Still reproducible (2017.11,HEAD(e5b660e))
On 2015-09-20 09:07:10, zef...@fysh.org wrote:
> Because a type object is an (undefined) instance of that type, it can
> in general be stored in a variable constrained to that type. We can
> demonstrate this with a subroutine taking a type parameter:
>
> $ ./perl6 -e 'sub aa (Mu:U ::T) { my T $a = T; say $a; }; aa(Any);
> aa(Int); aa(int)'
> (Any)
> (Int)
> (int)
>
> But if we try to do the same thing without the indirection of the
> subroutine, it behaves differently on native types:
>
> $ ./perl6 -e 'my Any $a = Any; say $a; my Int $b = Int; say $b; my int
> $c = int; say $c'
> (Any)
> (Int)
> Cannot unbox a type object
> in block <unit> at -e:1
>
> This indirection shouldn't make such a visible difference. Obviously
> there's some connection here to the specialised representation implied
> by the int type, and the fact that that representation can't
> accommodate
> the int type object. It's fine for the specialised representation to
> be used in only one of these two cases, but that should be an
> invisible
> implementation detail. Perhaps the specialised representation should
> only be used where the variable is constrained to *defined* ints.
>
> Another inconsistency arises in how assignment of a boxed value to the
> variable is treated:
>
> $ ./perl6 -e 'sub aa (Mu:U ::T) { my T $a = 3; say $a.WHAT; };
> aa(Any); aa(Int); aa(int)'
> (Int)
> (Int)
> Type check failed in assignment to '$a'; expected 'int' but got 'Int'
> in sub aa at -e:1
> in block <unit> at -e:1
> $ ./perl6 -e 'my Any $a = 3; say $a.WHAT; my Int $b = 3; say $b.WHAT;
> my int $c = 3; say $c.WHAT'
> (Int)
> (Int)
> (Int)
>
> -zefram
