What is it about Typescript's "typeof function" that meets the Generic Condition of "extends (...args: any[]) => any"?

Consider the code snippet below:

type Params<F extends (...args: any[]) => any> =
    F extends ((...args: infer A) => any)
    ? A
    : never;

const fn00 = (name: string, age: number, single: boolean) => true

type test07 = Params<typeof fn00>

How does typeof fn00 meet the Generic constraint of (...args: any[]) => any.

The typeof operator in TypeScript only returns strings representing certain types. These include:

"string" | "number" | "bigint" | "boolean" | "symbol" | "undefined" | "object" | "function"

For more information on the typeof operator, refer to the following documentation: https://www.typescriptlang.org/docs/handbook/advanced-types.html#typeof-type-guards

Answer №1

There are actually two different typeof operators. One is used in expressions, and it represents the JavaScript typeof operator. According to the documentation, this operator can return one of several possible values.

The second typeof operator is related to types and determines (during compile time) the TypeScript type of a value (such as a variable, class, or function). For example, the TypeScript type of fn00 is

(name: string, age: number, single: boolean) => boolean
. In the type declaration
type test07 = Params<typeof fn00>
, the typeof fn00 will be of this type.

If you were to use the typeof operator in an expression (like const test = typeof fn00), you would simply get function.

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