Don't widen return types of function expressions

This change courtesy @JsonFreeman who is trying it out

(context elided: widening of function expression return types)

The problem with this widening is observable in type argument inference and no implicit any. For type argument inference, we are prone to infer an any, where we should not:

function f<T>(x: T, y: T): T { }
f(1, null); // returns number
f(() => 1, () => null); // returns () => any, but should return () => number

So after we get to parity, I propose we do the following. We do not widen function expressions. A function is simply given a function type. However, a function declaration (and a named function expression) introduces a name whose type is the widened form of the type of the function. Very simple to explain and simple to implement.

I’ve been told that this would be a breaking change, because types that used to be any are now more specific types. But here are some reasons why it would be okay:

• In the places where you actually need the type to be any (because there are no other inference candidates), you would still get any as a result
• In places where there was a better (more specific) type to infer, you’d get the better type.
• With the noImplicitAny flag, you’d get fewer errors because there are actually fewer implicit anys

Questions:

Is a principle of design changes going forward to not switch from 'any' to a more precise type because it can be a breaking change?

Going with 'not a breaking change' here because this is unlikely to break working code, but we need to verify this.

Would this manufacture two types?

In essence, we already have two types: The original and the widened type. So by that measure this is not really a change

Has someone tried it?

Jason willing to try it out and report back

まず、2つのジェネリック呼び出しの例を再現し、noImplicitAny を使用した場合の動作を含めて、推論される型を確認します。関数式が固有の関数型を保持し、関数宣言と名前付き関数式が拡大された型を公開することを、推論と互換性の検証とともに確認できれば完了です。