• kameecoding@lemmy.worldEnglish
    31·
    1 year ago

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

    0 && 1 && false

    • EuphoricPenguin@normalcity.lifeEnglish
      2·
      1 year ago

      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@lemmy.worldEnglish
        1·
        1 year ago

        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.

        • EuphoricPenguin@normalcity.lifeEnglish
          1·
          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.