Re: isNumber( ) ??

[Philip Newton]

On Fri, 19 Oct 2001 18:05:54 +0100, [EMAIL PROTECTED] (Piers Cawley)
wrote:

>     #!/usr/bin/perl6
> 
>     sub is_number ($number) { return +$number eq 'NaN' }

s/eq/ne/ ?

Cheers,
Philip