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[Solved]Convert ASCII hex to int

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  • M Offline
    M Offline
    mpergand
    wrote on 6 Mar 2018, 13:12 last edited by
    #17
    QString line="80: 1C 02 FC 85 FF 4F FF 1D 89 17 08 10 01 00 00 00 3B 87 69 00 82 00 48 09 88 10 00 00 ";
    QStringList fields=line.split(':');
    qDebug()<<"en-tĂȘte:"<<fields[0];
    QByteArray values=QByteArray::fromHex(fields[1].toLatin1());
    qDebug()<<"data:"<<values;
    

    en-tĂȘte: "80"
    data :"\x1C\x02\xFC\x85\xFFO\xFF\x1D\x89\x17\b\x10\x01\x00\x00\x00;\x87i\x00\x82\x00H\t\x88\x10\x00\x00"

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    • L Offline
      L Offline
      Leopold
      wrote on 6 Mar 2018, 13:39 last edited by
      #18

      heureka
      VRonin's code works:
      QVector()
      "80: 1C 02 FC 85 FF 4F FF 1D 89 17 08 10 01 00 00 00 3B 87 69 00 82 00 48 09 88 10 00 00 "
      4
      QRegularExpressionMatch(Valid, has match: 0:(0, 3, "80:"), 1:(0, 2, "80"))
      "1C" 28
      "02" 2
      "FC" 252
      now I have to give eyh value to the different y
      but I think I can gigure that out.

      @mpergand
      this will divide first field from rest but how to divide the rest?
      y1= field1,y2=field2 and so on
      but I will test.

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      • M Offline
        M Offline
        mpergand
        wrote on 6 Mar 2018, 13:47 last edited by mpergand 3 Jun 2018, 13:47
        #19

        @Leopold said in [Solved]Convert ASCII hex to int:

        this will divide first field from rest but how to divide the rest?
        y1= field1,y2=field2 and so on
        but I will test.

        No need to split anything, you can access each byte in a QByteArray with at() or []
        See the doc ...

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        • L Offline
          L Offline
          Leopold
          wrote on 6 Mar 2018, 21:18 last edited by
          #20

          @VRonin
          the example works great for transforming from hex to dec but i get the same row what i have in the txt now as a vektor. In my code you see that I have different y Vektors.
          I need value of field1 in y1, field2 in y2 and so on.I can not figure out how to take the i into a loop with y(i).

          V 1 Reply Last reply 7 Mar 2018, 08:28
          0
          • L Leopold
            6 Mar 2018, 21:18

            @VRonin
            the example works great for transforming from hex to dec but i get the same row what i have in the txt now as a vektor. In my code you see that I have different y Vektors.
            I need value of field1 in y1, field2 in y2 and so on.I can not figure out how to take the i into a loop with y(i).

            V Offline
            V Offline
            VRonin
            wrote on 7 Mar 2018, 08:28 last edited by
            #21

            @Leopold Every time you call i.next(); it moves to the next field in order so it's quite easy to manage

            "La mort n'est rien, mais vivre vaincu et sans gloire, c'est mourir tous les jours"
            ~Napoleon Bonaparte

            On a crusade to banish setIndexWidget() from the holy land of Qt

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            • L Offline
              L Offline
              Leopold
              wrote on 7 Mar 2018, 08:33 last edited by
              #22

              I tried this:
              { // iterate over all the matches

              										const QRegularExpressionMatch hexMatch = i.next();
              										qDebug() <<  hexMatch.capturedRef(0) <<  hexMatch.capturedRef(0).toUInt(Q_NULLPTR,16); //capturedRef contains the part matched by the regular expression
                                                      y1.append(hexMatch.capturedRef(1).toUInt(Q_NULLPTR,16));
                                                       y2.append(hexMatch.capturedRef(2).toUInt(Q_NULLPTR,16));
                                                        y3.append(hexMatch.capturedRef(3).toUInt(Q_NULLPTR,16));
                                                         y4.append(hexMatch.capturedRef(4).toUInt(Q_NULLPTR,16));
                                                  }
              
              V 1 Reply Last reply 7 Mar 2018, 09:07
              0
              • L Leopold
                7 Mar 2018, 08:33

                I tried this:
                { // iterate over all the matches

                										const QRegularExpressionMatch hexMatch = i.next();
                										qDebug() <<  hexMatch.capturedRef(0) <<  hexMatch.capturedRef(0).toUInt(Q_NULLPTR,16); //capturedRef contains the part matched by the regular expression
                                                        y1.append(hexMatch.capturedRef(1).toUInt(Q_NULLPTR,16));
                                                         y2.append(hexMatch.capturedRef(2).toUInt(Q_NULLPTR,16));
                                                          y3.append(hexMatch.capturedRef(3).toUInt(Q_NULLPTR,16));
                                                           y4.append(hexMatch.capturedRef(4).toUInt(Q_NULLPTR,16));
                                                    }
                
                V Offline
                V Offline
                VRonin
                wrote on 7 Mar 2018, 09:07 last edited by
                #23

                @Leopold said in [Solved]Convert ASCII hex to int:

                { // iterate over all the matches

                Look at that comment. Do you see what you are doing wrong?

