Logical Assignment w/ Destructuring

💡 Ideas
1

I was wondering why we can't use Logical assignment operators with object destructuring.

Example...

const { prop &&= 1 } = object; // Sets prop to 1, if it is truthy.
const { prop ||= 1 } = object; // Sets prop to 1, if it is falsey.
const { prop ??= 1 } = object; // Sets prop to 1, if it is a null.

Let me know what you think. Thank you :grinning:.

2

My guess is that's to maintain consistency with other augmented assignment operators, none of which work in destructuring.

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3

Linking other thread where this was also asked
Logical assignment for default assignments

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4

The first “=” sign in const { prop = 1 } = object (or in function foo(arg = 1) { }) is not an assignment operator; rather it specifies a default value. Therefore, any extension of the assignment operator (“compound assignment” or “logical assignment”) does not apply.

Of course, it is possible to make sense of const { prop ||= 1 } = object. However, I’m wondering whether the cost of adding yet another complexity in the language is worth the gain in conciseness? For comparison, as of today, you can write:

let { prop } = object;
prop ||= 1;
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