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Apply function generics as part of type definition #40542
Description
Search Terms
generic typeof functions
Suggestion
The problem is with the code as below:
const id = <T>(val: T) => val
type IdType = typeof id
// type IdType = <T>(val: T) => T
type Ret = ReturnType<typeof id>
// type Ret = unknownIt would be nice to be able to do something like so:
const id = <T>(val: T) => val
type IdType<T> = (typeof id)<T>
// type IdType<T> = (val: T) => T
// Or by extension:
type Ret<T> = ReturnType<(typeof id)<T>>
// type Ret<T> = TUse Cases
This would be useful to use with the vriad/zod library, to implement different field validators depending on a parameter, like so:
// the dateSchema param could be a schema for an ISO 8601 string or a Date object, or even some custom date-like type
const fooSchema = <T>(dateSchema: z.ZodSchema<T>) => z.object({
name: z.string(),
date: dateSchema
})
type Foo<T> = z.infer<ReturnType<(typeof fooSchema)<T>>>
// expected result: type Foo<T> = { name: string, date: T }Checklist
My suggestion meets these guidelines:
- This wouldn't be a breaking change in existing TypeScript/JavaScript code
- This wouldn't change the runtime behavior of existing JavaScript code
- This could be implemented without emitting different JS based on the types of the expressions
- This isn't a runtime feature (e.g. library functionality, non-ECMAScript syntax with JavaScript output, etc.)
- This feature would agree with the rest of TypeScript's Design Goals.
Activity
Gozala commentedon Jun 9, 2021
I am understanding this proposal correctly that
IdTypeherewould be
type IdType<T> = (val:T) => Tas opposed totype IdType = <T>(val:T) => T? Or is idea here that you you should be able to qualify generic function type liketype IdType = <T>(val:T) => Tvia<T>suffix e.g.type IdNum = IdType<number> // (val:number) => number?jmacmahon commentedon Jun 29, 2021
I meant the latter, so like you say,
type IdNum = IdType<number> // (val:number) => number.I think the former would be incorrect, because the parameter
Tis not present when declaringtype IdType.ahejlsberg commentedon Jan 27, 2022
This feature is now implemented in #47607.