Issue with regular expressions in C++

Question

I tried to use the following regular expression, which already works in C#, in C++ as well but it\'s not working in C++.

Answer 1

This is similar sln's, but it's shorter and doesn't require a % part to match:

^(?:[^%]*(?:%\.?[0-9]*[a-z])?)*$

First - everything between ^ (start of line) and $ (end of line) is optional, so an empty string is accepted.

In an optional, non capturing group (?:...), match any number of anything but a %. Then, optionally, match a %, optionally followed by a ., and then any number of digits and finally a letter. Repeat this for any number of times.

(I, as the others answering, and as the regex provided in the question suggests, assume that OP doesn't mean "immediately preceded by a letter", but instead followed by one, right?)

See it here at regex101.

Answer 2

Here you go, all % in string must be compliant.
If so, match the entire string, if not, don't match
the string.

I suggest you do this with i.e. if ( regex_search( sTarget, sRx, sMatch, flags ) )
but regex_match() would do the same thing.

^(?:[^%]*%(?:\.[0-9]*)?[a-z])+[^%]*$

Expanded

 ^                             # BOS
 (?:                           # Cluster begin
      [^%]*                         # Not % characters
      %                             # % found
      (?: \. [0-9]* )?              # optional .###
      [a-z]                         # single a-z required
 )+                            # Cluster end, 1 to many times
 [^%]*                         # Not % characters
 $                             # EOS

Answer 3

A simple regular expression that works with VC++ and fits your description would be

  std::regex("^([^%]|%(\\.[0-9]+)?[a-zA-Z])*$", std::regex::extended)

Live demo

(Change [0-9]+ to [0-9]* if digits after %. should be optional).