                "La mort n'est rien, mais vivre vaincu et sans gloire, c'est mourir tous les jours"
                ~Napoleon Bonaparte

                On a crusade to banish setIndexWidget() from the holy land of Qt

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                • L Offline
                  L Offline
                  Leopold
                  wrote on 7 Mar 2018, 10:01 last edited by VRonin 3 Jul 2018, 10:03
                  #24

                  yes I see that it is a loop but if i write :

                   if(eingangMatch.hasMatch())
                  								{
                  									if(eingangMatch.capturedRef(1).compare("80",Qt::CaseInsensitive)==0)
                  									{ // Eingang==("80:")
                  										qDebug() << eingangMatch;
                  										for(QRegularExpressionMatchIterator i = hexRegExp.globalMatch(line);
                  
                                                              i.hasNext();
                  
                  
                                                              )
                  
                  
                  											const QRegularExpressionMatch hexMatch = i.next();
                                                              qDebug() << hexMatch.capturedRef(0).toUInt(Q_NULLPTR,16); //capturedRef contains the part matched by the regular expression
                                                              y1.append(hexMatch.capturedRef(1).toUInt(Q_NULLPTR,16));
                                                               y2.append(hexMatch.capturedRef(2).toUInt(Q_NULLPTR,16));
                                                                y3.append(hexMatch.capturedRef(3).toUInt(Q_NULLPTR,16));
                                                                 y4.append(hexMatch.capturedRef(4).toUInt(Q_NULLPTR,16));
                  
                  
                  
                                                  }
                  

                  the qDebug() << eingangMatch; and qDebug() << hexMatch.capturedRef(0).toUInt(Q_NULLPTR,16);
                  don't work and all my y vectors get "0"
                  "80: 1C 05 3A 85 FF 4F FF 1D 89 1E 08 10 01 00 00 00 3B 87 51 01 AF 00 58 09 88 10 00 00 "
                  QVector(0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0)
                  QVector(0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0)

                  V 1 Reply Last reply 7 Mar 2018, 10:07
                  0
                  • L Leopold
                    7 Mar 2018, 10:01

                    yes I see that it is a loop but if i write :

                     if(eingangMatch.hasMatch())
                    								{
                    									if(eingangMatch.capturedRef(1).compare("80",Qt::CaseInsensitive)==0)
                    									{ // Eingang==("80:")
                    										qDebug() << eingangMatch;
                    										for(QRegularExpressionMatchIterator i = hexRegExp.globalMatch(line);
                    
                                                                i.hasNext();
                    
                    
                                                                )
                    
                    
                    											const QRegularExpressionMatch hexMatch = i.next();
                                                                qDebug() << hexMatch.capturedRef(0).toUInt(Q_NULLPTR,16); //capturedRef contains the part matched by the regular expression
                                                                y1.append(hexMatch.capturedRef(1).toUInt(Q_NULLPTR,16));
                                                                 y2.append(hexMatch.capturedRef(2).toUInt(Q_NULLPTR,16));
                                                                  y3.append(hexMatch.capturedRef(3).toUInt(Q_NULLPTR,16));
                                                                   y4.append(hexMatch.capturedRef(4).toUInt(Q_NULLPTR,16));
                    
                    
                    
                                                    }
                    

                    the qDebug() << eingangMatch; and qDebug() << hexMatch.capturedRef(0).toUInt(Q_NULLPTR,16);
                    don't work and all my y vectors get "0"
                    "80: 1C 05 3A 85 FF 4F FF 1D 89 1E 08 10 01 00 00 00 3B 87 51 01 AF 00 58 09 88 10 00 00 "
                    QVector(0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0)
                    QVector(0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0)

                    V Offline
                    V Offline
                    VRonin
                    wrote on 7 Mar 2018, 10:07 last edited by VRonin 3 Jul 2018, 11:07
                    #25

                    @Leopold said in [Solved]Convert ASCII hex to int:

                    yes I see that it is a loop

                    I don't think you do...

                    int h=0;
                    for(QRegularExpressionMatchIterator i = hexRegExp.globalMatch(line);i.hasNext();++h){
                    const QRegularExpressionMatch hexMatch = i.next();
                    qDebug() << hexMatch.capturedRef(0).toUInt(Q_NULLPTR,16); 
                    switch(h){
                    case 1 : y1.append(hexMatch.capturedRef(0).toUInt(Q_NULLPTR,16)); break;
                    case 2 : y2.append(hexMatch.capturedRef(0).toUInt(Q_NULLPTR,16)); break;
                    case 3 : y3.append(hexMatch.capturedRef(0).toUInt(Q_NULLPTR,16)); break;
                    case 4 : y4.append(hexMatch.capturedRef(0).toUInt(Q_NULLPTR,16)); break;
                    default: break;
                    }
                    }

                    "La mort n'est rien, mais vivre vaincu et sans gloire, c'est mourir tous les jours"
                    ~Napoleon Bonaparte

                    On a crusade to banish setIndexWidget() from the holy land of Qt

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                    • L Offline
                      L Offline
                      Leopold
                      wrote on 7 Mar 2018, 10:44 last edited by
                      #26

                      i get error for the "i" no match for operator "++"
                      and the switch operator is not an integer
                      I copied the int i=0 as well
                      btw:my compiler is mingw9-32 from QT 5.6.2

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                      • L Offline
                        L Offline
                        Leopold
                        wrote on 7 Mar 2018, 10:53 last edited by
                        #27

                        the i was occupied by the "QRegularExpressionMatchIterator"
                        i changed int i to int h then ++h and switch h and it works.
                        Thank you great!

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                        7 Mar 2018, 10:44

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