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Exception in deleting an array

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  • D Offline
    D Offline
    ddze
    wrote on last edited by
    #1

    Hi, I have a strange exception that happens during the delete of an object.

    declaration

    class Something{
    public
    QList<QList<SomeItem *> > *getObject() const;
    private:
    QList<QList<SomeItem *>> *object;
    };
    

    definition

    // constructor
    Something::Something(){
    QList<QList<SomeItem *>> *object = 0;
    }
    // destructor
    Something::~Something(){
        if(object!=0){
            qDebug()<<"deleting objects";
            object->clear();
            delete object; object=0;  // Exception (SEH)
        }
    }
    

    I do not understand why the condition (object!= 0) is passed as a true? ( no other initialization aside from the origina where object is assigned 0. )

    1 Reply Last reply
    0
    • X Offline
      X Offline
      XavierLL
      wrote on last edited by XavierLL
      #2

      Hi,

      You are redeclaring object in the constructor. The constructor should be:

      Something::Something(){
       object = 0;
      }
      

      without the type, is a typical error ;)

      Hope it helps

      D 1 Reply Last reply
      2
      • X XavierLL

        Hi,

        You are redeclaring object in the constructor. The constructor should be:

        Something::Something(){
         object = 0;
        }
        

        without the type, is a typical error ;)

        Hope it helps

        D Offline
        D Offline
        ddze
        wrote on last edited by
        #3

        @XavierLL ,

        thank you so much.

        1 Reply Last reply
        0
        • X Offline
          X Offline
          XavierLL
          wrote on last edited by
          #4

          You're welcome :)

          1 Reply Last reply
          0
          • andrA Offline
            andrA Offline
            andr
            wrote on last edited by
            #5

            Note that you shouldn't have a QList<...> pointer member to start with. Just use the object directly as member. This also removes the need for the code you put in the constructor and destructor.

            D 1 Reply Last reply
            0
            • andrA andr

              Note that you shouldn't have a QList<...> pointer member to start with. Just use the object directly as member. This also removes the need for the code you put in the constructor and destructor.

              D Offline
              D Offline
              ddze
              wrote on last edited by
              #6

              @andr

              I must definitly disagree with you. Your suggestion depends on the nature of using it, how we use the object. The reassignment of the entire QList container to another QList invites a copy construction. By using a QList pointer I can just reassign it to an existing QList object somwhere else without copying it.

              kshegunovK 1 Reply Last reply
              0
              • D ddze

                @andr

                I must definitly disagree with you. Your suggestion depends on the nature of using it, how we use the object. The reassignment of the entire QList container to another QList invites a copy construction. By using a QList pointer I can just reassign it to an existing QList object somwhere else without copying it.

                kshegunovK Offline
                kshegunovK Offline
                kshegunov
                Moderators
                wrote on last edited by kshegunov
                #7

                @ddze
                You may disagree because you probably haven't read how QList manages its data. It uses implicit sharing, so the copy constructor reassigns a pointer and doesn't copy the data. The data is detached when a non-const method is invoked and only if more than one object references that data. So for all intents and purposes passing QList * around is pretty much pointless.

                Read and abide by the Qt Code of Conduct

                D 1 Reply Last reply
                0
                • kshegunovK kshegunov

                  @ddze
                  You may disagree because you probably haven't read how QList manages its data. It uses implicit sharing, so the copy constructor reassigns a pointer and doesn't copy the data. The data is detached when a non-const method is invoked and only if more than one object references that data. So for all intents and purposes passing QList * around is pretty much pointless.

                  D Offline
                  D Offline
                  ddze
                  wrote on last edited by ddze
                  #8

                  @kshegunov ,

                  good , now tell me this situation

                  QVector<QString*> vecItems1;    // ref count = 1
                  QVector<QString*> vecItems2;    //  ref count =1
                  
                  for(int c=5;c<20;++c){
                  QString s = QString::number(c);
                          vecItems.1.append(s);
                  }
                  vecItems2 = vecitems1;  // not copy constructor - both implicitly share same memory (refCount ==2) , 
                  
                  // here we come to call to copy constructor
                  for(int c=0;c<5;++c){
                  QString s = QString::number(c);
                          vecItems.2.append(s);
                  }
                   // vecItems1 -> changes its  refCount =1
                  //  vecItems2  ,called copy constructor refCount =1
                  
                  kshegunovK 1 Reply Last reply
                  0
                  • D ddze

                    @kshegunov ,

                    good , now tell me this situation

                    QVector<QString*> vecItems1;    // ref count = 1
                    QVector<QString*> vecItems2;    //  ref count =1
                    
                    for(int c=5;c<20;++c){
                    QString s = QString::number(c);
                            vecItems.1.append(s);
                    }
                    vecItems2 = vecitems1;  // not copy constructor - both implicitly share same memory (refCount ==2) , 
                    
