What you are seeing is a timing issue. Warnings get printed out via STDERR. Your print statements print out via STDOUT. These are two separate output streams. Sometimes one stream gets ahead of the other. $a has not yet become 2. The warning you are seeing is, in fact, coming from the $a +=1; statement.

You can see this for yourself by either finding line 6 (it is the one containing $a += 1;) or by changing all of the print statements to begin with print STDERR "...".

Best, beth

In reply to Re: _ in a number within quotes by ELISHEVA
in thread _ in a number within quotes by dilpane

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