You are 80% right, but it means more than $, you should also include @, %, &.

That is actually sort of comment for a person who is familiar with c/c++. In c/c++, there is a single de-ref operator *. It doesn't matter what the pointer points to, you can always use * to de-ref it.

However in Perl, we do not have such a single operator for de-ref. We have to use @, %, &. There is a big difference. In Perl, in order to de-ref a ref, you have to first know what the ref refs to. In c/c++, type doesn't matter at all, you can always cast a pointer to a certain type, which may have nothing to do with the original type, and then de-ref it, as long as don't corrupt the memory boundary. (In real life, you almost always deref a pointer to something meaningful in the context)
In Perl, if you do:

%a = (1,1,2,2);
$a = \%a;
print $$a;
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It gives you an error, and says that $a is not a SCALAR ref. You have to do:

%a = (1,1,2,2);
$a = \%a;
print %{$a};
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In c/c++, * also de-ref's pointers to functions, in Perl, we have to use &:

sub a {
    print "abc";
}

$a = \&a;
&{$a}; # call a, and prints abc
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