Create a TypeScript class object with specified constructor arguments

Question

Create a TypeScript class object with specified constructor arguments

I've been working on a function that is supposed to execute the init method of a class and then return an instance of that class. However, I'm running into issues with maintaining the constructor and class types.

This is what I have tried so far:

class Test {
  constructor(a: number, b: string, c?: number[]) {}

  protected init() {}
}

export function instantiateWithInit<Type extends new (args: any) => { init: () => any }>(
  clazz: new (args: ConstructorParameters<Type>) => InstanceType<Type>,
  ...args: ConstructorParameters<Type>
) {
  const instance = new clazz(args);
  instance.init();
  return instance;
}

instantiateWithInit(Test, "");

However, the return type only includes the init method and I'm also encountering an error related to parameter matching in the type constructor:

Argument of type 'typeof Test' is not assignable to parameter of type 'new (args: [args: any]) => { init: () => any; }'.
  Types of construct signatures are incompatible.
    Type 'new (a: number, b: string, c?: number[]) => Test' is not assignable to type 'new (args: [args: any]) => { init: () => any; }'.ts(2345)

javascript typescript constructor class-instance-variables

Answer 1

Answer №1

One challenge present in this scenario is the requirement for the function to only take a single argument in its constructor. By utilizing the spread operator, it becomes possible to resolve this particular issue.

In order to successfully implement your concept, I found it necessary to alter the function definition in the following manner:


class Example {
    constructor(x: number, y: string, z?: number[]) { }

    public start() { }
}

export function initializeWithStart<
    E extends { start(): any },
    X extends any[]
>(
    className: new (...args: X) => E,
    ...arguments: X
) {
    const instance = new className(...arguments);
    instance.start();
    return instance;
}

const xyz = initializeWithStart(Example, 7, "");
// ^ xyz is of type Example
// Additionally, the input arguments are validated (which is why the 7 was added)

I must acknowledge that TypeScript can pose challenges when using extends within function definitions.

Answer 2

One challenge present in this scenario is the requirement for the function to only take a single argument in its constructor. By utilizing the spread operator, it becomes possible to resolve this particular issue.

In order to successfully implement your concept, I found it necessary to alter the function definition in the following manner:


class Example {
    constructor(x: number, y: string, z?: number[]) { }

    public start() { }
}

export function initializeWithStart<
    E extends { start(): any },
    X extends any[]
>(
    className: new (...args: X) => E,
    ...arguments: X
) {
    const instance = new className(...arguments);
    instance.start();
    return instance;
}

const xyz = initializeWithStart(Example, 7, "");
// ^ xyz is of type Example
// Additionally, the input arguments are validated (which is why the 7 was added)

I must acknowledge that TypeScript can pose challenges when using extends within function definitions.

Create a TypeScript class object with specified constructor arguments

Answer №1

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