CODEWITHSUNDEEP
c#regexparsing
Robert
Robert

Reputation: 79

Parsing Error while using Regex

Let me explain the actual problem I am facing.

When my SearchString is C+, the setup of regular expression below works fine. However when my searchstring is C++, it throws me an error stating --

  parsing "C++" - Nested quantifier +.

Can anyone let me know how to get past this error?

RegExp = new Regex(Search_Str.Replace(" ", "|").Trim(), RegexOptions.IgnoreCase);

Upvotes: 1

Views: 1991

Answers (3)

Camilo Martin
Camilo Martin

Reputation: 37908

I could not help but notice @ghost's answer is wrong.

This snippet is wrong:

Regex regex = new Regex(@"C+\+");

Because the first + is not escaped. So the meaning still is "one or more C characters followed by a plus sign". The correct form would be @"C\+\+"

Upvotes: 1

Denis
Denis

Reputation: 6082

Plus + is a special symbol in regular expression. You must escape it. 

Regex regex = new Regex(@"C+\+");

Upvotes: 3

Jerry
Jerry

Reputation: 71598

First of all, I believe that you'll learn much more by looking at some regex tutorials instead of asking here at your current stage.

To answer your question, I would like to point out that + is a quantifier in regexp, and means 1 or more times the previous character (or group), so that C+ will match at least 1 C, meaning C will match, CC will match, CCC will match and so on. Your search with C+ is actually matching only C!

C++ in a regex will give you an error, at least in C#. You won't with other regex flavours, including JGsoft, Java and PCRE (++ is a possessive quantifier in those flavours).

So, what to do? You need to escape the + character so that your search will look for the literal + character. One easy way is to add a backslash before the +: \+. Another way is to put the + in square brackets.

This said, you can use:

C\+\+

Or...

C[+][+]

To look for C++. Now, since you have twice the same character, you can use {n} (where n is the number of occurrences) to denote the number of occurrences of the last character, whence:

C\+{2}

Or...

C[+]{2}

Which should give the same results.

Upvotes: 3

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