Nope. .. in scalar context is a flip-flop operator. A constant expression on either side is compared against $.; since you've not read any lines of input $. does not equal 1, so it's evaluating to a false value 0.

See perlop which explains all this in detail.

In reply to Re: Is this a bug, or expected behavior? by Fletch
in thread Is this a bug, or expected behavior? by fizbin

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