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= is right associative. Thus you have list = list (rightmost) before you get scalar = list = list. The list in question has 3 elements. Of course in this case, I think the behavior of an array in scalar context has more to do with the overall result than anything else. Note that if you replace the array declaration with any list of the form ($a,$b,...), you still get the count in $n2.

perldoc perldata, look for "List assignment in scalar context". Note that the extra parentheses around the rightmost assignment in the example are superfluous since they don't affect the order of operations.

Also, to those downvoting my nodes in this thread: if the reason that you're downvoting my nodes is due to the lack of technical merit of my posts, feel free to /msg me or reply.

In reply to Re^5: List assignment in scalar context by !1
in thread List assignment in scalar context by Grimnebulin

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