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Forum Update on Monday, May 27th 2025

Convert QVariantList to QList<Type> list

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  • S SGaist
    26 Mar 2015, 22:37

    QVariant has the == operator

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    ikim
    wrote on 26 Mar 2015, 22:43 last edited by
    #9

    @SGaist yes I could compare each QVariant in the list but that would mean iterating over the entire list to determine if the item exists in the second list. Is that my only option ?

    Thanks

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    0
    • S Offline
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      SGaist
      Lifetime Qt Champion
      wrote on 26 Mar 2015, 22:49 last edited by
      #10

      No, use the QSet features e.g.

      QSet<QVariant> commonSet = mySet1 & mySet2;
      

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      I 1 Reply Last reply 26 Mar 2015, 23:04
      0
      • S SGaist
        26 Mar 2015, 22:49

        No, use the QSet features e.g.

        QSet<QVariant> commonSet = mySet1 & mySet2;
        
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        ikim
        wrote on 26 Mar 2015, 23:04 last edited by
        #11

        @SGaist Thanks, so I can use a QSet<QVariant> out of the box ?

        other forums are suggesting that I need to define a qhash() function for the QVariant.

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        • M Offline
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          mcosta
          wrote on 26 Mar 2015, 23:58 last edited by
          #12

          QSet is based on a hash table; this means that internally it uses qHash()

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          I 1 Reply Last reply 27 Mar 2015, 08:34
          0
          • I ikim
            26 Mar 2015, 22:02

            Hi All

            I have a QVariantList which is passed as a parameter to my function.

            void myfunc(const QVariantList &list)
            {
              // I want to convert this to  QList<Type> newList
              // where Type is the type() of list.at(0)
              // the variant list consists of variants all of the same type 
            }
            

            What I want to do is

            QList<Type> newList;   //how do I declare this if I dont know the Type
              foreach(QVariant v, list) 
                newList << v.value();
            

            Then create a QSet from the newList

            Is there a way of easily achieving this

            Regards

            Ikim

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            MayEnjoy
            wrote on 27 Mar 2015, 05:48 last edited by
            #13

            @ikim why not use QVariant::Type method? even you can use it with c++ template.

            I am a engineer.

            I 1 Reply Last reply 27 Mar 2015, 08:40
            0
            • M mcosta
              26 Mar 2015, 23:58

              QSet is based on a hash table; this means that internally it uses qHash()

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              ikim
              wrote on 27 Mar 2015, 08:34 last edited by
              #14

              @mcosta Thanks but this gives a compile error

              QSet<QVariant> existingValues;
              existingValues.insert(1);

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              • M MayEnjoy
                27 Mar 2015, 05:48

                @ikim why not use QVariant::Type method? even you can use it with c++ template.

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                ikim
                wrote on 27 Mar 2015, 08:40 last edited by
                #15

                @MayEnjoy Thanks, but how would this enable me to achieve my goal ?

                M 1 Reply Last reply 28 Mar 2015, 07:15
                0
                • M Offline
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                  mcosta
                  wrote on 27 Mar 2015, 08:50 last edited by
                  #16

                  Which kind of error?

                  Once your problem is solved don't forget to:

                  • Mark the thread as SOLVED using the Topic Tool menu
                  • Vote up the answer(s) that helped you to solve the issue

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                  I 1 Reply Last reply 27 Mar 2015, 08:55
                  0
                  • M mcosta
                    27 Mar 2015, 08:50

                    Which kind of error?

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                    ikim
                    wrote on 27 Mar 2015, 08:55 last edited by
                    #17

                    @mcosta /usr/include/QtCore/qhash.h:882:24: error: no matching function for call to 'qHash(const QVariant&)'

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                    • M Offline
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                      mcosta
                      wrote on 27 Mar 2015, 09:11 last edited by
                      #18

                      mmmm, You're right.

                      So I suggest to create a template method

                      This code works for me

                      template <typename T>
                      QSet<T> mergeList(const QVariantList& l1, const QVariantList& l2) {
                          QSet<T> s1, s2;
                      
                          for (const QVariant& v: l1) {
                              s1.insert(v.value<T>());
                          }
                      
                          for (const QVariant& v: l2) {
                              s2.insert(v.value<T>());
                          }
                      
                          return s1 & s2;
                      }
                      

                      Used as

                          QVariantList l1, l2;
                          l1 << 1 << 2 << 3;
                          l2 << 1 << 3 << 5;
                      
                          QSet<int> s1 = mergeList<int> (l1, l2);
                          qDebug() << s1;
                      
                          l1.clear();
                          l2.clear();
                      
                          l1 << "Foo" << "Bar";
                          l2 << "Bar" << "Fred";
                      
                          QSet<QString> s2 = mergeList<QString> (l1, l2);
                          qDebug() << s2;
                      

                      Once your problem is solved don't forget to:

                      • Mark the thread as SOLVED using the Topic Tool menu
                      • Vote up the answer(s) that helped you to solve the issue

                      You can embed images using (http://imgur.com/) or (http://postimage.org/)

                      I 1 Reply Last reply 27 Mar 2015, 09:30
                      1
                      • M mcosta
                        27 Mar 2015, 09:11

                        mmmm, You're right.

