Use object parameters as arguments for function

Question

I am sure this has been answered before, but searching only yields results for passing an actual object as a function arguments, however I would like to destructure an object and use its parameters as input arguments for a function.

I am importing the function foo() from an external library and would like to supply its arguments using an object containing them. Here is foo.js (note that my use case has a lot more input arguments):

export function foo(arg1, arg2) {
  doStuff()
}

Here is my code; the following works :

import { foo } from 'foo'

const args = {
  arg1: "value1",
  arg2: "value2",
}

const { arg1, arg2 } = args
foo(arg1, arg2)

I would like to be a bit more concise and avoid repeating the variable names arg1 and arg2, along the lines of:

import { foo } from 'foo'

const args = {
  arg1: "value1",
  arg2: "value2",
}

foo(args) // doesn't work, apparently I have to do something more to the args object here

But this code doesn't work (arg1 gets the value of args and arg2 is undefined), I assume that this can be achieved, e.g. using the spread operator somehow, but how?

you need to get the properties before you call the function unless you change the singature of the function to accept an object for destructuring. — Nina Scholz
– Nina Scholz, Commented Jan 10, 2020 at 14:55
@ASDFGerte I don't have access to the declaration of foo/bar, I updated the question to reflect this — FerventHippo
– FerventHippo, Commented Jan 10, 2020 at 15:00
Then either use your own suggestion of just calling it by listing the properties as parameters, or write a wrapper, and use that instead (just make a new function function fooWrapped({ arg1, arg2 }){ return foo.call(this, arg1, arg2); }) — ASDFGerte
– ASDFGerte, Commented Jan 10, 2020 at 15:05

Alek Angelov · Accepted Answer · 2020-01-10 15:16:21Z

3

destructure the object in the function declaration like so:

function({arg1 = "", arg2 = 0}) {doStuff()}

and remember to always add default values :)

EDIT: what you're doing is impossible basically, because you can't use iterators like this foo(..{arg1, arg2). A solution to this would be unwrapping the object values using Object.values converting them into an array and then destructuring that.

function foo(arg1, arg2) {
  console.log(arg1, arg2);
}

function unwrap(object) {
  return Object.values(object);
}

foo(...unwrap({ x: "1", y: "2" }));

at this point tho you'd be better off using an array like so:

foo(...[1, 2]);

edited Jan 10, 2020 at 15:16

answered Jan 10, 2020 at 14:54

Alek Angelov

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2 Comments

FerventHippo Over a year ago

I do not actually have access to the function declaration. Let me update the question to clarify this

Alek Angelov Over a year ago

Ah then you have to wrap it, I'll update my answer accordingly :)

Nina Scholz · Accepted Answer · 2020-01-10 14:58:34Z

0

You could take a function for unwrapping the arguments object and spread the result for calling the wanted function without changing the function's signature.

// given function
function foo(arg1, arg2) {
    console.log(arg1, arg2);
}

const
    unwrap = ({ arg1, arg2 }) => [arg1, arg2], // get all properties in wanted order
    args = { arg1: "value1", arg2: "value2" };

foo(...unwrap(args));

answered Jan 10, 2020 at 14:58

Nina Scholz

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2 Comments

FerventHippo Over a year ago

Ok, this looks like it would work but seems similarly or more complex than my first suggestion (destructuring first). I would like to reduce the number of times that I type arg1, arg2 etc. to 1.

Nina Scholz Over a year ago

objects have no order. if you are sure about the order, you could take o => Object.values(o);

user11067476 · Accepted Answer · 2020-01-10 15:14:37Z

0

Use Map instead object to save insertion order:

const args = new Map();

args.set('arg1', 'value1');
args.set('arg2', 'value2');

foo(...args.values());

answered Jan 10, 2020 at 15:14

user11067476

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