Imagine you have a function like this:
const f = a => b => ... x => { return somevalue }
Is there a way to determine the type of just the final function
typeof x => { return somevalue }One idea is to create a type that uses recursion and conditional type checking to navigate through the closures until reaching the end. However, it's unclear if TypeScript supports recursive types for solving this specific issue.
Recursive conditional types are set to debut in TypeScript 4.1. In the meantime, you can access this feature by using typescript@next. Once available, you'll be able to define something like:
type LastFunc<T extends (...args: any) => any> =
T extends (...args: any) => infer R ?
R extends (...args: any) => any ? LastFunc<R> : T : never;
You can apply it to your function f as follows:
declare const somevalue: SomeValue;
const f = (a: any) => (b: any) => (c: any) => (d: any) => (x: any) => { return somevalue }
type LastFuncF = LastFunc<typeof f>; // type LastFuncF = (x: any) => SomeValue
In the interim, there are two ways to achieve similar results. The first is a workaround that may not be officially supported but gets the job done:
type LastFunc<T extends (...args: any) => any> =
T extends (...args: any) => infer R ? {
0: LastFunc<Extract<R, (...args: any) => any>>, 1: T
}[R extends (...args: any) => any ? 0 : 1] : never
The second method involves manually unrolling the loop to a fixed depth which can be tedious and repetitive:
type LastFunc<T extends (...args: any) => any> = T extends (...args: any) => infer R ? R extends (...args: any) => any ? LF0<R> : T : never;
// Additional LFX definitions...
If I were in your shoes, I'd recommend waiting for TypeScript 4.1's release. Best of luck with your coding endeavors!
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