correctly specify parameter values when calling a function in JavaScript

Question

I have a function foo defined as follows,

foo = function(a=1,b=2,c=3){
    console.log("your inputs are a:" +a+", b:"+b + ", c:"+c+".")
}

How do I use the default value of b and specify a and c only when calling the function?

For example, I would like to return a console log of your inputs are a:3, b:2, c:1. when calling foo(a=3,c=1).

However, by calling foo(a=3,c=1), the console log is your inputs are a:3, b:1, c:3.. The JavaScript thinks the c=1 should be the value of the second parameter b and the third parameter c is not given and will use its default value of 3. In addition, by calling this function, the object of a and c are also created.

The only thing you can do is leave off parameters at the end of the list; you can't omit a parameter in the middle. edit or do the object destructuring as suggested. — Pointy
– Pointy, Commented Jun 19, 2019 at 22:29

CertainPerformance · Accepted Answer · 2019-06-19 22:36:21Z

3

I'd pass an object instead, and destructure the arguments:

function foo ({ a=1,b=2,c=3 }){
    console.log("your inputs are a:" +a+", b:"+b + ", c:"+c+".")
}

foo({ a: 99, c: 99 });

In case the function may be called without any parameter, assign the default parameter to {} too:

function foo ({ a=1,b=2,c=3 } = {}){
    console.log("your inputs are a:" +a+", b:"+b + ", c:"+c+".")
}

foo();

You can also explicitly pass undefined for missing arguments:

function foo (a=1,b=2,c=3){
    console.log("your inputs are a:" +a+", b:"+b + ", c:"+c+".")
}

foo(99, undefined, 99);

edited Jun 19, 2019 at 22:36

answered Jun 19, 2019 at 22:29

CertainPerformance

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4 Comments

WCMC Over a year ago

I love your 1st and 2nd way. Interesting how the javascript can capture the unspecified parameter and find its default value. Though I still do not completely understand the mechanisms here. I just wanted to avoid using the 3rd way especially when there are a lot of parameters (with default values) in the function.

CertainPerformance Over a year ago

When there are lots of parameters, having multiple standalone parameters probably isn't a good idea, because you have to carefully keep in mind which order the parameters are supposed to be in - I'd prefer the object destructuring method regardless, when there are many "parameters", just for the sake of readability, even without the issue of default values.

CharlesEF Over a year ago

Should 'foo(99, undefined, 99);' be 'foo({99, undefined, 99});'? Or, since you are using all three parameters you don't need the {}?

CertainPerformance Over a year ago

When the function accepts multiple standalone parameters, you need to pass undefined to fill the "holes" in the parameters. When the function accepts a single object, the (attempted) destructured property will be undefined regardless, so specifying the missing parameters is unnecessary (as you can see by pressing "Run code snippet" in the answer)

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