I am currently using the object destructuring pattern with default parameters described in that answer to ES6 Object Destructuring Default Parameters.
(function test({a = "foo", b = "bar"} = {}) {
console.log(a + " " + b);
})();
I would like to be able to access the object parameter without assigning it to a variable in the function body and explicitly listing each key.
(function test({a = "foo", b = "bar"} = {}) {
const options = {a, b};
console.log(options);
})();
I tried naming the object argument, but the function looses the ability to resolve missing keys to their default value.
(function test(options = {a = "foo", b = "bar"} = {}) {
console.log(options);
})();
It seems to be ignoring the default parameters when destructuring into a named argument.
Is this part of the ES6 spec? Is there a way to achieve the desired behavior without additional code in the function body?
Edit: I removed a superfluous example that did not add context to the question.
Destructuring in JavaScript is used to unpack or segregate values from arrays or properties from object literals into distinct variables, thus it allows us to access only the values required.
Destructuring is a JavaScript expression that allows us to extract data from arrays, objects, and maps and set them into new, distinct variables. Destructuring allows us to extract multiple properties, or items, from an array at a time.
By writing var name = hero.name , you have to mention the name binding 2 times, and the same for realName . That's where the object destructuring syntax is useful: you can read a property and assign its value to a variable without duplicating the property name.
Honestly, I think you're overcomplicating this. Default parameters are not compulsory -- in this case your code can be cleaner without it.
I would simply take the object options
as the parameter and do the destructuring within the body of the function, after assigning default values.
function test(options) {
options = Object.assign({a: 'foo', b: 'bar'}, options);
let {a, b} = options;
console.log(options, a, b);
}
test(); // foo bar
test({a: 'baz'}); // baz bar
test({b: 'fuz'}); // foo fuz
test({c: 'fiz'}); // foo bar
With particular regard to your final snippet:
(function test(options = {a: "foo", b: "bar"}) {
console.log(options);
})({a: "baz"});
The problem is that a default parameter is used when the value passed is undefined
. Here, the value passed is {a: "baz"}
. That is not undefined
, so the default parameter is ignored. Objects are not merged automatically.
More broadly in answer to your question: there is no way of getting both an object and destructuring some of its properties in the parameters of a method. Frankly, I'm grateful, because function signatures can be hard enough to read at first glance as it is.
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