如何在构造函数上为静态属性定义类型

Question

I have a problem with typescript throwing an error when I try to assign a static property to a constructor function: Property 'wheels' does not exist on type '() => void'. How can I tell typescript that my Car object can have wheels property?

 function Car() { // do something } Car.wheels = 4 // throws: Property 'wheels' does not exist on type '() => void'. const audi = new Car() 

Above snippet can be tested on https://www.typescriptlang.org/play/index.html

Answer 1

function Car() {
    // do something
}

namespace Car { 
    export let wheels = 4
}

const audi = new Car()

See the handbook .

Answer 2

With a bit of type acrobatics, renaming and multiple variables you can use a type assertion to achieve what you want:

function Car_() {
    // do something
};

type TCar = typeof Car_ & { wheels: number };
const Car = (Car_ as TCar);

Car.wheels = 4; // throws: Property 'wheels' does not exist on type '() => void'.

const audi = new Car()

The solution suggested by @MattMcCutchen looks more legit though.