Ensure that the specified Class type must have a constructor with no arguments

When working with a function that takes a parameter representing a Class (not an object or instance, but the Class itself), or essentially a variable assigned to a Class.

The challenge is ensuring that the Class assigned to the parameter has a constructor with no arguments (only, as JS classes can have at most one constructor based on the spec, section 8.3 (8.3)). This is useful for creating a generic function that can instantiate (and return) objects of the specified class.

To implement the necessary type checking for parameter c:

function acceptsAClassParameter(c) {
    return new c();
}

class MyClassWithNoArgsConstructor { constructor() { ... } }
class MyClassWithArgsConstructor { constructor(foo) { ... } }

acceptsAClassParameter(MyClassWithNoArgsConstructor);
acceptsAClassParameter(MyClassWithArgsConstructor); // results in a type error

Answer №1

The TypeScript documentation for Generics showcases an illustration of factories referencing class types through their constructor functions:

function create<Type>(c: { new (): Type }): Type {
  return new c();
}

The crucial point is to adhere to this approach and avoid using a declaration that would mandate the constructor to not accept any parameters, such as: (...args: never[]) => any.

Illustrated through the following example:

type NoArgsConstructor<T> = new () => T;
function createInstanceWithNoArgsConstructor<T>(c: NoArgsConstructor<T>) {
    return new c();
}

class MyClassWithNoArgsConstructor { constructor() { ... } }
class MyClassWithBagConstraintstor { constructor(foo) { ... } }

createInstanceWithNoArgsConstructor(MyClassWithNoArgsConstructor);
createInstanceWithNoArgsConstructor(MyClassWithArgsConstructor); // Error: Types of construct signatures are incompatible.

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