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Start your free trialTaylor Thurston
2,909 PointsI have exactly what's written in the video (except I have tried it using both i and n. It tells me my logic is wrong.wtf
for i in 1...100 { if (i % 3 == 0) && (i % 5 == 0) { print("FizzBuzz") } else if (i % 3 == 0) { print("Fizz") } else if (i % 5 == 0) { print ("Buzz") } else { print(i) } }
That's my function. Runs in xcode just fine...
func fizzBuzz(n: Int) -> String {
// Enter your code between the two comment markers
for n in 1...100 {
if (n % 3 == 0) && (n % 5 == 0) {
print("FizzBuzz")
} else if (n % 3 == 0) {
print("Fizz")
} else if (n % 5 == 0) {
print ("Buzz")
} else {
print(n)
}
}
// End code
return "\(n)"
}
1 Answer
Brandon Mahoney
iOS Development with Swift Techdegree Graduate 30,149 PointsYou have 3 issues with your code.
- You are checking the Int 'n' and not a range of 1-100. Remember 'n' is a undetermined Int which will be passed in when the function is called. If you were running through a range of numbers every time when the function is called there would be no point to having the n parameter.
- You must return and not print
- The default is already taken care of for you so you can delete your last 'else'.
In cases like this especially when you start working on your own projects the best thing to do is go back slowly over what you are trying to do and check each detail. I find that actually talking through what you're doing out loud can also help, basically just explain what you are doing and why.
func fizzBuzz(n: Int) -> String {
// Enter your code between the two comment markers
if (n % 3 == 0) && (n % 5 == 0) {
return("FizzBuzz")
} else if (n % 3 == 0) {
return("Fizz")
} else if (n % 5 == 0) {
return ("Buzz")
}
// End code
return "\(n)"
}