Reputation: 23772
I try to define the type of the resolve
callback in the code below.
export interface PromiseToHandle<T> {
resolve: (result: T) => void // <----- The question is about that line
reject: (error: any) => void
promise: Promise<T>
}
export function promiseToHandle<T = any>(): PromiseToHandle<T> {
let resolve!: any;
let reject!: (error: any) => void;
const promise = new Promise<T>((resolveCb, rejectCb) => {
resolve = resolveCb;
reject = rejectCb;
});
return { promise, resolve, reject };
}
const pth1 = promiseToHandle<boolean>();
pth1.resolve(true); // OK
const pth2 = promiseToHandle<void>();
pth2.resolve(); // OK
It works the way I need, however the type is incorrect:
const pth3 = promiseToHandle<void>();
const voidResolve: () => void = pth3.resolve // Error: Type '(result: void) => void' is not assignable to type '() => void'.
export interface PromiseToHandle<T> {
resolve: T extends void ? () => void : (result: T) => void
reject: (error: any) => void
promise: Promise<T>
}
const pth4 = promiseToHandle<void>();
const voidResolve2: () => void = pth4.resolve // OK
The previous issue is solved. But now the main case doesn't work:
const pth5 = promiseToHandle<boolean>();
pth5.resolve(true); // Error: Argument of type 'true' is not assignable to parameter of type 'false & true'.
I don't know why but when a conditional type is used, then a union (boolean
is false | true
) is converted to an intersection (false & true
)!
Upvotes: 3
Views: 2201
Reputation: 250366
Your conditional type solution is good, you just need to disable the distribution behavior of conditional types
export interface PromiseToHandle<T> {
resolve: [T] extends [void] ? () => void : (result: T) => void
reject: (error: any) => void
promise: Promise<T>
}
Upvotes: 3