Swift: How do i insert item to array with specific order? - arrays

Question

I want to insert item at specific index in Swift. But an error has occurred: Index is out of range
var a = [String]()
a.insert("a", atIndex: 3)
a.insert("b", atIndex: 1)
a.insert("c", atIndex: 0)
a.insert("d", atIndex: 2)
print(a)
If i have to use append or insert sequentially, it will break the order i wanted.

Answer 1

The answer by #fiks does explain the problem and a the correct solution if you want to use Array.
Dictionary
Since you want to perform random insertions, using a Dictionary instead of Array does seem a more flexible choice.
var dict: [Int:String] = [:]
dict[3] = "a"
dict[1] = "b"
dict[0] = "c"
dict[2] = "d"
print(dict) // [2: "d", 3: "a", 1: "b", 0: "c"]
print(dict.sort { $0.0.0 < $0.1.0}.map { $0.1 } ) // ["c", "b", "d", "a"]

Answer 2

You are trying to insert an item in the empty array at index 3. This causes the crash.
You may want to initialise the array with default values:
var a = Array<String>(count: 10, repeatedValue: "")
a.insert("a", atIndex: 3)
print(a)
But then you are not inserting, you are jus modifying the value, so you could achieve the same in this way:
var a = Array<String>(count: 10, repeatedValue: "")
a[3] = "a"
print(a)

Answer 3

The problem with your code is that the second parameter of insert must be less than or equal to the array's endIndex.
When you want to insert an element at index 3 to an array that has zero length, you can't! Because if there is an element at index 3, there must be an element at index 0, 1, 2 as well! That's how arrays work: it's sequential.
I guess you want the elements at index 0, 1 and 2 to be some default value. Then use this extension
extension Array {
mutating func insert(element: Element, atIndex index: Int, withDefaultValue value: Element) {
if self.endIndex < index {
let diff = index - endIndex
for _ in 0...diff {
self.append(value)
}
self[endIndex - 1] = element
} else {
insert(element, atIndex: index)
}
}
}
It's pretty much self-explanatory.
This can also be an XY problem. You might just need a [Int: String] for this. But without further info, I don't know.