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hans wagner
4,872 PointsisDivisible function works in Playground but not is code challenge
Here is my code: func isDivisible (dividend: Int, divisor: Int) -> Bool? { if dividend % divisor == 0 { return true } else { return false } }
func isNotDivisible (dividend: Int, divisor: Int) -> Bool? { if dividend % divisor != 0 { return true } else { return false } }
if let answer = isDivisible(15, divisor: 3) { print("Divisible", terminator: "") } else { print("Not Divisible", terminator: "") }
if let result = isNotDivisible(15, divisor: 3) { print("Not Divisible", terminator: "") } else { print("Divisible", terminator: "") }
func isDivisible (dividend: Int, divisor: Int) -> Bool? {
if dividend % divisor == 0 {
return true
} else {
return nil
}
}
func isNotDivisible (dividend: Int, divisor: Int) -> Bool? {
if dividend % divisor != 0 {
return true
} else {
return nil
}
}
if let divisible = isDivisible(15, 3) {
print("Divisible", terminator: "")
} else {
print("Not Divisible", terminator: "")
}
if let result = isNotDivisible(15, 3) {
print("Not Divisible", terminator: "")
} else {
print("Divisible", terminator: "")
}
3 Answers
Steve Hunter
57,712 PointsThe question is looking for your method to return a Boolean, not an Optional - so if you amend your lines return nil to return false that may assist.
Also, you've duplicated pretty much the whole method. The two methods are a character, or so, different. How about calling isDivisible from inside isNotDivisible and negating the returned value?
func isDivisible(#dividend: Int, #divisor: Int) -> Bool {
if dividend % divisor == 0 {
return true
} else {
return false
}
}
func isNotDivisible(#dividend: Int, #divisor: Int) -> Bool {
return !isDivisible(dividend: dividend, divisor: divisor)
}
That's more DRY.
Hope that makes sense.
Steve.
hans wagner
4,872 PointsSteve, thanks for your quick reply. I like your DRY approach, as I recognized the two methods were very similar. Though that is what we were asked to do. Thanks again for your help.
Steve Hunter
57,712 PointsThe question asks for a method that returns a boolean depending whether the divisor and dividend are not divisible. How you do that is wholly up to you - there's no requirement to repeat code.
I think the Treehouse lessons make a strong point of refactoring to avoid repetition but it isn't always easy to see how to achieve it. Thankfully, this challenge will pass with both ways of achieving the result so the duplicated methd is fine, and it is only a few lines of code.
Steve.
hans wagner
4,872 PointsI appreciate striving being efficient in one's code. Thank you for encouraging me to pursue the best approach.