Is this logical operation:
const user = users && users[0] || null;
the same as this conditional / ternary operation:
const user = users ? users[0] : null;
? Assuming users is an array.
2 Answers
No, they are not the same. If users[0] is falsey, the || null will be selected in the first code:
const users = [0];
const user1 = users && users[0] || null;
const user2 = users ? users[0] : null;
console.log(user1, user2);The conditional operator is probably the better choice. But, even better, in modern JavaScript, you can use optional chaining instead, which is even more concise:
let users;
const user = users?.[0];
console.log(user);(though, note that it gives you undefined if the nested property doesn't exist, not null)
answered Oct 31, 2020 at 18:52
CertainPerformance
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1 Comment
jaredg
Ah interesting, I didn't know about the optional chaining. It's like a safe call in Kotlin.
Using users on its own isn't a very good way to do tests. Some languages like Go only allow tests against boolean values. To that end, it would be better to test against a boolean value if possible. Something like:
const user = users.length > 0 ? users[0] : null;
Comments
usersalways is an array, it is always thruthy and testing for it is pointless.