my @indicies = sort { $a[$a] <=> $a[$b] } 0..$#a;
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-- Randal L. Schwartz, Perl hacker

thank you very much. But why does this work??

my @indicies = sort { $a[$a] <=> $a[$b] } 0..$#a;

merlyn is sorting a list of integers (0..$#a) corresponding to the indices of the elements of your array @a. Recall that within sort code blocks, $a and $b are special variables representing the two variables to be ordered each time the block is evaluated. So here,  $a[$a] <=> $a[$b] means sort the indices of the array by the contents of its elements.

It's a bit confusing in this case because of the use of @a as your array variable and the use of $a and $b as special variables to sort.

HTH,

Josh

If you still have specific questions after reading "perldoc -f sort", please ask again.

-- Randal L. Schwartz, Perl hacker

my $i = 0;
my @sorted_indicies = map { $_->[1] } 
                      sort { $a->[0] <=> $b->[0] }
                      map { [ $_, $i++ ] } @array;
# OR 
my @sorted_indicies = map { $_->[1] } 
                      sort { $a->[0] <=> $b->[0] }
                      map { [ $array[$_], $_ ] } (0..$#array);
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-----------------------------------------------------
Dr. Michael K. Neylon - mneylon-pm@masemware.com || "You've left the lens cap of your mind on again, Pinky" - The Brain
"I can see my house from here!"
It's not what you know, but knowing how to find it if you don't know that's important

If you'll grant me an audience knowing that I know a bit about the Schwartzian Transform... {grin}

The ST makes sense only when the orderingfunction on each element to be sorted is expensive to compute, and should therefore be cached to prevent multiple calls (so the list has to be long enough as well). Since "array lookup" is a cheap (nearly free) orderingfunction, the ST is far too much overkill on this choice.

-- Randal L. Schwartz, Perl hacker