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iOS Swift 2.0 Collections and Control Flow Control Flow With Conditional Statements FizzBuzz

Carson Carbery
Carson Carbery
9,876 Points
Posted by Carson Carbery
Carson Carbery
9,876 Points

Working solution in playground doesn't work in code challenge screen

So here's my code again: Seems the solution I came up with, though it works doesn't fit in with the code challenge screen :-/

import Foundation

var randomNumber = Int(arc4random_uniform(UInt32(50)))

let divByThree = randomNumber % 3 let divByFive = randomNumber % 5

if divByThree == 0 && divByFive == 0 { print("FizzBuzz") } else if divByFive == 0 { print("Fizz") } else if divByThree == 0 { print("Buzz") } else { print(randomNumber) }

fizzBuzz.swift 
func fizzBuzz(n: Int) -> String {
  import Foundation

var randomNumber = Int(arc4random_uniform(UInt32(50)))

let divByThree = randomNumber % 3
let divByFive = randomNumber % 5

if divByThree == 0 && divByFive == 0 {
    return "FizzBuzz"
} else if divByFive == 0 {
    return "Fizz"
} else if divByThree == 0 {
    return "Buzz"
} else {
    return randomNumber
}

  return "\(n)"
}

1 Answer

jcorum
jcorum
71,830 Points

Carson, the Treehouse workspace is quite picky. You can do it "your way" in a Playground. But if you want to complete the challenge you have to do it the way Treehouse asks you to.

A couple of observations though. import statements must be at the very top of the code. You really don't need any random numbers. As he said, he was going to call your function inside a for in loop over a range of 1...100. The function returns a String, so all the return statements have to return Strings. Your last return statement in the if...else... tries to return an Int. I've also found that it really helps, when trying to find errors, to indent at the beginning of each code block, and to outdent at the end of each code block when I'm writing the code.

So here's what they want. You can compare it to yours.

func fizzBuzz(n: Int) -> String {

    if n % 3 == 0 && n % 5 == 0 {
        return "FizzBuzz"
    } else if n % 3 == 0 {
        return "Fizz"
    } else if n % 5 == 0 {
        return "Buzz"
    } else {
        return "\(n)"
    }
}