                    // here we come to call to copy constructor
                    for(int c=0;c<5;++c){
                    QString s = QString::number(c);
                            vecItems.2.append(s);
                    }
                     // vecItems1 -> changes its  refCount =1
                    //  vecItems2  ,called copy constructor refCount =1
                    
                    kshegunovK Offline
                    kshegunovK Offline
                    kshegunov
                    Moderators
                    wrote on last edited by kshegunov
                    #9

                    @ddze
                    What do you want to know? Yes, it causes copying to occur (on the first call to vecItems2.append(s)). But let me ask you, can you prevent data copying by using QVector<QString*> * so when that function finishes you have number 1-20 in vecItems1 and 5 more in vecItems2? I would think not ...

                    PS.
                    Btw, the same considerations apply for QString and QString *, as well as any other implicitly shared class.

                    Read and abide by the Qt Code of Conduct

                    D 1 Reply Last reply
                    0
                    • kshegunovK kshegunov

                      @ddze
                      What do you want to know? Yes, it causes copying to occur (on the first call to vecItems2.append(s)). But let me ask you, can you prevent data copying by using QVector<QString*> * so when that function finishes you have number 1-20 in vecItems1 and 5 more in vecItems2? I would think not ...

                      PS.
                      Btw, the same considerations apply for QString and QString *, as well as any other implicitly shared class.

                      D Offline
                      D Offline
                      ddze
                      wrote on last edited by
                      #10

                      @kshegunov said:

                      can you prevent data copying by using QVector<QString*> *

                      Of course that I can , by a simple dereferencing the pointer to different data structure.
                      I am surprised you do not know that.

                      P.S
                      By the way, yes you really helped me twice so I have a respect towards you, but stick to the subject without portraiting anyone (your claim I haven't read ... bla bla bla).

                      kshegunovK 1 Reply Last reply
                      0
                      • D ddze

                        @kshegunov said:

                        can you prevent data copying by using QVector<QString*> *

                        Of course that I can , by a simple dereferencing the pointer to different data structure.
                        I am surprised you do not know that.

                        P.S
                        By the way, yes you really helped me twice so I have a respect towards you, but stick to the subject without portraiting anyone (your claim I haven't read ... bla bla bla).

                        kshegunovK Offline
                        kshegunovK Offline
                        kshegunov
                        Moderators
                        wrote on last edited by
                        #11

                        @ddze

                        Of course that I can , by a simple dereferencing the pointer to different data structure.

                        Maybe I'm missing something, but perhaps you could explain what you mean by that?

                        but stick to the subject without portraiting anyone

                        It was not my intent, I'm sorry it was perceived like this. I just meant that many people (happens quite often in the forum) are unaware that Qt uses that technique (STL doesn't).

                        Read and abide by the Qt Code of Conduct

                        D 2 Replies Last reply
                        1
                        • kshegunovK kshegunov

                          @ddze

                          Of course that I can , by a simple dereferencing the pointer to different data structure.

                          Maybe I'm missing something, but perhaps you could explain what you mean by that?

                          but stick to the subject without portraiting anyone

                          It was not my intent, I'm sorry it was perceived like this. I just meant that many people (happens quite often in the forum) are unaware that Qt uses that technique (STL doesn't).

                          D Offline
                          D Offline
                          ddze
                          wrote on last edited by
                          #12

                          @kshegunov

                          "Of course that I can , by a simple dereferencing the pointer to different data structure."

                          Maybe I'm missing something, but perhaps you could explain what you mean by that?

                          here we go ...

                          QVector<QString*> *pVec=0;
                          QVector<QString*> *pVec2= new QVector<QString*>;
                          QVector<QString*> *pVec3= new QVector<QString*>;
                          QVector<QString*> *pVec_nn= new QVector<QString*>;
                          
                          pVec = pVec2; // no copy constructor
                          pVec = pVec2; // dereferencing , no copy constructor
                          pVec = pVec3; // dereferencing , no copy constructor
                          pVec = pVec_nn; // dereferencing , no copy constructor
                          

                          I hope it clarifies what I meant.

