What is the assumption made in “Learn You a Haskell” when deducing the kind?

Question

This question is not subjective. A very specific verb is used in the referenced book, and I\'d like to understand what the implication of that phrasing is, because I\'m afraid I

Answer 1

Good point. The author makes a needless assumtion. Perhaps just to make it easier to understand in his Type Foo chapter but people like yourself may rightfully question this.

Both t, k and p are type variables. As we see from yabba :: p it can live alone so it's like a constant function, as if it was a value instead of a type, it's type signature would say Int or Char, whatever... you name it. But since it is a type then it's kind signature is *.

However t type here takes a type variable k to construct a type (dabba :: t k) so we are sure that (no assumtion here) t has a kind signature like * -> * and k has *.

Once we know this... the type Barry t k p's kind signature is (* -> *) -> * -> * -> * which means it takes t then k and then p and give us Barry type.

Edit Make sure to read @luqui's comment below.