禁止跳过未定义的可选参数

Question

Can I (somehow?) forbid the skipping of optional parameter in typescript?

class MyList {
  constructor(
    public head?: number,
    public tail?: MyList
  ){}
}

const L0 = new MyList();              // <--- empty element list - good !
const L1 = new MyList(888);           // <--- single element list - good !
const L2 = new MyList(777, L0);       // <--- general list - good !
const L3 = new MyList(undefined, L1); // <--- forbid this 

I want to statically enforce the following property on my list:

• If head is undefined then tail is also undefined (and the list is empty)

Any TypeScript trick to achieve that? (This question is complementary to this question )

Answer 1

You can use something called overloading . This works for both methods and functions in TS. The basic idea is that you have single function/method implementation with all the possible arguments and you can specify different combintations of the function's arguments (just like for your case where you can have 0 arguments, only the first or both).

class MyList {
  constructor()
  constructor(head: number)
  constructor(head: number, tail: MyList)
  constructor(
    public head?: number,
    public tail?: MyList
  ){}
}

const L0 = new MyList(888);
const L1 = new MyList(777, L0);   
const L2 = new MyList(undefined, L1); // This will show error: Argument of type 'undefined' is not assignable to parameter of type 'number'.