2mo ago

if statement == false

41 comments

what would be the alternative? to always execute if the condition is true, but sometimes execute it even when false, for funsies?
- Hey, I like a little chaos in my codebase 😆
  javascript
  
  const funsies = () => (Math.round(Math.random() * 1000) % 2 == 0) if ( condition || funsies() ) { // do the thing }
  
  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
  
  is this AI?
- 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.
  
  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
- 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 statement to execute literally the same code.}
- Brb, making a truly "if" statement function in my products code base for funsies.

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.
- 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.
  
  If not pizza and not fart: pass
  
  I don't love the pizza fart variable naming convention, but it's better than foobar and I don't have a better suggestion 😅

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

No, they're not.
Let's assume they are. Let funky function be defined as:
undefined

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.
- I'm pretty sure that funky() would always return 0, as defined. I'll pseudocode that up:
  undefined
  
  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:
  undefined
  
  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.
  
  Yeah, I’m not sure what the original intent was here. If we’re missing something I’d like to know
  
  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.
- 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.
  
  I agree but it's also what the original meme is doing. I thought we were all shitposting here.
- Underrated comment.

Waiting for a programming language with iff Syntax

If and only if… haha unless…

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.

 undefined

    
    if ( test == false )   
        x = 1
    else
        x != 1

Not sure what you’re trying to achieve with that else block, as it affects nothing.
- okay I fixed it
  undefined
  
  if ( test == false ) x = 1 else assert( x != 1 )

Reminds me of this fun stack exchange q

IFF you use the universal quantifier

Don't you just use iif?

41 comments