find alphabetic substring - c++

Question

I have the following strings, from which i want to extract the alphabetic part (alphabetic substring) only which is greater than 1:
% d. i.p.p. attendu --> attendu
aprà ¨ s. expertise --> apr, expertise
n.c.p.c. condamner --> condamner
I am trying the following piece code :
#include <regex>
#include <iostream>
void main()
{
const std::string s = "% d. i.p.p. attendu";
std::regex rgx("[a-zA-Z]{2,20}");
std::smatch match;
if (std::regex_search(s.begin(), s.end(), match, rgx))
std::cout << "match: " << match[1] << '\n';
}
But I am having the following error when i run the code :
Terminate called after throwing an instance of 'std::regex_error' what(): regex_error
Can you please help me,
Thank you,
Hani.
Ok I managed to use boost since gcc's regex is an abomination.
#include <boost/regex.hpp>
void main()
{
const std::string s = "% d. i.p.p. tototo attendu";
boost::regex re("[a-zA-Z]{4,7}");
boost::smatch matches;
if( boost::regex_search( s, matches, re ) )
{
std::string value( matches[0].first, matches[0].second );
cout << value << " ";
}
}
Fine i found attendu but the output is only tototo. It's not incrementing
The return value is "tototo attendu" I was wondering if I can return each value at a time instead of 1 string

Answer 1

I was wondering if I can return each value at a time instead of 1 string
The only way of doing this seems to be via regex_iterator. Here’s an example using Boost:
#include <boost/regex.hpp>
#include <iostream>
int main() {
const std::string s = "% d. i.p.p. tototo attendu";
boost::regex rgx("([a-zA-Z]{2,20})");
boost::smatch match;
boost::sregex_iterator begin{s.begin(), s.end(), rgx},
end{};
for (auto&& i = begin; i != end; ++i)
std::cout << "match: " << *i << '\n';
}
This yields:
match: tototo
match: attendu
Two things:
The return type of main is always int. Your code shouldn’t even compile.
I’ve added parentheses around your (first, which was correct!) regular expression so that it creates a capture for each match. The iterators then iterate over each match in turn.