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 but I fixed it last month :-) — Bergi, Commented Feb 7, 2018 at 5:10
@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, Commented Feb 7, 2018 at 15:52

Bergi · Accepted Answer · 2018-02-07 05:12:14Z

21

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.

answered Feb 7, 2018 at 5:06

Bergi

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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
Commented Feb 7, 2018 at 6:14
Parenthesis? No, definitely not.
– Bergi
Commented Feb 7, 2018 at 6:20
5

I was trying to see if I can do this: func(...args || []). It works
– Sarsaparilla
Commented Jul 8, 2018 at 0:26
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
Commented Jul 8, 2018 at 15:53
IIUC, This is like saying is that it has a lower precedence than all operators. It applies to the fully-evaluated expression following it up to the closing ].
– einpoklum
Commented Sep 10, 2023 at 21:01

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