How do I pass only some values into args object?

Question

Is there a way to only pass in one argument into args and let other values be default in TypeScript? I dont want "args = {}" and declaring defaults within the function due to intellisense.

function generateBrickPattern (
    wallWidth: number,
    wallHeight: number,
    args = {
  maxBrickWidth: 100,
  maxBrickHeight: 50,
  minBrickWidth:  50,
  minBrickHeight: 25
}) {}

generateBrickPattern(500,500,{maxBrickWidth: 75}) //Prefered

generateBrickPattern(500,500,{maxBrickWidth: 75, 
                              maxBrickHeight: 50,  
                              minBrickWidth:  50,
                              minBrickHeight: 25}) //Not wanted

The prefered syntax gives the following error.

Argument of type '{ maxBrickWidth: number; }' is not assignable to parameter of type '{ maxBrickWidth: number; maxBrickHeight: number; minBrickWidth: number; minBrickHeight: number; }...'.

Answer 1

You can do this by destructuring args with default values:

function generateBrickPattern (
    wallWidth: number,
    wallHeight: number,
    {
        maxBrickWidth: maxBrickWidth = 100,
        maxBrickHeight: maxBrickHeight = 50,
        minBrickWidth: minBrickWidth = 50,
        minBrickHeight: minBrickHeight = 25
    } = {}
) {
    console.log(maxBrickWidth);
}

If you don't want to destructure you could merge the provided args with the defaults like this:

interface BrickPatternOptions {
    maxBrickWidth: number;
    maxBrickHeight: number;
    minBrickWidth: number;
    minBrickHeight: number;
}

function generateBrickPattern (
    wallWidth: number,
    wallHeight: number,
    args: Partial<BrickPatternOptions> = {}
) {
    const options: BrickPatternOptions = {
        maxBrickWidth: 100,
        maxBrickHeight: 50,
        minBrickWidth: 50,
        minBrickHeight: 25,
        ...args
    };

    console.log(options.maxBrickWidth);
}