                          kshegunovK 1 Reply Last reply
                          0
                          • kshegunovK kshegunov

                            @ddze

                            Of course that I can , by a simple dereferencing the pointer to different data structure.

                            Maybe I'm missing something, but perhaps you could explain what you mean by that?

                            but stick to the subject without portraiting anyone

                            It was not my intent, I'm sorry it was perceived like this. I just meant that many people (happens quite often in the forum) are unaware that Qt uses that technique (STL doesn't).

                            D Offline
                            D Offline
                            ddze
                            wrote on last edited by ddze
                            #13

                            @kshegunov ,

                            now I see why you replied.

                            I had written "The reassignment of the entire QList container to another QList invites a copy construction." Clearly the flaw in my explanation.

                            Not the reassignment, but the changing of the content in one of variables that implicitly share the data with others calls for a copy constructor.

                            1 Reply Last reply
                            0
                            • D ddze

                              @kshegunov

                              "Of course that I can , by a simple dereferencing the pointer to different data structure."

                              Maybe I'm missing something, but perhaps you could explain what you mean by that?

                              here we go ...

                              QVector<QString*> *pVec=0;
                              QVector<QString*> *pVec2= new QVector<QString*>;
                              QVector<QString*> *pVec3= new QVector<QString*>;
                              QVector<QString*> *pVec_nn= new QVector<QString*>;
                              
                              pVec = pVec2; // no copy constructor
                              pVec = pVec2; // dereferencing , no copy constructor
                              pVec = pVec3; // dereferencing , no copy constructor
                              pVec = pVec_nn; // dereferencing , no copy constructor
                              

                              I hope it clarifies what I meant.

                              kshegunovK Offline
                              kshegunovK Offline
                              kshegunov
                              Moderators
                              wrote on last edited by kshegunov
                              #14

                              @ddze

                              I hope it clarifies what I meant.

                              It does, but the last snippet is still very different from the one before that.

                              In the first one you do this:

                              QVector<QString*> vecItems2 = vecitems1;  // not copy constructor - both implicitly share same memory (refCount ==2)
                              

                              while in the second one you do this (pointers were converted to references for clarity):

                              QVector<QString*> & vecItems2 = vecitems1;
                              

                              See the difference? That's what I meant that in your first example no matter whether or not you use pointers you still are going to end up copying data. So your first example can be adapted in the described way and it won't call copying of data. So to be completely clear:

                              QVector<QString*> pVec2;
                              QVector<QString*> pVec3;
                              QVector<QString*> pVec_nn;
                              
                              QVector<QString*> & pVec = pVec2; // not causing copying (ever!)
                              pVec = pVec2; // not causing copying 
                              pVec = pVec3; // not causing copying
                              pVec = pVec_nn; // not causing copying
                              

                              Not the reassignment, but the changing of the content in one of variables that implicitly share the data with others calls for a copy constructor.

                              I do agree, but see above explanation.

                              Kind regards.

                              Read and abide by the Qt Code of Conduct

                              D 1 Reply Last reply
                              1
                              • kshegunovK kshegunov

                                @ddze

                                I hope it clarifies what I meant.

                                It does, but the last snippet is still very different from the one before that.

                                In the first one you do this:

                                QVector<QString*> vecItems2 = vecitems1;  // not copy constructor - both implicitly share same memory (refCount ==2)
                                

                                while in the second one you do this (pointers were converted to references for clarity):

                                QVector<QString*> & vecItems2 = vecitems1;
                                

                                See the difference? That's what I meant that in your first example no matter whether or not you use pointers you still are going to end up copying data. So your first example can be adapted in the described way and it won't call copying of data. So to be completely clear:

                                QVector<QString*> pVec2;
                                QVector<QString*> pVec3;
                                QVector<QString*> pVec_nn;
                                
                                QVector<QString*> & pVec = pVec2; // not causing copying (ever!)
                                pVec = pVec2; // not causing copying 
                                pVec = pVec3; // not causing copying
                                pVec = pVec_nn; // not causing copying
                                

                                Not the reassignment, but the changing of the content in one of variables that implicitly share the data with others calls for a copy constructor.

                                I do agree, but see above explanation.

                                Kind regards.

                                D Offline
                                D Offline
                                ddze
                                wrote on last edited by
                                #15

                                @kshegunov ,

                                agree with what you have written.

                                I think , I was in rush to reply to the one who objected on using the pointer to a container structure.

                                So I went on and hoping to convey the idea of me doing it rather than being careful in writing the correct way.

                                All the best to you Kshegunov.

                                1 Reply Last reply
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