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





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15















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

























  • 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
    12 hours ago






  • 3





    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
    12 hours ago






  • 1





    @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
    6 hours ago


















15















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

























  • 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
    12 hours ago






  • 3





    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
    12 hours ago






  • 1





    @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
    6 hours ago














15












15








15


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?






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







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




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






share|improve this question















share|improve this question













share|improve this question




share|improve this question








edited 12 hours ago









armitus

524114




524114










asked 13 hours ago









KorriKorri

35628




35628













  • 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
    12 hours ago






  • 3





    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
    12 hours ago






  • 1





    @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
    6 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
    12 hours ago






  • 3





    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
    12 hours ago






  • 1





    @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
    6 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
12 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
12 hours ago




3




3





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
12 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
12 hours ago




1




1





@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
6 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
6 hours ago












3 Answers
3






active

oldest

votes


















14














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
        10 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
        9 hours ago














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      3 Answers
      3






      active

      oldest

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      3 Answers
      3






      active

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      14














      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




























        14














        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


























          14












          14








          14







          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 12 hours ago









          AngewAngew

          135k11261354




          135k11261354

























              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 12 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
                        10 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
                        9 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
                        10 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
                        9 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 10 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
                        10 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
                        9 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
                        10 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
                        9 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
                      10 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
                      10 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
                      9 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
                      9 hours ago




















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                      Ida-Boy-Ed-Garten

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                      Die Lage des Ida-Boy-Ed-Gartens, rot markiert Der Ida-Boy-Ed-Garten Der Ida-Boy-Ed-Garten, Blick von der Wakenitzmauer durch das Tor in der Stadtmauer Der Ida-Boy-Ed-Garten ist eine Straße der Lübecker Altstadt. Inhaltsverzeichnis 1 Lage 2 Geschichte 3 Literatur 4 Weblinks Lage | Der etwa 90 Meter lange Ida-Boy-Ed-Garten befindet sich an der nördlichen Spitze der Altstadtinsel, vor dem Burgtor. Die in einem Bogen durch eine Grünanlage hangaufwärts führende Straße verbindet die Wakenitzmauer mit der Burgtorbrücke. Geschichte | Der heutige Ida-Boy-Ed-Garten wurde ursprünglich als Verlängerung der Wakenitzmauer angelegt. 1903 wurde in den erhaltenen Teil der mittelalterlichen Stadtmauer, an dem die Wakenitzmauer bei der Einmündung der Kaiserstraße begann, ein Tor mit einem großen Bogen für die Durchführung der Straße und zwei kleineren seitlichen Öffnungen für Fußgänger gebrochen, um für die neu errichteten Wohnhäuser in diesem Bereich ...
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