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iOS

Posted by Buka Cakrawala
Buka Cakrawala
5,042 Points

FizzBuzz Generator in Swift Basics

for number in 1...20 {
    if (number % 3 == 0) && (number % 5 == 0){
        println("FizzBuzz")
    }   else if (number % 3 ==0) {
        println("Fizz")
    }
}

This is what i wrote in xcode and I did the same thing as what the video has been told me, but why does the xcode always left a red mark or a wrong mark in the println function always suggesting me to change to print. PLEASE HELP ME. thank you

1 Answer

Steve Hunter
Steve Hunter
57,712 Points

Hi there,

If you change println to print it'll work fine.

Some of the courses are for Swift v1.* but now Xcode is using Swift 2.*. Between the two releases, a couple of things changed; print became the new println, for example. Just use print instead, and you'll be fine.

Steve.