I don't understand most of your post (too many ...s that aren't clear), but you should probably stay away from the word "type" when you are talking about Perl variables. "Scalar" doesn't really mean anything. If anything, it describes the cardinality of the contained value, but certainly not the "type".

Even values stored in variables have very flexible "types" that I wouldn't really call a "type".

If you store "42 is a number" in a scalar, you still don't have a type. Maybe it's a string, maybe it's an integer, maybe it's a boolean. Types only show up when you apply an operator. If you say "42 is a number" - 42, then the result is 0 because "-" coerces the string into an integer (actually a SvPv to an SvPvIv, and then looks at the integer part of the data structure). This is different from other languages that choose the operator based on the type of the value (look at how SBCL chooses the right + operator to used based on the types of the contained values. the exact opposite of Perl).

As an aside, this coercion isn't an implementation detail; Perl exposes it all to you. If you "use overload", you can have "boolify" subroutine called when someone does "!$your_object", a stringify subroutine that's called when someone does "$your_object =~ /.../" or "qq{$your_object}", a "numify" subroutine that's called when someone does "$your_object - 42", etc.

Anyway, not sure what my point is... but don't use the word "type" to talk about values or variables in Perl. It just doesn't make sense. Perl is totally different from other languages here, and the word "type" has too much incompatible baggage associated with it.

okay, in summary, I wanted to say.

In Perl $ means scalar, @ means array and % means hash

Those are variable types, you learn them by looking at the variable syntax or name

The value pointed to by those TYPED variables can be more complex, you can have integers, floats, strings, references, reference to a string, reference to array, reference to hash, array of strings, array of references etc .....

but the variable type won't tell you about these complex structures, it will only tell you that the value is $ scalar or @ array or % hash ... and in that there is no confusion, all the variables you mentioned above are typed scalar and can only point to values of type scalar

I understand your problem, since a scalar can be a reference to an array

you want to think that $ref = [] points to an array but it doesn't it points to scalar which happens to be a reference to an array!

you can for example do this

my $ref = [];

$ref->[0]= 0;

say $ref->[0];

$ref = {};

$ref ->{0} = 1;

say $ref->{0};

but you can't do this

my $ref = [];

$ref->[0]= 0;

say $ref->[0];

## $ref = {};

$ref ->{0} = 1;

say $ref->{0};

because in $ref ->{0} = 1 the assignment is to value referenced to by te scalar, and this value type is not a hash and can't have a hash member {0}