perlrocks has asked for the wisdom of the Perl Monks concerning the following question:

Hi everyone, I have the following piece of code: 
my @set_S;
my @nset_S;
my @epsilons;
my $r=0;

for $r (0..99){
    $set_S[$r] = [$r];
}

$q = 2;
$sigma_square = 0.1 * $q;
for my $err (0..$#set_S){
    $epsilons[$err] = rand($sigma_square + 1);
}
my $t=0;
for my $s (0..$#set_S){
    print "@{$set_S[$s]}\t$epsilons[$s]\t";
    @{$set_S[$s]} = @{$set_S[$s]} + $epsilons[$s];
    print @{$set_S[$s]}, "\n";
    $t++;
}

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The output looks like: 
0       0.16702880859375        1.16702880859375
1       0.54862060546875        1.54862060546875
2       0.75128173828125        1.75128173828125
3       0.67547607421875        1.67547607421875
4       0.58304443359375        1.58304443359375
5       0.98953857421875        1.98953857421875
6       0.07342529296875        1.07342529296875
7       0.43919677734375        1.43919677734375
8       0.70689697265625        1.70689697265625

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Apparently, my question is :Why is the addition wrong?? Why is perl turning the first columns to 1's when adding them to the second column?? How do i solve this? I appreciate your prompt response. Thanks

Replies are listed 'Best First'.
Re: Turning numbers to 1!
by moritz (Cardinal) on Jul 03, 2008 at 20:38 UTC
    @{$set_S[$s]} + $epsilons[$s];

    The first term here is an array. If you use an array in scalar context, it evaluates to the number of items in the array, which happens to be 1 in your case.

    If you just want to add the current index to the value, why not just say $s + $epsilons[$s]? Or if you want to access the first item of @{$set_S[$s]}, you can say $set_S[$s][0] + $epsilons[$s].

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[d/l]
[select]
      Thanks a lot pal. Now i know why! Pheww! :-)
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Re: Turning numbers to 1!
by alexm (Chaplain) on Jul 03, 2008 at 20:51 UTC
    Instead of 
        $set_S[$r] = [$r];

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    you can just say 
        $set_S[$r] = $r;

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    and then 
        print "$set_S[$s]\t$epsilons[$s]\t";
    $set_S[$s] = $set_S[$s] + $epsilons[$s];
    print $set_S[$s], "\n";

    [download]
    I don't see any reason to enclose $r on an anonymous array reference, at least in this context.
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[d/l]
[select]
Re: Turning numbers to 1!
by jds17 (Pilgrim) on Jul 03, 2008 at 20:43 UTC
    In the line 
    @{$set_S[$s]} = @{$set_S[$s]} + $epsilons[$s];

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    the addition creates a scalar context for the right-hand side, in which 
    @{$set_S[$s]} = 1.

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    To get what you want, write 
    @{$set_S[$s]} = $set_S[$s] + $epsilons[$s];

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    instead.
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[d/l]
[select]
      $set_S[$s] + $epsilons[$s];

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      You forgot that $set_S[$s] is an array reference; moritz pointed out the right way to access the value.
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[d/l]
[select]
        Thanks, you are right, I focused on the scalar context and did not finish to examine the expression correctly.
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