obj later, w/o a cast, as the ptr stored would have "(in construction)T*" not plain T*.

    – curiousguy
    13 hours ago

















  • Note there is a difference in scope between a child class constructor body and the parent class constructor: the latter has been executed completely before you even start initializing the child class's members (if any), let alone enter the child class constructor body.

    – rubenvb
    Apr 8 at 14:58






  • 4





    Nice question. One way would be to make a dummy class with pure virtual function weak_from_this and inherit yours from it. This will make it a hard compile error.

    – SergeyA
    Apr 8 at 14:59






  • 5





    @SergeyA Why didn't you post that as an answer? All other people here seem to conclude that it's not possible so either your comment is wrong and misleading or they are wrong and you should show how it can be achieved.

    – Bakuriu
    Apr 8 at 21:17











  • @Bakuriu well, I did not have the energy to polish it to the full blown answer. It is possible that it is not a workable solution.

    – SergeyA
    Apr 9 at 14:07











  • Simply based on the fact the static type of an object doesn't (and cannot) depend on the fact its construction is completed. Note that if that were the case, you wouldn't be able in a ctor of T to store this in a global obj register, and use the obj later, w/o a cast, as the ptr stored would have "(in construction)T*" not plain T*.

    – curiousguy
    13 hours ago
















Note there is a difference in scope between a child class constructor body and the parent class constructor: the latter has been executed completely before you even start initializing the child class's members (if any), let alone enter the child class constructor body.

– rubenvb
Apr 8 at 14:58





Note there is a difference in scope between a child class constructor body and the parent class constructor: the latter has been executed completely before you even start initializing the child class's members (if any), let alone enter the child class constructor body.

– rubenvb
Apr 8 at 14:58




4




4





Nice question. One way would be to make a dummy class with pure virtual function weak_from_this and inherit yours from it. This will make it a hard compile error.

– SergeyA
Apr 8 at 14:59





Nice question. One way would be to make a dummy class with pure virtual function weak_from_this and inherit yours from it. This will make it a hard compile error.

– SergeyA
Apr 8 at 14:59




5




5





@SergeyA Why didn't you post that as an answer? All other people here seem to conclude that it's not possible so either your comment is wrong and misleading or they are wrong and you should show how it can be achieved.

– Bakuriu
Apr 8 at 21:17





@SergeyA Why didn't you post that as an answer? All other people here seem to conclude that it's not possible so either your comment is wrong and misleading or they are wrong and you should show how it can be achieved.

– Bakuriu
Apr 8 at 21:17













@Bakuriu well, I did not have the energy to polish it to the full blown answer. It is possible that it is not a workable solution.

– SergeyA
Apr 9 at 14:07





@Bakuriu well, I did not have the energy to polish it to the full blown answer. It is possible that it is not a workable solution.

– SergeyA
Apr 9 at 14:07













Simply based on the fact the static type of an object doesn't (and cannot) depend on the fact its construction is completed. Note that if that were the case, you wouldn't be able in a ctor of T to store this in a global obj register, and use the obj later, w/o a cast, as the ptr stored would have "(in construction)T*" not plain T*.

– curiousguy
13 hours ago





Simply based on the fact the static type of an object doesn't (and cannot) depend on the fact its construction is completed. Note that if that were the case, you wouldn't be able in a ctor of T to store this in a global obj register, and use the obj later, w/o a cast, as the ptr stored would have "(in construction)T*" not plain T*.

