in reply to Stupid, yet simple, sort question

Okay...I got 3 answers...all three sound right...but I don't know which to use..I need the most efficient one.. 
#1:

sort{ $a->{users} ,<=> $b->{users} } @locations

#2

for my $loc ( sort { $a->{users} <=> $b->{users} } @locations )

#3

foreach $loc (reverse sort { int($a) <=> int($b) } @locations)

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Which one is the most efficient one? I need one that can take at least 3,000 different ranks...thats the stress limit of my script and it must be able to handle the stress limit..

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Re: Re: Stupid, yet simple, sort question
by davorg (Chancellor) on Oct 24, 2001 at 20:48 UTC

    None of them will work, as they all assume that you're using hard references - which you aren't.

    Unless I've completely misunderstood your code.

    --
    <http://www.dave.org.uk>

    "The first rule of Perl club is you don't talk about Perl club."

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      As noted in chatter, davorg was correct about Kage using symbolic referencees but was incorrect in thinking that the syntax $a->{key} does not work when $a isn't a "hard" reference (that is, when $a simply contains the name of some global hash variable).

      Of course, use strict would prevent it from working, but that also prevents ${$a}{key} from working.

              - tye (but my friends call me "Tye")
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        Let the record show that tye is 100% correct and my assumption that $a->{key} implied hard references was 100% wrong.

        Let the record further show, however, that in my opinion allowing that syntax to work with symbolic references is a very bad idea :)

        --
        <http://www.dave.org.uk>

        "The first rule of Perl club is you don't talk about Perl club."

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