C++11 Regular Expression

Syed Babu wrote:

regex r1("<.*>.*</\\1>");
regex r2("<(.*)>(.*)</(\\1)>");
smatch s1,s2;
string str = "Hello,world<book>C++</book>";
regex_search( str,s1,r1);
regex_search( str,s2,r2);

Both the seraches return the result as true but s1.size is not equal  to s2.size. What does this mean?

Parentheses introduce capture groups. match_results object represents an  array of submatches - one (at index zero) for the regular expression as  a whole, and one for each capture group.

In your example, s1.size() should be 1, with s1[0] ==  "<book>C++</book>". s2.size() should be 4: s2[0] ==  "<book>C++</book>", s2[1] == "book", s2[2] == "C++", s2[3]  == "book".

What significance the paranthesis apart from "capture group"?

None. Their significance is precisely in indicating a capture group.

string data = "<Person>\n""<First>Nico</First>\n""<Last>Jos</Last>\n""</Person>";
regex r("<(.*)>(.*)</(\\1)>");
sregex_iterator pos( data.cbegin( ), data.cend( ),r )
sregex_iterator end;
for( ; pos != end ; ++pos )
{
}

The above code works fine. But passing the regex as temporary object  gives assert failure while iterating the iterator. What is
the reason for the assert?

The documentation for regex_iterator constructor states: "The second  constructor initializes... the stored value pregex with &re". So it  stores a pointer to regex object. Naturally, said regex object must  outlive regex_iterator, for the pointer to remain valid.

Igor Tandetnik