What is the method for utilizing a function's input type specified as "typeof A" to output the type "A"?

Question

What is the method for utilizing a function's input type specified as "typeof A" to output the type "A"?

Check out this example from my sandbox:

class A {
    doSomething() {}
}
class B {}

const collection = {
    a: new A(),
    b: new B()
}

const findInstance = <T>(list: any, nonInstance: T): T => {
    for (const item in list) {
        if (list[item] instanceof (nonInstance as any)) {
            return list[item] as any
        }
    }
    throw new Error('Unable to locate instance')
}

const a: A = findInstance(collection, A)

a.doSomething()

The type error I am encountering is:

Property 'doSomething' is missing in type 'typeof A' but required in type 'A'.

Instead of receiving typeof A, what I need is A.

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typescript

Answer 1

Answer №1

Modify the nonInstance type declaration to be new(...args: any[]) => T instead of just T.

Here is the updated function signature for reference:

const findInstance = <T>(list: any, nonInstance: new(...args: any[]) => T): T => {
    // ...
}

This change signifies that you are specifying a constructor for type T rather than an instance of type T itself.

For more information, check out this related question on Stack Overflow: Using a generic type argument with `typeof T`

You can also refer to the TypeScript Handbook for another example: https://www.typescriptlang.org/docs/handbook/generics.html#using-class-types-in-generics

Answer 2

Modify the nonInstance type declaration to be new(...args: any[]) => T instead of just T.

Here is the updated function signature for reference:

const findInstance = <T>(list: any, nonInstance: new(...args: any[]) => T): T => {
    // ...
}

This change signifies that you are specifying a constructor for type T rather than an instance of type T itself.

For more information, check out this related question on Stack Overflow: Using a generic type argument with `typeof T`

You can also refer to the TypeScript Handbook for another example: https://www.typescriptlang.org/docs/handbook/generics.html#using-class-types-in-generics

Answer 3

Answer №2

Have you thought about utilizing 'new' in this expression? It seems unlikely because you named your variable nonInstance... A is the type of the class itself and does not have the doSomething method (unless static keyword is used, but that might not be the best solution)... Therefore, adding new() => T could indicate to the compiler that you are looking for an instance,

const findInstance = <T>(list: any, instance: new () => T): T => {
  for (const item in list) {
    if (list[item] instanceof (instance as any)) {
      return list[item] as any;
    }
  }
  throw new Error('Unable to locate desired object');
};

Answer 4

Have you thought about utilizing 'new' in this expression? It seems unlikely because you named your variable nonInstance... A is the type of the class itself and does not have the doSomething method (unless static keyword is used, but that might not be the best solution)... Therefore, adding new() => T could indicate to the compiler that you are looking for an instance,

const findInstance = <T>(list: any, instance: new () => T): T => {
  for (const item in list) {
    if (list[item] instanceof (instance as any)) {
      return list[item] as any;
    }
  }
  throw new Error('Unable to locate desired object');
};

What is the method for utilizing a function's input type specified as "typeof A" to output the type "A"?

Answer №1

Answer №2

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