• raoul@lemmy.sdf.orgEnglish
    55·
    6 days ago

    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.

  • juliebean@lemm.eeEnglish
    61·
    6 days ago

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

    • Admiral Patrick@dubvee.orgEnglish
      41·
      5 days 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.worksEnglish
        5·
        5 days 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.comEnglish
          2·
          4 days 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

        • Admiral Patrick@dubvee.orgEnglish
          19·
          6 days 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?

        • Admiral Patrick@dubvee.orgEnglish
          2·
          5 days 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.eeEnglish
            2·
            18 hours 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?

            • Admiral Patrick@dubvee.orgEnglish
              1·
              10 hours ago

              It’ll give you 1 ~= true or 0 ~= undefined, but I typically use Typescript which prefers actual booleans to boolean-ish

              • juliebean@lemm.eeEnglish
                1·
                4 hours ago

                huh. interesting. i wonder what number it’s actually storing for false then?

    • bstix@feddit.dkEnglish
      17·
      6 days ago

      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.worldEnglish
        1·
        4 days ago

        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.dkEnglish
          1·
          4 days ago

          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.worldEnglish
      7·
      6 days ago

      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.

    • unmagical@lemmy.mlEnglish
      1·
      6 days ago

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

  • stevedice@sh.itjust.worksEnglish
    10·
    6 days ago

    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.

    • nickwitha_k (he/him)@lemmy.sdf.orgEnglish
      14·
      6 days ago

      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.

      • stevedice@sh.itjust.worksEnglish
        5·
        6 days ago

        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 (he/him)@lemmy.sdf.orgEnglish
          2·
          6 days ago

          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.worksEnglish
            3·
            6 days ago

            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”

    • ninja@lemmy.worldEnglish
      7·
      6 days ago

      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.

  • dohpaz42@lemmy.worldEnglish
    6·
    6 days ago

    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.