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iOS

Posted by Andis Slaitas
Andis Slaitas
3,430 Points

FizzBuzz function - something I wrote, would like to share.

The function takes 4 arguments: range of values - start and end, and the values for fizz and buzz conditions. Returns an array of values.

func fizzBuzz(start: Int, end: Int, fizzValue: Int, buzzValue: Int) -> [String] {
    var numbersCollection: [String] = []

    for number in start...end {
        switch number {
            case 0: numbersCollection.append("\(number)")
            case _ where number % (fizzValue * buzzValue) == 0: numbersCollection.append("FizzBuzz")
            case _ where number % fizzValue == 0: numbersCollection.append("Fizz")
            case _ where number % buzzValue == 0: numbersCollection.append("Buzz")
            default: numbersCollection.append("\(number)")
        }
    }
    return numbersCollection
}

1 Answer

Tommy Choe
Tommy Choe
38,156 Points
Tommy Choe
Tommy Choe
38,156 Points

Hey Andis, I love it man! If I may offer one suggestion though: Add a case to handle 0. 0 shows up as "fizzbuzz".

Andis Slaitas
Andis Slaitas
3,430 Points

Hey and thanks! I updated the code to handle 0. However zero is a somewhat dubious case. Obviously 0 divides with any number (except 0) without a remainder. Does that mean 0 is a multiple of 3 and/or 5?

Tommy Choe
Tommy Choe
38,156 Points

Good job, looks like you fixed it.

You're right though 0 is one of several dubious cases. When the left operand is less than the right operand, the compiler simply returns the left operand. If you want to see how it works you should test it out on the playground. That's what I did to find out how the compiler reacted to abnormal situations.

In your case, the only reason why 0 does not reach the default statement is because you're first case statement tests to see if the result of number % (fizzValue * buzzValue) == 0, which is true. Since the left operand is less than the right, the expression simply returns a 0 (so true).

Let me know if I wasn't clear on something.