Issue
I tried to create IsAny generic based on this.
And my IsAny generic seems to work fine.
But when I use it in another generic(IsUnknown) it is broken:
const testIsUnknown2: IsUnknown<any> = true; // problem here, must be false
But when I change IsAny generic to commented one, – it works fine again. So why is it happening? Because I do not see any difference in both IsAny generics, but it seems my generic is not working inside other generics.
// type IsAny<T> = 0 extends (1 & T) ? true : false;
type IsAny<T> = unknown extends T ? (T extends object ? true : false) : false;
const testIsAny1: IsAny<any> = true;
const testIsAny2: IsAny<unknown> = false;
const testIsAny3: IsAny<string> = false;
const testIsAny4: IsAny<object> = false;
// unknown is only assignable to two types: unknown and any
type IsUnknown<T> = unknown extends T ? (IsAny<T> extends true ? false : true) : false;
const testIsUnknown1: IsUnknown<unknown> = true;
const testIsUnknown2: IsUnknown<any> = true; // problem here, must be false
const testIsUnknown3: IsUnknown<object> = false;
const testIsUnknown4: IsUnknown<number> = false;
Solution
The issue is that your isAny<any> yields boolean, which means both of these assignments will pass:
const testIsAny1: IsAny<any> = true;
const testIsAny2: IsAny<any> = false;
For the other isAny definition, testIsAny2 fails.
Regarding your definition of isAny, it fails because the conditional type T extends object ? true : false returns both alternatives for any. I couldn’t find any documentation about this, but you can fix it the same way you can prevent conditional types from distributing over unions, by surrounding both sides with square brackets (docs):
type IsAny<T> = unknown extends T ? ([T] extends [object] ? true : false) : false;
Answered By – Oblosys
Answer Checked By – Terry (BugsFixing Volunteer)