                        So I suggest to create a template method

                        This code works for me

                        template <typename T>
                        QSet<T> mergeList(const QVariantList& l1, const QVariantList& l2) {
                            QSet<T> s1, s2;
                        
                            for (const QVariant& v: l1) {
                                s1.insert(v.value<T>());
                            }
                        
                            for (const QVariant& v: l2) {
                                s2.insert(v.value<T>());
                            }
                        
                            return s1 & s2;
                        }
                        

                        Used as

                            QVariantList l1, l2;
                            l1 << 1 << 2 << 3;
                            l2 << 1 << 3 << 5;
                        
                            QSet<int> s1 = mergeList<int> (l1, l2);
                            qDebug() << s1;
                        
                            l1.clear();
                            l2.clear();
                        
                            l1 << "Foo" << "Bar";
                            l2 << "Bar" << "Fred";
                        
                            QSet<QString> s2 = mergeList<QString> (l1, l2);
                            qDebug() << s2;
                        
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                        ikim
                        wrote on 27 Mar 2015, 09:30 last edited by
                        #19

                        @mcosta Thanks for your effort, but this still doesn't solve the problem as I dont know the Type at compile time

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                        1
                        • M Offline
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                          mcosta
                          wrote on 27 Mar 2015, 10:46 last edited by
                          #20

                          Hi,

                          I'm not a C++ super-guru but I don't think you can do it because the compiler must know the type of each variable at compile time.

                          A solution could be implement qHash(const QVariant &v) and use QSet<QVariant>

                          Once your problem is solved don't forget to:

                          • Mark the thread as SOLVED using the Topic Tool menu
                          • Vote up the answer(s) that helped you to solve the issue

                          You can embed images using (http://imgur.com/) or (http://postimage.org/)

                          I 1 Reply Last reply 27 Mar 2015, 10:49
                          1
                          • M mcosta
                            27 Mar 2015, 10:46

                            Hi,

                            I'm not a C++ super-guru but I don't think you can do it because the compiler must know the type of each variable at compile time.

                            A solution could be implement qHash(const QVariant &v) and use QSet<QVariant>

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                            ikim
                            wrote on 27 Mar 2015, 10:49 last edited by
                            #21

                            @mcosta thanks for your time, I think you may be right :)

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                            • I ikim
                              27 Mar 2015, 08:40

                              @MayEnjoy Thanks, but how would this enable me to achieve my goal ?

                              M Offline
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                              MayEnjoy
                              wrote on 28 Mar 2015, 07:15 last edited by
                              #22

                              @ikim said:

                              @SGaist yes I could compare each QVariant in the list but that would mean iterating over the entire list to determine if the item exists in the second list. Is that my only option ?

                              Thanks

                              I think @mclark 's answer is the better one.

                              I am a engineer.

                              1 Reply Last reply
                              0
                              • S Offline
                                S Offline
                                SGaist
                                Lifetime Qt Champion
                                wrote on 28 Mar 2015, 21:32 last edited by
                                #23

                                You can also implement qHash for the type you need to support and thus it will be used automatically in all your software.

                                #include <QtDebug>
                                
                                inline uint qHash(const QVariant &key, uint seed = 0)
                                {
                                    switch (key.userType()) {
                                        case QVariant::Int:
                                            return qHash(key.toInt(), seed);
                                
                                        case QVariant::UInt:
                                            return qHash(key.toUInt(), seed);
                                
                                       // add all cases you want to support;
                                
                                    }
                                
                                    return 0;
                                }
                                
                                int main( int argc, char * argv[] )
                                {
                                    QApplication app( argc, argv );
                                    QVariantList vl1;
                                    QVariantList vl2;
                                
                                    for (int i = 0 ; i < 5 ; ++i) {
                                        vl1 << i;
                                    }
                                
                                    for (int i = 0 ; i < 3 ; ++i) {
                                        vl2 << i;
                                    }
                                
                                    qDebug() << (vl1.toSet() & vl2.toSet());
                                
                                    return 0;
                                }
                                

                                Interested in AI ? www.idiap.ch
                                Please read the Qt Code of Conduct - https://forum.qt.io/topic/113070/qt-code-of-conduct

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