How can I make TypeScript deconstruct function parameter, assign default value, and optional whole object type at same time?

Question

Please see this minimum function

function addX({ num = 1 }) {
  return num + 1;
}

The above function has done two things

Now, I can use the function like this:

add({ num: 2 }); // valid
add({}); // valid

However, I can't use it like this

add(); // Expected 1 arguments, but got 0

I can't find a way to tell TypeScript the whole object is optional

// Invalid
function add({ num = 1 }?) {
  return num + 1;
}

// Invalid
function add({ num = 1 }?: { num: number }) {
  return num + 1;
}

How can I achieve it?

Karthick Vinod · Accepted Answer · 2020-05-30 09:11:53Z

1

How about this

function addX({ num = 1 } = {num: 1}) {
  return num + 1;
}

addX() // returns 2

addX({}) // returns 2

addX({num: 2}) // returns 3

answered May 30, 2020 at 9:11

Karthick Vinod

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2 Comments

Neat! I think this is the only way.

Even easier as function addX({ num = 1 } = {}) since you already have a default value for num