if statement == false

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.

This is true for purely procedural programming languages, but many on the functional side do have an if-else with a truth-table, for example:

let x = if a < b {
    3
} else {
    5
};

It should also be noted that even procedural languages typically have the ternary operator, which is the same thing in unreadable:

let x = a < b ? 3 : 5;

Hey, I like a little chaos in my codebase 😆

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

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.

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.~

Even in language it forms a conditional sentence. If A then B

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

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.

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.

Underrated comment.

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