当值为布尔类型时使用!!（双重否定）语法是否属于冗余写法？

Great question! Let's break this down clearly:

First, let's confirm what !!IS_DEV does when IS_DEV is a strict boolean:

• The first ! inverts the boolean value: !true becomes false, !false becomes true
• The second ! inverts it again: !false becomes true, !true becomes false

So when IS_DEV is guaranteed to be only true or false, !!IS_DEV will always evaluate to exactly the same value as IS_DEV itself. There's no functional difference between const sandboxMode = !!IS_DEV; and const sandboxMode = IS_DEV; here—no change to the value, type, or runtime behavior.

Is this a case of redundant syntax? #

Yes, absolutely. Since IS_DEV is already a boolean, wrapping it in double negation doesn't serve any practical purpose—it's just unnecessary extra syntax that adds no value.

Why did ESLint fix it this way? #

ESLint's rule that flagged your original ternary (IS_DEV ? true : false) is built to eliminate redundant boolean conversions. For cases where the input might be a truthy/falsy non-boolean value (like 0, "", or an object), !!value is a concise way to coerce it to a strict boolean. But in your specific scenario where IS_DEV is guaranteed to be a boolean, this fix introduces a new, unnecessary redundancy.

What's the best approach here? #

If you're certain IS_DEV will always be a boolean (true or false), stick with const sandboxMode = IS_DEV;—it's cleaner, more readable, and avoids unnecessary runtime operations.

内容的提问来源于stack exchange，提问作者Luis Febro