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)?
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.
#!/usr/bin/perl -wn
print if /^(f(o[^o]|[^o])|[^f])*$/;
#!/usr/bin/env python
import fileinput, re, sys
from itertools import ifilter
re_not_foo = re.compile(r"^(f(o[^o]|[^o])|[^f])*$")
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 "^\(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