if statement == false
from tatterdemalion@programming.dev to science_memes@mander.xyz on 05 Dec 21:07
https://programming.dev/post/22510734

#science_memes

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juliebean@lemm.ee on 05 Dec 21:19 next collapse

what would be the alternative? to always execute if the condition is true, but sometimes execute it even when false, for funsies?

ptz@dubvee.org on 05 Dec 21:23 next collapse

Hey, I like a little chaos in my codebase 😆

const funsies = () => (Math.round(Math.random() * 1000) % 2 == 0)
if ( condition || funsies() ) {
    // do the thing
}
modality@lemmy.myserv.one on 05 Dec 23:00 next collapse

is this AI?

ptz@dubvee.org on 05 Dec 23:04 collapse

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?

[deleted] on 06 Dec 05:59 next collapse

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sorrybookbroke@sh.itjust.works on 06 Dec 06:19 collapse

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 on 07 Dec 11:41 collapse

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

unmagical@lemmy.ml on 05 Dec 21:26 next collapse

Brb, making a truly “if” statement function in my products code base for funsies.

[deleted] on 05 Dec 21:36 next collapse

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bstix@feddit.dk on 05 Dec 22:34 next collapse

The “only” part implies exclusivity, which may be false, because other things might run the code anyway.

IF “I can see the sun” THEN “It’s day.”

Nothing wrong about that. However if we make it exclusive:

IF AND ONLY IF “I can see the sun” THEN “It’s day.”

That’s obviously wrong. I can actually not keep the day away by sitting with closed eyes in my mothers basement with the curtains shut.

“Only if” might make sense in a legal contract, but there’s no way a piece of code can stop other pieces of code from calling the same functions.

adj16@lemmy.world on 07 Dec 16:10 collapse

But that’s not how if statements in code work. So what you’ve said isn’t wrong, but the premise of this meme is completely off

bstix@feddit.dk on 07 Dec 17:18 collapse

Yes sure. Code is logical stepwise. By including the “only if” it implies that other stuff is taken into account, which it isn’t at that moment in code.

I mean, I don’t need to extend the implications of an IF statement. It already does exactly what it says.

Anyway fuzzy logic does exist for people who want “sometimes if”. It’s useful in certain cases. I’ve only ever considered it in music production, where things very often get to the point of complexity where it makes a (sometimes) useful difference.

General_Effort@lemmy.world on 05 Dec 23:00 collapse

In normal parlance, “if and only if” rules out that something could also happen as a result of other circumstances. EG, if you fall out of a plane, you will lose your glasses. But there are other conditions that would lead to the same result.

In code, the alternative would be to have a different if statement that executes identical code. Or *cough* you could use a jump instruction to execute literally the same code.

kbal@fedia.io on 05 Dec 21:27 next collapse

    if ( test == false )   
        x = 1
    else
        x != 1
dohpaz42@lemmy.world on 05 Dec 21:52 collapse

Not sure what you’re trying to achieve with that else block, as it affects nothing.

kbal@fedia.io on 06 Dec 00:17 collapse

okay I fixed it

    if ( test == false )
        x = 1
    else
        assert( x != 1 )
stevedice@sh.itjust.works on 05 Dec 21:49 next collapse

No, they’re not.

Let’s assume they are. Let funky function be defined as:

int funky() {
    a=0
    b=1
    if ( a==1 ) {
        b=1
    }
    return(a)
}

Since a==1 if, and only if, b=1, in particular a==1 if b=1. We have b=1, therefore a==1. It follows funky will always return 1 but… it doesn’t. QED.

pancake@lemmygrad.ml on 05 Dec 22:03 next collapse

Underrated comment.

ninja@lemmy.world on 05 Dec 22:27 next collapse

Translating structured logic into spoken language is iffy. (I’m sorry. I couldn’t help myself)

The code reads to match OP if stated as: “If and only if the value of ‘a’ equals 1 then set the value of ‘b’ to equal one.” Placing the conditional at the beginning of the sentence maintains the correct dependency.

stevedice@sh.itjust.works on 06 Dec 00:18 collapse

I agree but it’s also what the original meme is doing. I thought we were all shitposting here.

nickwitha_k@lemmy.sdf.org on 05 Dec 22:39 collapse

I’m pretty sure that funky() would always return 0, as defined. I’ll pseudocode that up:

funky takes no args, returns int {
  a is assigned the value 0
  b is assigned the value 1

  test if a is equal to 1, if it is {
    b is assigned the value 1
  }

  return a
}

The if in your function can never be reached, without some weird manipulation of the value of a that breaks variable scoping in most syntaxes.

I think that I see your logic but it is syntactically incorrect:

    if ( a==1 ) {
        b=1
    }

In most syntaxes, this is a conditional execution and value assignment. That is, the code in curly braces only gets executed, if the conditional evaluates as true. If the conditional evaluates as true, the code is executed, assigning the value 1 to the variable b.

