C++11 RegEx, non-greedy

Question

I have a little bit of a problem with a C++11 RegEx and I think it is about greedynes.

Here is a little sample.

#include <stdio.h>
#include <string>
#include <regex>

int main (void)
{
  std::string in="{ab}{cd}[ef]{gh}[ij][kl]";  // the input-string

  std::regex rx1 ("(\\{.+?})(.*)", std::regex::extended);       // non-greedy?
  std::smatch match;

  if (regex_match (in, match, rx1))
  {
    printf ("\n%s\n", match.str(1).c_str());
  }

  return 0;
}

I would expect

{ab}

for output. But I got

{ab}{cd}[ef]{gh}

I would expect the result I get, if I do it greedy but not with the ? after the .+. Should make it non-greedy, right?

So whats the problem in my idea? Thanks for help!

Chris

Why do you escape the opening curly brace with `\\{` but you do not escape the closing curly brace? — Bobulous, Nov 18 '16 at 10:02
@Bobulous that's a regex special char which helps to repeat the previous pattern.. — Avinash Raj, Nov 18 '16 at 10:03
I think you should not use `std::regex::extended`, it makes your regex POSIX ERE compliant, that does not support lazy quantifiers. — Wiktor Stribiżew, Nov 18 '16 at 10:03
Ah, okay, the opening brace is considered a special character by regex, but a closing brace is not. Did not know that. — Bobulous, Nov 18 '16 at 10:05

score 5 · Accepted Answer · answered Nov 18 '16 at 10:06

5

You need to remove the std::regex::extended, it makes your regex POSIX ERE compliant, and that regex flavor does not support lazy quantifiers.

std::regex rx1("(\\{.+?})(.*)");

See the C++ demo

answered Nov 18 '16 at 10:06

Wiktor Stribiżew

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C++11 RegEx, non-greedy

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