The behavior is different for two code that looks the same

[kgtkr]

TypeScript Version: 3.4.2

Search Terms:
conditional type

Code

OK

type F<X> = <T>() => T extends X ? 1 : 2;

declare const x: F<any>;
const y: F<number> = x;

Error

declare const x: <T>() => T extends any ? 1 : 2;
const y: <T>() => T extends number ? 1 : 2 = x;

Expected behavior:
The two codes do the same thing

Actual behavior:
Different behavior

Playground Link:

Related Issues:

[jack-williams]

I think the effect of the type alias F is causing reduction in the body of the alias, in particular, I think it's simplifying the return type to the constraint of the conditional 1 | 2. Once this happens, the type parameter X no longer appears in the type and the two signatures are related.

type F<X> = <T>() => T extends X ? 1 : 2;
declare const x: F<number>;
const y: F<string> = x; // no error

type F2<X> = <T>() => T extends X ? 1 : 2;
const y2: F2<string> = x; // error

Haven't got as far as tracking down where exactly this behaviour comes from, but I'm guessing there is some instantiation business going on.

[kgtkr]

Related code

[RyanCavanaugh]

These samples are quite different and it's not apparent why they must behave identically.