How does this particular scenario of default params and destructuring work?

Question

I was trying some things today and came across a behaviour I would like to understand.

var b = ({a = 1, b = 1, c = 1}) => a + b + c;

b(); // throws error.

But if it is defined like this

var b = ({a = 1, b = 1, c = 1} = 0) => a + b + c;

b() // returns 3
b([]) // returns 3

Shouldn’t this be an error? Did zero somehow become an object here? Is it somehow equivalent to the following?

var b = ({a = 1, b = 1, c = 1} = {}) => a + b + c; // this is possible I guess.

My question is not how regular destrcuturing and default params work, but only how this particular scenario is being evaluated.

Can some one explain this to me?

Answer 1

({a = 1, b = 1, c = 1} = something) => {}

just means that something must be an object or can be converted to one, i.e. it can’t be null or undefined.¹

So, in the case of 0, it proceeds to grab the a, b and c properties of 0, i.e. (0).a, (0).b, (0).c, all of which are undefined, hence all of them default to 1, their provided default value.

0 can of course be coerced to a Number object. That’s why you can do (0).toString() or {toString} = 0. That’s exactly what happens here.

It’s generally not equivalent to using {} as a default, since that would use an empty object’s properties (both own properties and the ones on the prototype chain), not the number’s properties.

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1: The most reduced form of this “structural verification” is ({} = something). For destructuring onto arrays, it is ([] = something) and it means that the something must also be iterable. Those empty destructuring assignments, by the way, don’t create any variables, they just do the structure check.