TypeScript Constructor Overload with Empty Constructor

Question

Why is it not allowed to have separate constructor definitions in TypeScript?
To have e.g. two constructors, I need to write my code like this.

constructor(id: number)
constructor(id: number, name?: string, surname?: string, email?: string) {
    this.id = id;
    this.name = name;
    this.surname = surname;
    this.email = email;
}

Thereby I need to put ? after the parameters that are not required in the first constructor.

Why can't I write it like this?

constructor(id: number) {
    this.id = id;
}

constructor(id: number, name: string, surname: string, email: string) {
    this.id = id;
    this.name = name;
    this.surname = surname;
    this.email = email;
}

So that for both constructors all parameters are mandatory.

Moreover, if I need to have an empty constructor things get even weirder, since I need to mark every parameter with a ?.

constructor()
constructor(id?: number, name?: string, surname?: string, email?: string) {
    this.id = id;
    this.name = name;
    this.surname = surname;
    this.email = email;
}

Why does TypeScript differs from common languages like C# or Python here?

I would expect it to work like this.

constructor() {

}
constructor(id: number, name: string, surname: string, email: string) {
    this.id = id;
    this.name = name;
    this.surname = surname;
    this.email = email;
}

So you can pass none parameter or must pass all parameters.

question from:https://stackoverflow.com/questions/35998629/typescript-constructor-overload-with-empty-constructor

Answer 1

Because your constructor implementation is called by all your overload constructors. (Technically, at runtime there's only one constructor function that gets called with the various overload argument signatures.)

Imagine it like this:

overload_constructor(id:string) {
    implementation_constructor(id);
}

implementation_constructor(id:string, name?:string, age?:number) {
    // ...
}

Thinking of it this way, overload_constructor could not call implementation_constructor unless name and age are optional.

Also see Basarat's answer, the implementation isn't exposed for public usage by the type checker (though at runtime it's the "real" constructor used in JS). If you want to only allow (), (id), or (id, name, surname, email) as the only valid call signatures you would do it like this:

constructor()
constructor(id: number)
constructor(id: number, name: string, surname: string, email: string)
constructor(id?: number, name?: string, surname?: string, email?: string) {
    this.id = id;
    this.name = name;
    this.surname = surname;
    this.email = email;
}

Note that in the implementation all parameters are optional, but that signature is not exposed when compiling and you can only use these these calls:

new Foo()
new Foo(1)
new Foo(1, "a", "b", "c")

Not, for example:

new Foo(1, "a")