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

Let's put it in English

I want to say: $operator is something more than *just* a space or spaces. So saying as you suggest: "$operator does not match a space". Wouldn't do that.

You are right about the plus sign and I have removed it. Thanks. So now I have:

$operator !~ /[^\s]/
But I am still seeing the same problem with $;

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Re^3: printing the subscript separator
by cog (Parson) on Feb 09, 2005 at 19:16 UTC
    $operator !~ /^\s*$/

    I used a plus, but I should have used an asterisk :-)

    How about now? Does it work? :-)