Typescript check for the ‘any’ type

Yeah, you can test for any:

type IfAny<T, Y, N> = 0 extends (1 & T) ? Y : N; 
type IsAny<T> = IfAny<T, true, never>;
type A = IsAny<any> // true
type B = IsAny<number> // never
type C = IsAny<unknown> // never
type D = IsAny<never> // never

The explanation for this is in this answer. In short, any is intentionally unsound, and violates the normal rules of types. You can detect this violation because it lets you do something crazy like assign 0 to 1.

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