Specifying a function which takes an 'options' objcet with default values

Question

I would like to implement a function which takes an options object with default values.

I know how to do that, but I want to impose that any field not specified by the user will attain the default value specified in the function header.

Here is what I have so far:

function func(options: {x: number; y: string;} = {x: 1, y: "1"}) {
    const x: number = options.x != undefined ? options.x : 1;
    const y: string = options.y != undefined ? options.y : "1";
    console.log(x, y);
}

This works fine when I test this function on the output Javascript file:

func();
func({});
func({x: 0});
func({y: "2"});
func({x: 3, y: "4"});

Which results with:

1 '1'
1 '1'
0 '1'
1 '2'
3 '4'

However, it feels a little hacky (in particularly, the fact that I need to specify each one of the default values in two different places).

Is there a known design-pattern for this in Typescript?

Answer 1

No need to have default values in two places, you can shorten it up like this:

function func({ x = 1, y = "1" }: { x?: number; y?: string; } = {}) {
    console.log(x, y);
}

func();
func({});
func({ x: 0 });
func({ y: "2" });
func({ x: 3, y: "4" });

Or even shorter, if you don't need an explicit type:

function func({ x = 1, y = "1" } = {}) {
    console.log(x, y);
}

Results:

1 '1'
1 '1'
0 '1'
1 '2'
3 '4'