– curiousguy
13 hours ago












3 Answers
3






active

oldest

votes


















15














I am afraid the answer is "no, it's not possible to protect against this at compile-time." It's always difficult to prove a negative, but consider this: if it was possible to protect a function this way, it would probably have been done for weak_from_this and shared_from_this in the standard library itself.






share|improve this answer






























    5














    No there is no way. Consider:



    void call_me(struct widget*);

    struct widget : std::enable_shared_from_this<widget>
    widget()
    call_me(this);


    void display()
    shared_from_this();

    ;

    // later:

    void call_me(widget* w)
    w->display(); // crash



    The thing is there is a reason you want to check for not calling shared_from_this in the constructor. Think about that reason. It's not that shared_from_this cannot be called, it's because it's return value has no way of being assigned yet. It is also not because it will never be assigned. It's because it will be assigned later in the execution of the code. Order of operation is a runtime property of your program. You cannot assert at compile time for order of operation, which is done at runtime.






    share|improve this answer






























      4














      Not as such, but - if performance is not an issue, you could add a flag which indicates construction is complete, and use that to fail at run-time with such calls:



      class A 

      // ... whatever ...
      public:
      A()
      // do construction work
      constructed = true;


      foo()
      if (not constructed)
      throw std::logic_error("Cannot call foo() during construction");

      // the rest of foo


      protected:
      bool constructed false ;



      You could also make these checks only apply when compiling in DEBUG mode (e.g. with conditional compilation using the preprocessor - #ifndef NDEBUG) so that at run time you won't get the performance penalty. Mind the noexcepts though.



      An alternative to throwing could be assert()'ing.






      share|improve this answer

























      • Since apparently there isn't compile-time solution which doesn't make code less readable, I decided to go with assert(!wptr.expired()). I think it's a little bit more fitting than exception, because there is no way to recover from this situation.

        – Korri
        Apr 8 at 17:23











      • @Korri remember that asserts are usually compiled out in release builds, so nothing will happen. An exception however is not, so it would still terminate the program (if not caught and swallowed) in a release build. You could do both; first assert then throw, or just throw.

        – Jesper Juhl
        Apr 8 at 18:38












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      3 Answers
      3






      active

      oldest

      votes








      3 Answers
      3






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      15














      I am afraid the answer is "no, it's not possible to protect against this at compile-time." It's always difficult to prove a negative, but consider this: if it was possible to protect a function this way, it would probably have been done for weak_from_this and shared_from_this in the standard library itself.






      share|improve this answer



























        15














        I am afraid the answer is "no, it's not possible to protect against this at compile-time." It's always difficult to prove a negative, but consider this: if it was possible to protect a function this way, it would probably have been done for weak_from_this and shared_from_this in the standard library itself.






        share|improve this answer

























          15












          15








          15







          I am afraid the answer is "no, it's not possible to protect against this at compile-time." It's always difficult to prove a negative, but consider this: if it was possible to protect a function this way, it would probably have been done for weak_from_this and shared_from_this in the standard library itself.






          share|improve this answer













          I am afraid the answer is "no, it's not possible to protect against this at compile-time." It's always difficult to prove a negative, but consider this: if it was possible to protect a function this way, it would probably have been done for weak_from_this and shared_from_this in the standard library itself.







          share|improve this answer












          share|improve this answer



          share|improve this answer










          answered Apr 8 at 14:59









          AngewAngew

          135k11261354




          135k11261354























              5














              No there is no way. Consider:



              void call_me(struct widget*);

              struct widget : std::enable_shared_from_this<widget>
              widget()
              call_me(this);


              void display()
              shared_from_this();

              ;

              // later:

              void call_me(widget* w)
              w->display(); // crash



              The thing is there is a reason you want to check for not calling shared_from_this in the constructor. Think about that reason. It's not that shared_from_this cannot be called, it's because it's return value has no way of being assigned yet. It is also not because it will never be assigned. It's because it will be assigned later in the execution of the code. Order of operation is a runtime property of your program. You cannot assert at compile time for order of operation, which is done at runtime.






              share|improve this answer



























                5














                No there is no way. Consider:



                void call_me(struct widget*);

                struct widget : std::enable_shared_from_this<widget>
                widget()
                call_me(this);


                void display()
                shared_from_this();

                ;

                // later:

                void call_me(widget* w)
                w->display(); // crash



                The thing is there is a reason you want to check for not calling shared_from_this in the constructor. Think about that reason. It's not that shared_from_this cannot be called, it's because it's return value has no way of being assigned yet. It is also not because it will never be assigned. It's because it will be assigned later in the execution of the code. Order of operation is a runtime property of your program. You cannot assert at compile time for order of operation, which is done at runtime.






