Operator precedence for JS spread and rest operators?

Question

I was trying to find them on MDN's Operator Precedence table (https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Operators/Operator_Precedence#Table) but unless they are a subcategory of an existing operator type, I don't see them. I couldn't find any other obvious documentation about it.

It had been part of that table [since 2014](https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Operators/Operator_Precedence$compare?to=613269&from=612097) but I fixed it [last month](https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Operators/Operator_Precedence$compare?to=1346329&from=1346230) :-) — Bergi, Feb 07 '18 at 05:10
@Bergi: thank you! I can’t believe someone added that to the table o_O but who knows, things were still in flux in 2014 — Felix Kling, Feb 07 '18 at 15:52

Bergi · Accepted Answer · 2018-02-07T05:12:14.477

22

Spread syntax is not an operator and therefore does not have a precedence.

It is part of the array literal and function call (and object literal) syntax.

Similarly, rest syntax is part of the array destructuring and function parameter (and object destructuring) syntax.

edited Feb 07 '18 at 05:12

answered Feb 07 '18 at 05:06

Bergi

1

I guess that was the source of my confusion. Thanks for the quick answer! Hopefully it is more Googleable now. I had someone mention it in code review and suggest parentheses and it made me question how it got evaluated. – Scotty Waggoner Feb 07 '18 at 06:14
Parenthesis? No, definitely not. – Bergi Feb 07 '18 at 06:20
4

I was trying to see if I can do this: `func(...args || [])`. It works – Sarsaparilla Jul 08 '18 at 00:26
3

@HaiPhan Yes, you can do `func(...anyExpression)` where the only limitation to *any expression* is that it must not be a comma operator expression (which would count as an argument delimiter instead). – Bergi Jul 08 '18 at 15:53

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