why is that in some cases I have to use $ and some cases @?

$array refers to an element of an array. It must be followed by an index, like $array[3] and $array[$i].

@array referes to the entire array. For example scalar(@array) returns the number of elements in the array, and @a = @b copies the array.

@array can also be used to return an array slice. For example, @s = @array[1, 5 .. 7, $i] is the same as @s = ($array[1], $array[5], $array[6], $array[7], $array[$i]).

Basically, use $ when the return type is a scalar, and @ when the return type is a list/array.

Therefore $matrix[$ab][3]{_PARTNERS} is the appropriate choice here.

Examples:
$count = scalar(@matrix);
$count = @matrix;           # scalar() is optional since we're assigni
+ng to a scalar.
$count = $#matrix + 1;
$first_row = $matrix[0];    # this is a reference, rememeber..
@first_row = @$first_row;   # dereference it.
@first_row = @{$first_row}; # dereference it.
@first_row = @{$matrix[0]}; # dereference it.
$count_of_first_row = scalar(@first_row);     # scalar() is optional h
+ere.
$count_of_first_row = scalar(@{$matrix[0]});  # scalar() is optional h
+ere.
$first_field_of_second_row = $matrix[1][0];
$first_field_of_second_row = $matrix[1]->[0]; # this is what you are r
+eally doing.

[download]
Lets say if I have a variable that is supposed to look like this: Evan_PARTNERS and I'm trying to declar it by using the matrix variable

I don't understand the question. but since you're dealing with a single element of your datastructure, you'd be using $, not @.

In reply to @array vs $array by ikegami
in thread Confused about "unitialized value" warning by xspikx

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