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)

CodePudding user response：

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'.