A colleague of mine was having trouble getting the \b metacharacter in a regular expression to work. Of course, when somebody asks a question like that, you first have to establish what their definition of "work" is. Fortunately, he provided some examples:

Regex.IsMatch("foo", @"\b" + @"foo" + @"\b") true
Regex.IsMatch("%1" , @"\b" + @"%1"  + @"\b") false
Regex.IsMatch("%1" , @"\b" + @"\%1" + @"\b") false
Regex.IsMatch("%1" , @"\b" + @"\%1" + @"\b") false
Regex.IsMatch("%1" , @"..") true
Regex.IsMatch("%1" , @"%1") true

"The last two entries are just sanity checks to make sure I didn't make some stupid mistake like passing the parameters in the wrong order. I want to search for a string that contains %1 with word boundaries on either side, something I would normally use \b for. Is there something special about the % character? Notice that the match succeeds when I look for the word foo."

Everything is working as it should. Recall that the \b metacharacter matches when there is a \w on one side and a \W on the other, where the beginning and end of the string are treated as if they were \W.

The string %1 therefore breaks down as

virtual \W beginning of string
\W % is not an alphanumeric or _
\w 1 is a digit
virtual \W end of string

The only points where \b would match are immediately before and after the 1, since those are the transition points between \w and \W and vice versa. In particular, the location immediately before the percent sign does not match since it is surrounded by \W on both sides.

My colleague responded, "D'oh! I keep forgetting that % won't act like a \w just because I want it to."