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iOS

Posted by Chris Freeman
Chris Freeman
1,077 Points

Extra credit - Given a range from 1 to 100. Write a loop which prints out wether a number is odd or even.

can anyone tell me why my code doesn't work please?

import UIKit

for number in 1...100
    if (number % 2 == 0) {
        println("\(number) Even")
    } else if (number % 2 != 0) {
        println("\(number) Odd")
    } else {
        println(number)
}

1 Answer

Steve Hunter
Steve Hunter
57,712 Points

You haven't wrapped the code after the for loop with curly braces.

Plus; what's all the tests? I don't think they're needed. The initial test has a binary outcome x % 2 == 0 is either true or false. There's no need for testing if x % 2 != 0; if the first test fails this is the only other outcome. So, you just need your first if testing for number % 2 == 0 printing out "Even", then an else to print out "Odd".

Steve.

Steve Hunter
Steve Hunter
57,712 Points

My code looks like:

for number in 1...100 {  // <- you missed that curly brace
  if (number % 2 == 0){ // binary test - only two outcomes
    println("\(number) is even")
  } else {
    println("\(number) is odd")
  }
} // <- close it here
Chris Freeman
Chris Freeman
1,077 Points

Thanks Steve, this has helped me a lot.

Steve Hunter
Steve Hunter
57,712 Points

No problem! Glad it helped. :-)