The answer is that the [^a] phrase means any character other than 'a', but it still must match one character. Since the string 'at' only has 2 characters, there is no character at all to match.

So what is actually called for is a negative-lookbehind: /(?<!a)at/ that ensures 'at' isn't preceded by an 'a' character, it will still match 'at' when it begins in the first character position of a string.

/[^a]?at/ Would behave as you expected the original regex to behave. Update: ugh, this was a little optimistic, it will do the wrong thing with 'aat' now.

Just note the difference between these two regular expressions:

[^a]at
[download]

Your regex, will match any expression with three characters where the first one is not 'a' and the rest is 'at'

[^a]?at
[download]

The regex you would expect, will match any expression ending with 'at' and starting AT MOST (repetition operator ?) by any character not being 'a'

Not the suitable place to post it since it seems I'm answering to Loops, sorry. The response was directed to anaconda

Why not 'at'?

The Perl Tutorial answers your questions as Both [...] and  [^...] must match a character

The special character ^ in the first position of a character class denotes a negated character class, which matches any character but those in the brackets. Both ... and ^... must match a character, or the match fails.

Thank loop and vinoth both. Yes I noticed ... must match a character in the tutorial but didn't understand ^... must match a character. Now I see. Thanks!

What tutorial (link)?