Why did index go out of range

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Hello i'm doing the simple FizzBuzz challenge. I figured out a solution to the problem. But I don't understand why my first attempt went out of range or what is happening behind the scenes.

To stop it from going out of range I changed

array.remove(at: i)

to

array.remove(at: i - 1)

Below is my entire project.
First attempt
    var array = [String]()


for i in 1...100{



           array.append("\(i)")

           if i % 3 == 0 && i % 5 == 0 {

               array.remove(at: i)
               array.append("FizzBuzz")
           }else if i % 3 == 0{

               array.remove(at: i)
               array.append("Fizz")

           } else if i % 5 == 0{

               array.remove(at: i)
               array.append("Buzz")

           }


       }
print(array)
Second attempt
var array = [String]()

for i in 1...100{

           array.append("\(i)")

           if i % 3 == 0 && i % 5 == 0 {

               array.remove(at: i - 1 )
               array.append("FizzBuzz")
           }else if i % 3 == 0{

               array.remove(at: i - 1)
               array.append("Fizz")

           } else if i % 5 == 0{

               array.remove(at: i - 1)
               array.append("Buzz") 
           }
       }
print(array)

Because arrays are zero index and you're starting at 1 in your for loop, so when you get to three and try to retrieve the third element (which indicates they're four elements in the array) swift will throw an error. The simple solution will be to start at 0

for i in 0...100{

print(i)

           array.append("\(i)")

           if i % 3 == 0 && i % 5 == 0 {

               array.remove(at: i)
               array.append("FizzBuzz")
           }else if i % 3 == 0{

               array.remove(at: i)
               array.append("Fizz")

           } else if i % 5 == 0{

               array.remove(at: i)
               array.append("Buzz")

           }



       }
print(array)

}

Help: Index out of range � YourBasic Go, If the user does not provide any input, the inp array is empty. This means that even the index 0 is out of range, i.e. inp[0] can't be� I.e., an array of length = 1, would have an values at index = 0, an array of length = 2 would have values at index = 0 and index = 1, etc Looking at the picture and values you pass above to the function variables, you start your for loop at the passed value of start(1), it tries to get the chars[1] value, when it only has a chars[0] value.

The remove(at:) function removes the element at the specified position. For example, if I have an array like [1,2,3,4] and I call array.remove(at: 2) the resulting array would be [1,2,4].

It is important to understand that the function removes the element at the specified INDEX location. Not the values itself. Swift arrays are zero indexed meaning, the first element is at location 0.

Now with this knowledge, I suggest you go ahead and debug your first code with some print statements and you be able to quickly understand why your code is going out of range. Hopefully through this experiment you can understand some basic concepts.

Good luck!

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You are saying

for i in 1...100{

But array indices start at 0. So on every loop the item you just added has index i-1. There is no item with index i yet.

To put it another way, consider this part of your code:

       if i % 3 == 0 && i % 5 == 0 {
           array.remove(at: i)

Here you are prevaricating on the meaning of i. In the first line, it is what number i is. But in the second line, it is what index i is at. And because you started with 1, those two numbers are apart by 1. For example, the first time thru the loop, i is 1, but the value 1 is at index 0.

How can I skip the index out of range error? - Getting Help, One common error every array user ever faced is “Fatal error: Index out of range”.this error was caused by accessing an index that array� You are reducing the length of your list l as you iterate over it, so as you approach the end of your indices in the range statement, some of those indices are no longer valid.

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Comments
  • Arrays start at index 0
  • @jacob welcome to stackoverflow, how about adding else and appending "i" only if it do not go through any of the conditions above.
  • Swift 4.x or later you can use i.isMultiple(of: 3). Btw it would be much easier/faster to simply assign a new value using subscript instead of remove / append. array[i] = "FizzBuzz"
  • Yeah, you can use that as well. I personally prefer to see use of the modulus operator as it shows more arithmetic thinking than calling to the library but both approaches are solid and get the job done.
  • a switch would be more appropriate for this job for i in 0...100 { array.append(.init(i)) switch i { case let x where x.isMultiple(of: 3) && x.isMultiple(of: 5): array[i] = "FizzBuzz" case let x where x.isMultiple(of: 3): array[i] = "Fizz" case let x where x.isMultiple(of: 5): array[i] = "Buzz" default: break } }
  • this answer is not right, as per the FizzBuzz challenge, it should start from 1-100, so the attempt 2 in the question will give the right answer.
  • @prasad Actually the solution does work. I just checked it. Iterating by zero index matches each multiple of three and five and you can remove from the array at the index number cleanly. Attempt 2 does give the correct answer. You should check it on your machine to see for yourself.