TypeScript Expert — Series 2

Preview — 3 of 10 questions

What type does TypeScript infer for result?

javascript

function first<const T extends readonly unknown[]>(arr: T): T[0] {
  return arr[0];
}
const result = first(['a', 'b', 'c']);

Aresult is typed string — the const modifier on the type parameter has no effect on how T is inferred here

BCompile-time error — const can't be applied to a type parameter, only to variable declarations

Cresult has literal type "a" because const inference preserves the tuple.

Dresult's type is left as the unresolved expression readonly ["a","b","c"][0]

What type does Result resolve to?

javascript

type UnionToIntersection<U> = (U extends unknown ? (k: U) => void : never) extends (k: infer I) => void ? I : never;
type Result = UnionToIntersection<{ a: string } | { b: number }>;

AResult is { a: string } | { b: number }, unchanged

BResult is never

CCompile-time error — infer can't be used to extract a function's parameter type

DResult is { a: string } & { b: number }.

With strictFunctionTypes: true enabled, does Money's compare (parameter narrowed to Money instead of unknown) satisfy Comparator?

javascript

interface Comparator {
  compare(other: unknown): number;
}
class Money implements Comparator {
  constructor(public cents: number) {}
  compare(other: Money): number { return this.cents - other.cents; }
}

AYes. The method syntax keeps parameter checks bivariant under strictFunctionTypes.

BNo — strictFunctionTypes makes all function-parameter checks strictly contravariant, everywhere, with no exceptions, and Money's narrower parameter type is rejected

CNo — methods and properties are checked completely identically under strictFunctionTypes

DYes, but only because Money is a class — the same code would fail if Comparator were implemented via a plain object literal instead

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