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submitted 1 week ago by tatterdemalion@programming.dev to c/science_memes@mander.xyz

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[-] ptz@dubvee.org 40 points 1 week ago* (last edited 1 week ago)

Hey, I like a little chaos in my codebase 😆

const funsies = () => (Math.round(Math.random() * 1000) % 2 == 0)
if ( condition || funsies() ) {
    // do the thing
}

[-] sorrybookbroke@sh.itjust.works 6 points 1 week ago* (last edited 1 week ago)

Wait, why the the *1000 or mod 2? Won't that give a 50\50 chance or the same as Math.round(Math.random).

No shade, and I may be wrong myself I am very tired

[-] mexicancartel@lemmy.dbzer0.com 2 points 1 week ago

Probably to make the fractional random value between 0 and 1 to become an integer so that you can divisibility check for even with mod 2

[-] modality@lemmy.myserv.one 3 points 1 week ago

is this AI?

[-] ptz@dubvee.org 19 points 1 week ago* (last edited 1 week ago)

Nope. I'm staunchly against "AI". If the code sucks, it's because I wrote bad code.

Edit: Oh, or did you mean is that how "AI" works?

[+] sorrybookbroke@sh.itjust.works 1 points 1 week ago* (last edited 1 week ago)

[deleted]

[-] ptz@dubvee.org 3 points 1 week ago* (last edited 1 week ago)

In Javascript, Math.random() produces a float value between 0 and 1, so I multiply by 1000 and round it to get a larger integer. The value %2 == 0 is a non-library way of performing isEven() on the random integer (value % 2 is 0 if even, 1 if odd, and the ==0 makes it return a boolean). When used in the if statement, it's essentially a coin flip.

[-] juliebean@lemm.ee 3 points 1 week ago

does javascript not allow you to interpret integers as booleans in a conditions directly? seems it'd be simpler to just do math.round(math.random()), which should still get you true (1) or false (0) in equal likelihood. or am i missing something?