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.
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