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in reply to Re: Save all but line 32!
in thread Save all but line 32!


Very nice. ++

I like the twisted logic but the best part is the bare $ at the end. I had to run it through B::Deparse to see what was happening: 

    $ perl -MO=Deparse  -pe '$_=$'

    LINE: while (defined($_ = <ARGV>)) {
        $_ = $;
    }
    continue {
        die "-p destination: $!\n" unless print $_;
    }

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It shows that in this case $ gets interpreted as $; the subscript separator. However, this means that your code inserts an extra character in the file as shown by this:     perl -pe '$.-32or$_=$' file | cat -A

But I'm still not sure why perl inserts that semicolon. It doesn't in other cases: 

    $ perl -MO=Deparse  -pe '$_=$.'

    LINE: while (defined($_ = <ARGV>)) {
        $_ = $.
    }
    continue {
        die "-p destination: $!\n" unless print $_;
    }

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Anyone have an explanation for this last point?

Update: blakem's answer below is right. This last case is a perl 5.005 issue with B::Deparse.

--
John.

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Re3: Save all but line 32!
by blakem (Monsignor) on Sep 19, 2002 at 08:08 UTC
    I seem to get the semicolon with your second case: 
    % perl5.6.1 -MO=Deparse  -pe '$_=$.'
   LINE: while (defined($_ = <ARGV>)) {
       $_ = $.;
   }
   continue {
       die "-p destination: $!\n" unless print $_;
   }

% perl5.8.0 -MO=Deparse  -pe '$_=$.'
   LINE: while (defined($_ = <ARGV>)) {
       $_ = $.;
   }
   continue {
       die "-p destination: $!\n" unless print $_;
   }

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    Update: Response to above update: with 5.00503 I get: 
    % perl5.00503 -MO=Deparse  -pe '$_=$.'
   LINE: while (defined($_ = <ARGV>)) {
       $_ = $.
   }
   continue {
       die "-p destination: $!\n" unless print $_;
   }

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    -Blake

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