CODEWITHSUNDEEP
c++regexqtqregexp
Devolus
Devolus

Reputation: 22094

Regular Expression gives wrong matches in QRegEx

I have the following function which should give me the matched string from a regular expression:

QString selectByPattern(QString const &oValue, QString const &oPattern, bool bRegularExpression) const
{
    QString s;
    QRegExp regex;

    // default
    regex.setPatternSyntax(QRegExp::RegExp);

    regex.setPattern(oPattern);
    int i = oValue.indexOf(regex);
    int l = regex.matchedLength();

    if(i == -1 || l < 1)
        return s;

    s = oValue.mid(i, l);

    return s;
}

I'm using the string *VALUE* as input for testing this. Now i get the following results:

 oPattern = "[A-Z]"
 Output = "V"

which is correct. For this pattern it gives me the first upper case character and a length of 1.

But when I use this:

 oPattern = "[A-Z]*"
 Output = ""

and I don't understand why. The index returned is 0 and length is also 0. First of all, of course the index is wrong, because it should be 1. And then I don't understand the meaning of 0 as length. According to the documentation (http://qt-project.org/doc/qt-4.8/qregexp.html#pattern) Either there is a match, then the index is a positive value, and in that case I would expect that matchedLength() should also return some value `> 0``.

Am I missunderstanding something here, or is this a problem in Qt 5.2.1?

Upvotes: 0

Views: 300

Answers (1)

vks
vks

Reputation: 67988

[A-Z]*

can matched the string VALUE and an empty string as well. See demo. The second match will be of 0 length as after VALUE is matched nothing is left.

You can try [A-Z]+

Your index is 0, because positions in strings start with 0.

string:   "VALUE"
           ^   ^
position:  0   4

So the match (pos=0, len=3) would be the following VAL, (pos=0, len=1) is V and (pos=0, len=0) is the empty string.

Upvotes: 3

Related Questions