CODEWITHSUNDEEP
javascript
Léo Schneider
Léo Schneider

Reputation: 2295

What is the point of using the double logical not "!!" operator in JavaScript?

I quite regularly see some code like this:

const a = !!b && b.c === true;

I suppose the idea is to not have the a variable nullable, but in that case what is the difference with this code:

const a = b?.c === true

Is there a fundamental difference in between the two?

Upvotes: 2

Views: 1092

Answers (3)

LeGEC
LeGEC

Reputation: 51850

Both expressions have the same result.

The crucial difference is: compatibility

In pure JavaScript, the ?. operator is really recent: See the Browser compatibility chart on MDN. My current browser, for example, (today is 2020-03-11, and my Linux system is running Firefox 73) does not support the b?.c syntax.

The b?.c === true version could simply not be written before ES2020 and will not work as is on your client's browser if he/she hasn't updated to recent builds to this day: "recent" would mean "bleeding edge"...

As mentioned by jonrsharpe in his comment, the ?. operator is also available through transpiled languages (TypeScript, CoffeeScript, Babel, etc.), with various dates of support.

Upvotes: 0

Ch3micaL
Ch3micaL

Reputation: 323

This is more a JavaScript side of a problem...

The && operator returns the first falsy value, so 0, '', undefined, NaN or null would be the value of const a. If you want a boolean then the !! syntax is the most common way to ensure it being a Boolean.

If I'm not completely wrong on this the optional chaining (?.) just stops the execution on undefined or null values and returns undefined.

Upvotes: 1

CertainPerformance
CertainPerformance

Reputation: 370779

If the situation is such that b, if it exists (isn't undefined or null), will be an object, then no, there isn't any difference between those two.

The largest reason why you probably see someVar && someVar.someProp (or !!someVar && someVar.someProp) and variations is that optional chaining is pretty new syntax. It will only exist in recently-updated codebases (running TypeScript 3.7 or above).

But if a variable may be falsy, but is not necessarily an object - for example, if it's 0, NaN, false, or the empty string, then those constructs are not equivalent. Optional chaining with ?. will short-circuit to undefined only if expression to the left of the ?. is null or undefined. Other falsy values will continue to have their properties be evaluated as normal.

Upvotes: 0

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