in reply to if modifier of a print statement

I strongly suspect you're being bitten by operator precedence.

print $_ . "\n" and next if /no_proc/;

[download]
--

See the Copyright notice on my home node.

Perl training courses

Replies are listed 'Best First'.
Re^2: if modifier of a print statement
by tprocter (Sexton) on Jun 30, 2009 at 11:14 UTC
    davorg is right. This is a precedence issue.

    using 'and' instead of '&&' lowers the operator precedence.

    The first one is interpreted as:
    print $_ . ("\n" && next) if /no_proc/;

    The fixed one is intepreted as:
    (print $_ . "\n") and next if /no_proc/;

    Note that the brackets include the print statement, not just the parameters.
[reply]
[d/l]
[select]
      The first one is interpreted as:
      print $_ . ("\n" && next) if /no_proc/;

      The  . operator has higher precedence than the  && operator so it is actually interpreted as: 

      $ perl -MO=Deparse,-p -e' print $_ . "\n" && next if /no_proc/; '
(/no_proc/ and print((($_ . "\n") && next)));
-e syntax OK

      [download]
[reply]
[d/l]
[select]

In Section Seekers of Perl Wisdom