Re: [Boost-users] Boost-regex: Weird behaviour with non-greedy matching operator in regex_replace in boost 1.40?

Subject: Re: [Boost-users] Boost-regex: Weird behaviour with non-greedy matching operator in regex_replace in boost 1.40?
From: Steven Watanabe (watanabesj_at_[hidden])
Date: 2009-09-23 11:58:11

AMDG

Florian Schwarz wrote:
> the non-greedy matching seems to have a weird behaviour (maybe 
> non-determeterministic?) to work if the pattern is preceeded with 
> something.

Boost.Regex doesn't guarantee that it finds the longest possible match.
perl does exactly the same thing:

$ perl -e '$x = "hhallo"; $x =~ s/h(.*?)o?/$1/g; print $x;'
allo

> E.g. when I want to get all characters in a string except the last on, 
> if its an 'o', I would use
>      regex matchExpr("(.*?)o?");
> So if I write
>      string text("hallo");
>      regex matchExpr("(.*?)o?");
>      string valueExpr("$1");
>      string result;
>      regex_replace(back_inserter(resul), text.begin(), text.end(), 
> matchExpr, valueExpr);
>      cout << "Match \"" << result << "\"" << endl;
> it will print the expected "hall". If I now use instead
>      string text(" hallo");
>      regex matchExpr(" (.*?)o?");
> it will print "hallo". And for
>      string text("hhallo");
>      regex matchExpr("h(.*?)o?");
>      string valueExpr("$1");
>      regex_replace(back_inserter(resul), text.begin(), text.end(), 
> matchExpr, valueExpr);
>      cout << "Match \"" << result << "\"" << endl;
> it will print "allo" which seems even more strange to me.

In this case, the regex matches each 'h'.

In Christ,
Steven Watanabe