Line Oddskool wrote:
John Maddock wrote:
The way to find out which sub-expression matched is simply:
match_results<something> what; ... for(unsigned i = 1; i < what.size(); ++i) { if(what[i].matched) std:cout << "sub-expression " << i << " matched " << what[i] << std::endl; }
First, thanks for the reply, it's nearly what I need ;)
Actually, either I misunderstood something or is this code giving me just the first part in the regex that matches ?
e.g. if I try
<code>
std::string myString ="jayjay"; std::string myRegexSearch = "(ay)|(j)(?=[aeiouy])";
boost:regex* myRegexp = new boost::regex (myRegexSearch, boost::regex::normal);
boost::cmatch what;
boost::regex_search (myString , what, *myRegexp);
for(unsigned _ = 1; _ < what.size(); ++_) { if(what[_].matched) printf("rule [%d] matched '%s'\n",_,what[_]); } delete myRegexp;
</code>
The output is
rule [2] matched 'jayjay'
where I expected it to tell me "rule 1" and "rule 2" matched!
Am I missing something?
Well I'm amazed your computer didn't explode in a ball of flame when you passed a structure (boost::sub_match) by value to printf :-) Hint: match_results::operator[] does *not* return a null-terminated string. But apart from that if you want to search through all the occurances of the expression in the text, then use regex_iterator: std::string mystring = whatever; boost::regex e(myregex); boost::sregex_iterator i(e, mystring.begin(), mystring.end()), j; while(i != j) { for(unsigned sub = 1; sub < what.size(); ++sub) { if((*i)[sub].matched) std::cout << (*i)[sub] << std::endl; } } HTH, John.