                share|improve this answer

























                  5












                  5








                  5







                  No there is no way. Consider:



                  void call_me(struct widget*);

                  struct widget : std::enable_shared_from_this<widget>
                  widget()
                  call_me(this);


                  void display()
                  shared_from_this();

                  ;

                  // later:

                  void call_me(widget* w)
                  w->display(); // crash



                  The thing is there is a reason you want to check for not calling shared_from_this in the constructor. Think about that reason. It's not that shared_from_this cannot be called, it's because it's return value has no way of being assigned yet. It is also not because it will never be assigned. It's because it will be assigned later in the execution of the code. Order of operation is a runtime property of your program. You cannot assert at compile time for order of operation, which is done at runtime.






                  share|improve this answer













                  No there is no way. Consider:



                  void call_me(struct widget*);

                  struct widget : std::enable_shared_from_this<widget>
                  widget()
                  call_me(this);


                  void display()
                  shared_from_this();

                  ;

                  // later:

                  void call_me(widget* w)
                  w->display(); // crash



                  The thing is there is a reason you want to check for not calling shared_from_this in the constructor. Think about that reason. It's not that shared_from_this cannot be called, it's because it's return value has no way of being assigned yet. It is also not because it will never be assigned. It's because it will be assigned later in the execution of the code. Order of operation is a runtime property of your program. You cannot assert at compile time for order of operation, which is done at runtime.







                  share|improve this answer












                  share|improve this answer



                  share|improve this answer










                  answered Apr 8 at 15:37









                  Guillaume RacicotGuillaume Racicot

                  16.5k53872




                  16.5k53872





















                      4














                      Not as such, but - if performance is not an issue, you could add a flag which indicates construction is complete, and use that to fail at run-time with such calls:



                      class A 

                      // ... whatever ...
                      public:
                      A()
                      // do construction work
                      constructed = true;


                      foo()
                      if (not constructed)
                      throw std::logic_error("Cannot call foo() during construction");

                      // the rest of foo


                      protected:
                      bool constructed false ;



                      You could also make these checks only apply when compiling in DEBUG mode (e.g. with conditional compilation using the preprocessor - #ifndef NDEBUG) so that at run time you won't get the performance penalty. Mind the noexcepts though.



                      An alternative to throwing could be assert()'ing.






                      share|improve this answer

























                      • Since apparently there isn't compile-time solution which doesn't make code less readable, I decided to go with assert(!wptr.expired()). I think it's a little bit more fitting than exception, because there is no way to recover from this situation.

                        – Korri
                        Apr 8 at 17:23











                      • @Korri remember that asserts are usually compiled out in release builds, so nothing will happen. An exception however is not, so it would still terminate the program (if not caught and swallowed) in a release build. You could do both; first assert then throw, or just throw.

                        – Jesper Juhl
                        Apr 8 at 18:38
















                      4














                      Not as such, but - if performance is not an issue, you could add a flag which indicates construction is complete, and use that to fail at run-time with such calls:



                      class A 

                      // ... whatever ...
                      public:
                      A()
                      // do construction work
                      constructed = true;


                      foo()
                      if (not constructed)
                      throw std::logic_error("Cannot call foo() during construction");

                      // the rest of foo


                      protected:
                      bool constructed false ;



                      You could also make these checks only apply when compiling in DEBUG mode (e.g. with conditional compilation using the preprocessor - #ifndef NDEBUG) so that at run time you won't get the performance penalty. Mind the noexcepts though.



                      An alternative to throwing could be assert()'ing.






                      share|improve this answer

























                      • Since apparently there isn't compile-time solution which doesn't make code less readable, I decided to go with assert(!wptr.expired()). I think it's a little bit more fitting than exception, because there is no way to recover from this situation.

