why is type T = (x: T) => void valid whereas T = T isnt?

[zpdDG4gta8XKpMCd]

nightly build Nov 27, 2016:

type U = U; // problem: Type alias 'U' circularly references itself.
type T = (value: T) => void; // ok, but why?

[HerringtonDarkholme]

https://github.com/Microsoft/TypeScript/blob/master/doc/spec.md#310-type-aliases

Note that object type literals, function type literals, and constructor type literals do not depend on types referenced within them and are therefore permitted to circularly reference themselves through type aliases.

[DanielRosenwasser]

The restriction that a type alias can't be referenced by itself at the top level has been the behavior since we implemented type aliases; however, you might recall that a while back, we started allowing type aliases to be referenced from within an object type.

If we desugar T's declaration to use object type, we can that under the current rules, it's allowed.

type T = {
    (value: T) => void
};

Overall this is actually pretty reasonable - even Haskell can't express something like this:

type Wat = Wat

<interactive>:1:1: error:
    Cycle in type synonym declarations:
      <interactive>:1:1-14: type Wat = Wat

However, it has no problem provided that the type can be expanded, or "unrolled" one level at a time. In other words, as long as you have some sort of box-y thing containing the type:

newtype Box a = Box a

the type system will be okay.

[zpdDG4gta8XKpMCd]

is there any chance for recursive types in unions?

type Expression = Binary <Expression> | Unary <Expression>

[mhegazy]

is there any chance for recursive types in unions?

tracked by #6230