## Why did index go out of range

index out of range
goroutine panic: runtime error: index out of range
golang range
golang index out of range 3 with length 3
go catch index out of range
golang check if index exists
slice index out of range
golang check if index exists in slice

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)

##### 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!

panic: runtime error: index out of range in Go, Is this the type of thing that you are trying to do, since this is what I am understanding from your question? package main import "fmt" func� Another reason this error may come up is when the refresh rate that is selected is higher than what the monitor supports. The monitor can't "sync" with the video card, so it just displays an "Out of Range" error. Also, if the resolution is set at a reasonable amount, make sure that the monitor is plugged in all the way.

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.

Say Goodbye to “Index out of range” — Swift, Low is the index of the array where to start the… Println(sl) // output = panic: runtime error: index out of range [0] with length 0. To resolve an "input signal out of range" error on your computer monitor, follow the steps below. Run Device Manager in Safe Mode If the problem is caused by the display settings or device drivers, boot into Safe Mode and change the settings in Device Manager. Reboot the computer into Safe Mode.

GO: slices explained — Part 1. In this GO (golang) tutorial I will…, Could someone help me out with what is going wrong and what I can do to fix it? edit: I initialised the so_far variable like so: so_far = "-" *� 2nd Monitor out of range I have been using a 2nd monitor for a few months and out of no where my 2nd keeps displaying "out of range 60hz/67.4khz" (while also making my main laptop screen go black) every time i plug in my HDMI converter to VGA.

Python error: string index out of range python, In array/slice element indexing and subslice operations, Go runtime will check whether or not the involved indexes are out of range. If an index is out of� Now we get the “Network Out of Range” message and her book cannot connect YET she can go to school and friends homes it still connects fine to the their wifi networks. I note that I have found other postings that have reported similar issues of connecting wifi to some networks where other networks get the “Network Out Of Range” issue.

• 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"`
• 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 } }`