On Fri, Jan 30, 2004 at 10:33:27AM -0500, Jeff 'japhy' Pinyan wrote: > Well, that only works if $a[0] is [1,2,3] and $b[1] is $a[0] -- that is, > the EXACT SAME reference. It won't work if $b[1] is its own [1,2,3]. Hmm, right. Not so good. I had thought of that, and thought I tested it in that situation, but I can't reproduce those tests now, so obviously I messed something up. Your method looks somewhat nicer than mine, anyway, so I may steal that :)
cheers, -- Robin <[EMAIL PROTECTED]> JabberID: <[EMAIL PROTECTED]> Hostes alienigeni me abduxerunt. Qui annus est? PGP Key 0x776DB663 Fingerprint=DD10 5C62 1E29 A385 9866 0853 CD38 E07A 776D B663
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