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I'm trying to figure out if there's a way to use object destructuring of default parameters without worrying about the object being partially defined. Consider the following:

(function test({a, b} = {a: "foo", b: "bar"}) {
  console.log(a + " " + b);
})();

When I call this with {a: "qux"}, for instance, I see qux undefined in the console when what I really want is qux bar. Is there a way to achieve this without manually checking all the object's properties?

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1 Answer 1

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266

Yes. You can use "defaults" in destructuring as well:

(function test({a = "foo", b = "bar"} = {}) {
  console.log(a + " " + b);
})();

This is not restricted to function parameters, but works in every destructuring expression.

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11 Comments

Nifty! It seems like setting the defaults on the destructuring side (the way you're doing it) would always be preferred to doing it on the default parameter side (like I was). Would you agree with that? Can you think of anything to watch out for?
I don't think one is "preferred" over the other. It just does something else.
@AlanHamlett No, destructuring only deals with undefined - passing null must be done explicitly, and therefore leads to an error. To deal with both null and undefined, use the standard b == null ? "bar" : b inside the function.
@Bergi Could you explain what the difference between your answer and test({a="foo", b="bar"}) is?
@YonggooNoh See here and here
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