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

Posted by james white
james white
78,399 Points

FizzBuzz casting

I tried this code:

  for n in 1...100 {
    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
    }
  }

..and it gave me this error:

swift_lint.swift:13:16: error: cannot convert return expression of type 'Int' to return type 'String'
        return n
               ^

So I figured that meant I had to do some type casting.

This StackOverFlow thread was helpful in figuring out how to do the string conversion:

http://stackoverflow.com/questions/24161336/convert-int-to-string-in-swift

So based on the above StackOverFlow thread where it had this line:

var myString = String(x)

I then decided to try this code:

  for n in 1...100 {
    if (n % 3 == 0) && (n % 5 == 0) {
        return "FizzBuzz"
    } else if (n % 3 == 0) {
        return "Fizz"
    } else if (n % 5 == 0) {
        return "Buzz"
    } else {
        var myString = String(x)
        return myString
    }
  }

..which gave me this Bummer:

Bummer! Double check your logic for Fizz values and make sure you're printing the correct string!

So it wants printing? Huh?! :confused:

I thought Step 3 said:

Step 3: Change all your print statements to return statements.

But if it wants print statements then I'll add back the print statements:

  for i in 1...100 {
    if (n % 3 == 0) && (n % 5 == 0) {
        print("FizzBuzz")
        return "FizzBuzz"
    } else if (n % 3 == 0) {
        print("Fizz")
        return "Fizz"
    } else if (n % 5 == 0) {
        print("Buzz")
        return "Buzz"
    } else {
        print(i)
        var myString = String(i)
        return myString
    }
  }

Note: I also figured (re-reading the challenge instructions very carefully),

I would change back some of the 'n' occurrences back to 'i'

and I somehow got it to pass (don't really know why though? --but whatever.. )

1 Answer

Pasan Premaratne
STAFF
Pasan Premaratne
Treehouse Teacher
Pasan Premaratne
Pasan Premaratne
Treehouse Teacher

james white,

This course simply builds on top of everything taught so far and since we haven't covered casting the challenge doesn't specifically look for that.

The starter code for the challenge already returns the else case as a string which is why you're running into the error. 

return "\(n)"

It uses simple string interpolation instead of casting.