I just started using Boost::regex today and am quite a novice in Regular Expressions too. I have been using "The Regulator" and Expresso to test my regex and seem satisfied with what I see there, but transferring that regex to boost, does not seem to do what I want it to do. Any pointers to help me a solution would be most welcome. As a side question are there any tools that would help me test my regex against boost.regex?
using namespace boost;
using namespace std;
vector<string> tokenizer::to_vector_int(const string s)
{
regex re("\d*");
vector<string> vs;
cmatch matches;
if( regex_match(s.c_str(), matches, re) ) {
MessageBox(NULL, L"Hmmm", L"", MB_OK); // it never gets here
for( unsigned int i = 1 ; i < matches.size() ; ++i ) {
string match(matches[i].first, matches[i].second);
vs.push_back(match);
}
}
return vs;
}
void _uttokenizer::test_to_vector_int()
{
vector<string> __vi = tokenizer::to_vector_int("0<br/>1");
for( int i = 0 ; i < __vi.size() ; ++i ) INFO(__vi[i]);
CPPUNIT_ASSERT_EQUAL(2, (int)__vi.size());//always fails
}
Update (Thanks to Dav for helping me clarify my question): I was hoping to get a vector with 2 strings in them => "0" and "1". I instead never get a successful regex_match() (regex_match() always returns false) so the vector is always empty.
Thanks '1800 INFORMATION' for your suggestions. The to_vector_int()
method now looks like this, but it goes into a never ending loop (I took the code you gave and modified it to make it compilable) and find "0","","","" and so on. It never find the "1".
vector<string> tokenizer::to_vector_int(const string s)
{
regex re("(\d*)");
vector<string> vs;
cmatch matches;
char * loc = const_cast<char *>(s.c_str());
while( regex_search(loc, matches, re) ) {
vs.push_back(string(matches[0].first, matches[0].second));
loc = const_cast<char *>(matches.suffix().str().c_str());
}
return vs;
}
In all honesty I don't think I have still understood the basics of searching for a pattern and getting the matches. Are there any tutorials with examples that explains this?
See Question&Answers more detail:os