Question

In some regex flavors, [negative] zero-width assertions (look-ahead/look-behind) are not supported.

This makes it extremely difficult (impossible?) to state an exclusion. For example ‘every line that does not have ‘foo’ on it’, like this:

^((?!foo).)*$ 

Can the same thing be achieved without using look-around at all (complexity and performance concerns set aside for the moment)?

Answer 1

UPDATE: It fails ‘with two ff before oo’ as @Ciantic pointed out in the comments.

---

^(f(o[^o]|[^o])|[^f])*$ 

NOTE: It is much much easier just to negate a match on the client side instead of using the above regex.

The regex assumes that each line ends with a newline char if it is not then see C++’s and grep’s regexs.

Sample programs in Perl, Python, C++, and grep all give the same output.

• perl

  #!/usr/bin/perl -wn print if /^(f(o[^o]|[^o])|[^f])*$/; 

• python

  #!/usr/bin/env python import fileinput, re, sys from itertools import ifilter  re_not_foo = re.compile(r'^(f(o[^o]|[^o])|[^f])*$') for line in ifilter(re_not_foo.match, fileinput.input()):     sys.stdout.write(line) 

• c++

  #include <iostream> #include <string> #include <boost/regex.hpp>  int main() {   boost::regex re('^(f(o([^o]|$)|([^o]|$))|[^f])*$');   //NOTE: '|$'s are there due to `getline()` strips newline char    std::string line;   while (std::getline(std::cin, line))      if (boost::regex_match(line, re))       std::cout << line << std::endl; } 

• grep

  $ grep '^\(f\(o\([^o]\|$\)\|\([^o]\|$\)\)\|[^f]\)*$' in.txt 

Sample file:

foo 'foo' abdfoode abdfode abdfde abcde f  fo foo fooo ofooa ofo ofoo 

Output:

abdfode abdfde abcde f  fo ofo