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


What are the differences between a pointer variable and a reference variable in C++?Can I call a constructor from another constructor (do constructor chaining) in C++?Throwing exceptions from constructorsHow do I call ::std::make_shared on a class with only protected or private constructors?Calling a base member in constructor in multiple inheritance in C++Equality-compare std::weak_ptrClass inheritance: Constructor and member functions of class not recognized by compilerHow does shared_ptr<T> detect that T derives from enable_shared_from_this<T>?enable_shared_from_this derived class methods are undefined referenceDefault move constructor with mutex member






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12















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
    8 hours ago






  • 1





    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
    8 hours ago











  • @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
    1 hour ago

















12















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
    8 hours ago






  • 1





    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
    8 hours ago











  • @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
    1 hour ago













12












12








12


2






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






share|improve this question















share|improve this question













share|improve this question




share|improve this question








edited 8 hours ago









armitus

524114




524114










asked 8 hours ago









KorriKorri

34128




34128












  • 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
    8 hours ago






  • 1





    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
    8 hours ago











  • @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
    1 hour 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
    8 hours ago






  • 1





    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
    8 hours ago











  • @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
    1 hour 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
8 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
8 hours ago




1




1





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
8 hours ago





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
8 hours ago













@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
1 hour ago





@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
1 hour ago












3 Answers
3






active

oldest

votes


















10














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






























    4














    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 ...

      A()
      // do construction work
      constructed = true;


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

      // the rest of foo


      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
        5 hours ago











      • @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
        4 hours ago












      Your Answer






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      3 Answers
      3






      active

      oldest

      votes








      3 Answers
      3






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      10














      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



























        10














        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

























          10












          10








          10







          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 8 hours ago









          AngewAngew

          135k11260354




          135k11260354























              4














              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














                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












                  4








                  4







                  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 7 hours ago









                  Guillaume RacicotGuillaume Racicot

                  16.3k53872




                  16.3k53872





















                      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 ...

                      A()
                      // do construction work
                      constructed = true;


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

                      // the rest of foo


                      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
                        5 hours ago











                      • @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
                        4 hours ago
















                      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 ...

                      A()
                      // do construction work
                      constructed = true;


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

                      // the rest of foo


                      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
                        5 hours ago











                      • @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
                        4 hours ago














                      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 ...

                      A()
                      // do construction work
                      constructed = true;


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

                      // the rest of foo


                      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 ...

                      A()
                      // do construction work
                      constructed = true;


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

                      // the rest of foo


                      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










                      answered 6 hours ago









                      einpoklumeinpoklum

                      37k28132263




                      37k28132263












                      • 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
                        5 hours ago











                      • @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
                        4 hours ago


















                      • 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
                        5 hours ago











                      • @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
                        4 hours ago

















                      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
                      5 hours ago





                      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
                      5 hours ago













                      @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
                      4 hours ago






                      @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
                      4 hours ago


















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