It does NOT imply that the assignment of the value 1 to the variable b is a conditional requiring the assignment of the value 1 to the variable b.

Remember: = in most programming is NOT an equality symbol but a value-assigment symbol. It would be nice if people creating the initial syntaxes used something else that is harder to confuse but they didn’t.

ikilledlaurapalmer@lemmy.world on 05 Dec 22:56 next collapse

Yeah, I’m not sure what the original intent was here. If we’re missing something I’d like to know

stevedice@sh.itjust.works on 06 Dec 00:16 collapse

Yes, I know, that’s the point. Funky is specifically constructed to always return 0. Then we assume “if” and “if, and only if” are equivalent and by following that assumption to its logical conclusion, we deduce that funky returns 1. Therefore, our assumption was incorrect because 0≠1. It follows that “if” isn’t equivalent to “if, and only if”. Also, it’s just a shitpost.

nickwitha_k@lemmy.sdf.org on 06 Dec 03:23 collapse

If reading the code as non-programming logic, that conclusion makes sense, yes. However, if, in most syntaxes, is a type of flow control. What it wraps has no meaning to the if statement itself. Reading it through the lens of an interpreter/compiler makes it clear. The statement is approximately:

If and only if a is equal to 1, do the thing {
  The thing is: assign the variable b with the value 1
}

To one not familiar with how programs are executed, it would make sense that the return value could be 1. But understanding how flow control works in programming, makes this interpretation a challenge.

stevedice@sh.itjust.works on 06 Dec 05:38 collapse

I don’t think you’re picking up what I’m putting down. I’m not arguing that the return value can be 1, I’m well aware that it can’t — I wrote the function so that it will always return 0. It only returns 1 if we make an incorrect assumption (and mix up semantics with formal logic, but that’s another conversation), the incorrect assumption being “if is equivalent to if, and only if

nickwitha_k@lemmy.sdf.org on 07 Dec 04:27 collapse

Sorry! I sometimes get carried away on correctness.

stevedice@sh.itjust.works on 07 Dec 20:12 collapse

I mean, making an assumption and arriving to a contradiction is as correct as a proof gets.

dohpaz42@lemmy.world on 05 Dec 21:55 next collapse

An if statement with one condition is an if and only if statement. The moment you add a second (or more) condition (regardless of && or ||), then it’s no longer if and only if.

GammaGames@beehaw.org on 05 Dec 22:00 next collapse

Reminds me of this fun stack exchange q

reallykindasorta@slrpnk.net on 05 Dec 23:30 next collapse

IFF you use the universal quantifier

raoul@lemmy.sdf.org on 05 Dec 23:51 next collapse

The boolean operator ‘If and only if’ do not have a relation with the program instruction ‘if’.

The programatic ‘if’ is a jump, not a boolean operator. It do not have truth table.

In logic:, if and iff can be seen like functions taking two booleans and returning a boolean

  • ‘if a then b’ (noted a -> b): return true if a is false or b is true. Example: ‘if I eat pizza then I fart’ This is true even if I fart all the time (if b is true, we do not care about the value of a) as long as I fart when eating pizza (if a is true, b must be also true)

  • ‘a <-> b’ is equivalent to ‘a -> b and b -> a’: the two should be true at the same time. I can only fart will eating pizza and cannot fart otherwise.

[deleted] on 06 Dec 05:30 next collapse

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Hoimo@ani.social on 06 Dec 12:43 collapse

So in programming, you’d write ‘if’ as:
not pizza or fart where the farting is irrelevant until the pizza is involved.

While ‘iff’ would be:
pizza equals fart where pizza means fart and no pizza means no fart.

I actually wrote iff as (not pizza and not fart) or (pizza and fart) before, and I’m pretty sure that’s the way I wrote an iff in production code in the past, but your comment made me realize that “they should be true at the same time” can be tested really easily with equality.

captain_aggravated@sh.itjust.works on 07 Dec 04:44 next collapse

If not pizza and not fart: pass

problematicPanther@lemmy.world on 07 Dec 10:05 collapse

If pizza then fart else !fart

Venator@lemmy.nz on 08 Dec 09:31 collapse

I don’t love the pizza fart variable naming convention, but it’s better than foobar and I don’t have a better suggestion 😅

BreadOven@lemmy.world on 06 Dec 01:55 next collapse

Don’t you just use iif?

Hammocks4All@lemmy.ml on 06 Dec 02:22 next collapse

If and only if… haha unless…

sik0fewl@lemmy.ca on 07 Dec 09:50 next collapse

I hope this memer is not a programmer or logician, but ideally neither.

eunieisthebus@feddit.org on 07 Dec 15:23 collapse

Waiting for a programming language with iff Syntax