Due to the mechanism of introspection, it is easily possible to tell the type of the data living in a variable (a variable in Perl 6 is often referred as a container). To do that, call the predefined WHAT method on a variable. Even if it is a bare scalar, Perl 6 treats it internally as an object; thus, you may call some methods on it.
For scalars, the result depends on the real type of data residing in a variable. Here is an example (parentheses are part of the output):
my $scalar = 42; my $hello-world = "Hello, World"; say $scalar.WHAT; # (Int) say $hello-world.WHAT; # (Str)
For those variables, whose names start with the sigils @ and %, the WHAT method returns the strings (Array) and (Hash).
Try with arrays:
my @list = 10, 20, 30; my @squares = 0, 1, 4, 9, 16, 25; say @list.WHAT; # (Array) say @squares.WHAT; # (Array)
Now with hashes:
my %hash = 'Language' => 'Perl'; my %capitals = 'France' => 'Paris'; say %hash.WHAT; # (Hash) say %capitals.WHAT; # (Hash)
The thing, which is returned after a WHAT call, is a so-called type object. In Perl 6, you should use the === operator to compare these objects.
For instance:
my $value = 42; say "OK" if $value.WHAT === Int;
There’s an alternative way to check the type of an object residing in a container — the isa method. Call it on an object, passing the type name as an argument, and get the answer:
my $value = 42; say "OK" if $value.isa(Int);