4

Hey, let's say I have this regex: (test[0-9])+

And that I match it against: test1test2test3test0

const bool ret = boost::regex_search(input, what, r);

for (size_t i = 0; i < what.size(); ++i)
    cout << i << ':' << string(what[i]) << "\n";

Now, what[1] will be test0 (the last occurrence). Let's say that I need to get test1, 2 and 3 as well: what should I do?

Note: the real regex is extremely more complex and has to remain one overall match, so changing the example regex to (test[0-9]) won't work.

rcollyer
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Andreas Bonini
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3 Answers3

11

I think Dot Net has the ability to make single capture group Collections so that (grp)+ will create a collection object on group1. The boost engine's regex_search() is going to be just like any ordinary match function. You sit in a while() loop matching the pattern where the last match left off. The form you used does not use a bid-itterator, so the function won't start the next match where the last match left off.

You can use the itterator form:
(Edit - you can also use the token iterator, defining what groups to iterate over. Added in the code below).

#include <boost/regex.hpp> 
#include <string> 
#include <iostream> 

using namespace std;
using namespace boost;

int main() 
{ 
    string input = "test1 ,, test2,, test3,, test0,,";
    boost::regex r("(test[0-9])(?:$|[ ,]+)");
    boost::smatch what;

    std::string::const_iterator start = input.begin();
    std::string::const_iterator end   = input.end();

    while (boost::regex_search(start, end, what, r))
    {
        string stest(what[1].first, what[1].second);
        cout << stest << endl;
        // Update the beginning of the range to the character
        // following the whole match
        start = what[0].second;
    }

    // Alternate method using token iterator 
    const int subs[] = {1};  // we just want to see group 1
    boost::sregex_token_iterator i(input.begin(), input.end(), r, subs);
    boost::sregex_token_iterator j;
    while(i != j)
    {
       cout << *i++ << endl;
    }

    return 0;
}

Output:

test1
test2
test3
test0

6

Boost.Regex offers experimental support for exactly this feature (called repeated captures); however, since it's huge performance hit, this feature is disabled by default.

To enable repeated captures, you need to rebuild Boost.Regex and define macro BOOST_REGEX_MATCH_EXTRA in all translation units; the best way to do this is to uncomment this define in boost/regex/user.hpp (see the reference, it's at the very bottom of the page).

Once compiled with this define, you can use this feature by calling/using regex_search, regex_match and regex_iterator with match_extra flag.

Check reference to Boost.Regex for more info.

Vitus
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3

Seems to me like you need to create a regex_iterator, using the (test[0-9]) regex as input. Then you can use the resulting regex_iterator to enumerate the matching substrings of your original target.

If you still need "one overall match" then perhaps that work has to be decoupled from the task of finding matching substrings. Can you clarify that part of your requirement?

Steve Townsend
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