Why is this using printf and not print? Perhaps the line you're printing (in $_) contains some formatting that printf is trying to interpret, and that causes the failure.
UPDATE: This gets the message you're getting:
printf '%n';
According to sprintf, the %n format is "special: *stores* the number of characters output so far into the next variable in the parameter list".
So! What's the text on line 20804003?
In reply to Re: line number ($.) problem ?
by kyle
in thread line number ($.) problem ?
by 5mi11er
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