Re: 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)
[download]

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..

Comment on Re: Stupid, yet simple, sort question Download Code

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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."	[reply]
(tye)Re: Stupid, yet simple, sort question by tye (Sage) on Oct 24, 2001 at 23:19 UTC
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 $a simply contains the name of some global hash variable). Of course, use strict would prevent it from working, but that also prevents `${$a}{key}` from working. - tye (but my friends call me "Tye")	[reply] [d/l] [select]
Re: (tye)Re: Stupid, yet simple, sort question by davorg (Chancellor) on Oct 25, 2001 at 12:15 UTC
Let the record show that tye is 100% correct and my assumption that `$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."	[reply]


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