[Solved]Convert ASCII hex to int

[mpergand]

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-tête:"<<fields[0];
QByteArray values=QByteArray::fromHex(fields[1].toLatin1());
qDebug()<<"data:"<<values;

en-tête: "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"

[Leopold]

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.

[mpergand]

@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 ...

[Leopold]

@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).

[VRonin]

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

[Leopold]

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));
                                    }

[VRonin]

@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?

[Leopold]

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)

[VRonin]

@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;
}
}

[Leopold]

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

[Leopold]

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!