There is related misbehaviour that causes variables to accept values that they should reject:
> my $v = 5 5 > my $s = $v.VAR 5 > my Int:D $x = $s 5 > $x.WHAT.say (Scalar) > $v = Int > my Any:U $x = $s > $x.DEFINITE True The general pattern is that if a variable has a type constraint that includes a definiteness qualifier, and an attempt is made to assign a Scalar value to the variable, then the type constraint gets applied to the value contained in the Scalar rather than to the Scalar itself. If the type check against that value passes, then the Scalar value is assigned to the variable, regardless of whether the Scalar itself satisfies the type constraint. Type constraints that do not have definiteness qualifiers work as expected, and so do type constraints on subroutine parameters. -zefram
