If .. or ... in used scalar context, and one of the arguments is a numerical literal (as is the case here), there's an implicit comparison against $. is made. So, in the given examples, after matching the left hand side of ..., $. equals 1. So, before the right hand side is tested, $. equals 2. Hence, ...0 or ...1 will never match, but ...2 will.

In reply to Re^2: What's the "...0" mean? by JavaFan
in thread What's the "...0" mean? by tel2

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