in reply to Re: printing the subscript separator
in thread printing the subscript separator

Perhaps you mean $operator !~ /^\s*$/?

Nope. That if statement is suppose to filter values that are space characters (not zero or more space char).

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Re^3: printing the subscript separator
by RazorbladeBidet (Friar) on Feb 09, 2005 at 18:25 UTC
    Perhaps you could be more descriptive :)

    Update: That's zero or more space characters from start to end... in other words... either nothing (no chars) or all whitespace.

    my $var = '';
    That will not be caught by /^\s+$/
      But it would be caught by the if statement since it checks if $operator is not defined.

      Update: I was thinking of 'false'.

        An empty string is defined; only undef is undefined.

        freebie:~ 663> perl -le 'print "defined" if defined ""' defined
        It is defined.
        $operator = '';

        Contains no whitespace, yet it's defined... not sure where that falls in your original criteria (my guess is you don't want it)