how to remove properties and promisify method via mapped type in TypeScript

Question

Here is the code

class A {
    x = 0;
    y = 0;
    visible = false;
    render() {
        return 1;
    }
}

type RemoveProperties<T> = {
    readonly [P in keyof T]: T[P] extends Function ? T[P] : never//;
};

type JustMethodKeys<T> = ({ [P in keyof T]: T[P] extends Function ? P : never })[keyof T];
type JustMethods<T> = Pick<T, JustMethodKeys<T>>;


type IsValidArg<T> = T extends object ? keyof T extends never ? false : true : true;

type Promisified<T extends Function> =
    T extends (...args: any[]) => Promise<any> ? T : (
        T extends (a: infer A, b: infer B, c: infer C, d: infer D, e: infer E, f: infer F, g: infer G, h: infer H, i: infer I, j: infer J) => infer R ? (
            IsValidArg<J> extends true ? (a: A, b: B, c: C, d: D, e: E, f: F, g: G, h: H, i: I, j: J) => Promise<R> :
            IsValidArg<I> extends true ? (a: A, b: B, c: C, d: D, e: E, f: F, g: G, h: H, i: I) => Promise<R> :
            IsValidArg<H> extends true ? (a: A, b: B, c: C, d: D, e: E, f: F, g: G, h: H) => Promise<R> :
            IsValidArg<G> extends true ? (a: A, b: B, c: C, d: D, e: E, f: F, g: G) => Promise<R> :
            IsValidArg<F> extends true ? (a: A, b: B, c: C, d: D, e: E, f: F) => Promise<R> :
            IsValidArg<E> extends true ? (a: A, b: B, c: C, d: D, e: E) => Promise<R> :
            IsValidArg<D> extends true ? (a: A, b: B, c: C, d: D) => Promise<R> :
            IsValidArg<C> extends true ? (a: A, b: B, c: C) => Promise<R> :
            IsValidArg<B> extends true ? (a: A, b: B) => Promise<R> :
            IsValidArg<A> extends true ? (a: A) => Promise<R> :
            () => Promise<R>
        ) : never
    );



var a = new A() as JustMethods<A>  // I want to JustMethod && Promisified
a.visible // error
var b = a.render() // b should be Promise<number>

How to implement it ? I want to remove visible and promisify render method , how to composite Promisified and JustMethods ?

Titian Cernicova-Dragomir · Answer 1 · 2018-07-19T13:49:02.720

You need to use a mapped type that takes just the methods of the type using JustMethodKeys and uses Promisified on each property

class A {
    x = 0;
    y = 0;
    visible = false;
    render() {
        return 1;
    }
}

type JustMethodKeys<T> = ({ [P in keyof T]: T[P] extends Function ? P : never })[keyof T];

type IsValidArg<T> = T extends object ? keyof T extends never ? false : true : true;

type Promisified<T extends Function> =
    T extends (...args: any[]) => Promise<any> ? T : (
        T extends (a: infer A, b: infer B, c: infer C, d: infer D, e: infer E, f: infer F, g: infer G, h: infer H, i: infer I, j: infer J) => infer R ? (
            IsValidArg<J> extends true ? (a: A, b: B, c: C, d: D, e: E, f: F, g: G, h: H, i: I, j: J) => Promise<R> :
            IsValidArg<I> extends true ? (a: A, b: B, c: C, d: D, e: E, f: F, g: G, h: H, i: I) => Promise<R> :
            IsValidArg<H> extends true ? (a: A, b: B, c: C, d: D, e: E, f: F, g: G, h: H) => Promise<R> :
            IsValidArg<G> extends true ? (a: A, b: B, c: C, d: D, e: E, f: F, g: G) => Promise<R> :
            IsValidArg<F> extends true ? (a: A, b: B, c: C, d: D, e: E, f: F) => Promise<R> :
            IsValidArg<E> extends true ? (a: A, b: B, c: C, d: D, e: E) => Promise<R> :
            IsValidArg<D> extends true ? (a: A, b: B, c: C, d: D) => Promise<R> :
            IsValidArg<C> extends true ? (a: A, b: B, c: C) => Promise<R> :
            IsValidArg<B> extends true ? (a: A, b: B) => Promise<R> :
            IsValidArg<A> extends true ? (a: A) => Promise<R> :
            () => Promise<R>
        ) : never
    );

type PromisifyMethods<T> = { 
    // We take just the method key and Promisify them, 
    // We have to use T[P] & Function because the compiler will not realize T[P] will always be a function
    [P in JustMethodKeys<T>] : Promisified<T[P] & Function>
}

//Usage
declare var a : PromisifyMethods<A>  
a.visible // error
var b = a.render() // b is Promise<number>

Edit

Since the original question was answered typescript has improved the possible solution to this problem. With the addition of Tuples in rest parameters and spread expressions we now don't need to have all the overloads for Promisified:

type JustMethodKeys<T> = ({ [P in keyof T]: T[P] extends Function ? P : never })[keyof T];


type ArgumentTypes<T> = T extends (... args: infer U ) => any ? U: never;
type Promisified<T> = T extends (...args: any[])=> infer R ? (...a: ArgumentTypes<T>) => Promise<R> : never;

type PromisifyMethods<T> = { 
    // We take just the method key and Promisify them, 
    // We have to use T[P] & Function because the compiler will not realize T[P] will always be a function
    [P in JustMethodKeys<T>] : Promisified<T[P]>
}

//Usage
declare var a : PromisifyMethods<A>  
a.visible // error
var b = a.render("") // b is Promise<number> , render is render: (k: string) => Promise<number>

Not only is this shorter but it solves a number of problems

Optional parameters remain optional
Argument names are preserved
Works for any number of arguments

@Florian I don't think this is what you are looking for. It sounds like what you want are decorators. This question is strictly related to changing the signatures of functions on a class, it has no runtime behavior. But ask a question and someone will answer it :-) — Titian Cernicova-Dragomir, Sep 12 '18 at 18:02

score 0 · Answer 2 · answered Aug 26 '19 at 11:48

Similar to another answer, but here's what I arrived at

type Method = (...args: any) => any;
type KeysOfMethods<T> = ({ [P in keyof T]: T[P] extends Method ? P : never })[keyof T];
type PickMethods<T> = Pick<T, KeysOfMethods<T>>;
type Promisify<T> = T extends Promise<infer U> ? Promise<U> : Promise<T>;
type PromisifyMethod<T extends Method> = (...args: Parameters<T>) => Promisify<ReturnType<T>>;
type PromisifyMethods<T> = { [P in keyof T]: T[P] extends Method ? PromisifyMethod<T[P]> : T[P] };

type PromisifiedMethodsOnly<T> = PickMethods<PromisifyMethods<T>>

how to remove properties and promisify method via mapped type in TypeScript

2 Answers2

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