Comment on if statement == false

stevedice@sh.itjust.works ⁨11⁩ ⁨months⁩ ago

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

• nickwitha_k@lemmy.sdf.org ⁨11⁩ ⁨months⁩ ago

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

          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.

  source

  • ikilledlaurapalmer@lemmy.world ⁨11⁩ ⁨months⁩ ago

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

    source
  • stevedice@sh.itjust.works ⁨11⁩ ⁨months⁩ 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 of 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.

    source

    • nickwitha_k@lemmy.sdf.org ⁨11⁩ ⁨months⁩ 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.

      source

      • stevedice@sh.itjust.works ⁨11⁩ ⁨months⁩ 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”

        source

        • -> View More Comments
• ninja@lemmy.world ⁨11⁩ ⁨months⁩ ago

  [deleted]

  source

  • stevedice@sh.itjust.works ⁨11⁩ ⁨months⁩ ago

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

    source