Closed bbrk24 closed 2 months ago
Does T.prototype.length
actually make sense? T.foo
is a namespace dereference, so I don't think there's any sense of prototype chain.
In type position, T::length
currently compiles to T["length"]
.
Oops, right, I see.
With coffeePrototype,
T::#
works as an expression but not a type.