Re: scalar to list conversion

[Steve Grazzini]

On Sat, Jun 28, 2003 at 11:04:33AM +0100, Rob Dixon wrote:
> John W. Krahn wrote:
> >
> > @s = split ' ', $s;
> 
>   @s = split for $s

    (*s,*_) = \($_,@s);

    split;

-- 
Steve

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