On Wed, Oct 09, 2002 at 02:16:52PM -0700, nkuipers wrote: > When I first saw this, I worked out the assignment where $a and $b each > get a 7 and the third 7 is discarded, then $x gets the last value in the > list of $a and $b because of the comma operator for list literals in a > scalar context...or does the presence of vars in the list mean that this > isn't a list literal, more like an array in behavior?
You go into a sort of ramble here, but I think this is where your confusion lies. You're trying to equate list assignment with either how an array behaves, or how a list behaves. List assignment is neither an array nor a list, it's a list assignment, so it behaves differently in different contexts. > So really, I guess I don't understand either side of the assignment and > how it behaves in scalar context due to $x. :) So I guess this could end > up being something I just memorize, but I'd rather understand what's > happening... Well, it's both; this is something you'll have to memorize, and with that, understand what's happening when a list assignment is encountered. Michael -- Administrator www.shoebox.net Programmer, System Administrator www.gallanttech.com -- -- To unsubscribe, e-mail: [EMAIL PROTECTED] For additional commands, e-mail: [EMAIL PROTECTED]