                        – Korri
                        Apr 8 at 17:23











                      • @Korri remember that asserts are usually compiled out in release builds, so nothing will happen. An exception however is not, so it would still terminate the program (if not caught and swallowed) in a release build. You could do both; first assert then throw, or just throw.

                        – Jesper Juhl
                        Apr 8 at 18:38














                      4












                      4








                      4







                      Not as such, but - if performance is not an issue, you could add a flag which indicates construction is complete, and use that to fail at run-time with such calls:



                      class A 

                      // ... whatever ...
                      public:
                      A()
                      // do construction work
                      constructed = true;


                      foo()
                      if (not constructed)
                      throw std::logic_error("Cannot call foo() during construction");

                      // the rest of foo


                      protected:
                      bool constructed false ;



                      You could also make these checks only apply when compiling in DEBUG mode (e.g. with conditional compilation using the preprocessor - #ifndef NDEBUG) so that at run time you won't get the performance penalty. Mind the noexcepts though.



                      An alternative to throwing could be assert()'ing.






                      share|improve this answer















                      Not as such, but - if performance is not an issue, you could add a flag which indicates construction is complete, and use that to fail at run-time with such calls:



                      class A 

                      // ... whatever ...
                      public:
                      A()
                      // do construction work
                      constructed = true;


                      foo()
                      if (not constructed)
                      throw std::logic_error("Cannot call foo() during construction");

                      // the rest of foo


                      protected:
                      bool constructed false ;



                      You could also make these checks only apply when compiling in DEBUG mode (e.g. with conditional compilation using the preprocessor - #ifndef NDEBUG) so that at run time you won't get the performance penalty. Mind the noexcepts though.



                      An alternative to throwing could be assert()'ing.







                      share|improve this answer














                      share|improve this answer



                      share|improve this answer








                      edited Apr 9 at 7:29

























                      answered Apr 8 at 16:58









                      einpoklumeinpoklum

                      37.4k28135264




                      37.4k28135264












                      • Since apparently there isn't compile-time solution which doesn't make code less readable, I decided to go with assert(!wptr.expired()). I think it's a little bit more fitting than exception, because there is no way to recover from this situation.

                        – Korri
                        Apr 8 at 17:23











                      • @Korri remember that asserts are usually compiled out in release builds, so nothing will happen. An exception however is not, so it would still terminate the program (if not caught and swallowed) in a release build. You could do both; first assert then throw, or just throw.

                        – Jesper Juhl
                        Apr 8 at 18:38


















                      • Since apparently there isn't compile-time solution which doesn't make code less readable, I decided to go with assert(!wptr.expired()). I think it's a little bit more fitting than exception, because there is no way to recover from this situation.

                        – Korri
                        Apr 8 at 17:23











                      • @Korri remember that asserts are usually compiled out in release builds, so nothing will happen. An exception however is not, so it would still terminate the program (if not caught and swallowed) in a release build. You could do both; first assert then throw, or just throw.

                        – Jesper Juhl
                        Apr 8 at 18:38

















                      Since apparently there isn't compile-time solution which doesn't make code less readable, I decided to go with assert(!wptr.expired()). I think it's a little bit more fitting than exception, because there is no way to recover from this situation.

                      – Korri
                      Apr 8 at 17:23





                      Since apparently there isn't compile-time solution which doesn't make code less readable, I decided to go with assert(!wptr.expired()). I think it's a little bit more fitting than exception, because there is no way to recover from this situation.

                      – Korri
                      Apr 8 at 17:23













                      @Korri remember that asserts are usually compiled out in release builds, so nothing will happen. An exception however is not, so it would still terminate the program (if not caught and swallowed) in a release build. You could do both; first assert then throw, or just throw.

                      – Jesper Juhl
                      Apr 8 at 18:38






                      @Korri remember that asserts are usually compiled out in release builds, so nothing will happen. An exception however is not, so it would still terminate the program (if not caught and swallowed) in a release build. You could do both; first assert then throw, or just throw.

                      – Jesper Juhl
                      Apr 8 at 18:38


















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