You are correct, $a would equal 3 in my last bit of code, because the sort will return a list and then store it in a referenced array which is then dereferenced and cast into scalar context which finally returns the size of the array. Got it? Good.
I dont think I implied anything else though. I am not under the impression that ZZamboni wanted anything other than the size of the sorted array, but maybe I am wrong.

If you wanted the last element of the sorted array of course you would do (sort @a)[-1] I certainly would not guess that scalar(sort @a) would return the last element, so I am glad that it doesnt, but apparently you think it should. I suppose you could write your own sort routine to do that in scalar context, but I like it the way it is. Just my opinion though.

In reply to RE: RE: Re: scalar doesn't work values returned by a function? by perlmonkey
in thread scalar doesn't work values returned by a function? by ZZamboni

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