this post was submitted on 27 Jul 2023
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I think we need all support we can get to fight Google on this, so I welcome Brave here actually.

Use this link to avoid going to Twitter:

https://nitter.kavin.rocks/BrendanEich/status/1684561924191842304

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[–] kameecoding 5 points 1 year ago* (last edited 1 year ago) (1 children)

sure mate, just tell me the result of the following without trying it out.

0 && 1 && false

[–] [email protected] 3 points 1 year ago* (last edited 1 year ago) (1 children)

If I remember correctly, 0 and 1 are considered falsy and truthy respectively, so it should be falsy and truthy and false which I believe would return false.

Tried it out to double-check, and the type of the first in the sequence is what ultimately is returned. It would still function the same way if you used it in a conditional, due to truthy/falsy values.

[–] kameecoding 2 points 1 year ago* (last edited 1 year ago) (1 children)

yes, that is a solid logic, one that I also applied and expected to be the result.

that is until a Vue component started complaining that I am passing in a number for a prop that expects a boolean.

turns out the result of that code is actually: 0, because javascript

of course if you flip it and try

false && 0 && 1

then you get false, because that's what you really want in a language, where && behaves differently depending on what is on what side.

[–] [email protected] 2 points 1 year ago* (last edited 1 year ago)

I was incorrect; the first part of my answer was my initial guess, in which I thought a boolean was returned; this is not explicitly the case. I checked and found what you were saying in the second part of my answer.

You could use strict equality operators in a conditional to verify types before the main condition, or use Typescript if that's your thing. Types are cool and great and important for a lot of scenarios (used them both in Java and Python), but I rarely run into issues with the script-level stuff I make in JavaScript.