How to initialize Swift array with nil - swift

Question

I am trying to have nil for zeroth index element and rest will have value of Generic type T which is Comparable.
So when I initialise it will nil it works as expected
struct Container<T: Comparable> {
var container = [T?]()
init() {
container.append(nil)
}
}
but when I do it with an integer say 0, I get ambiguous reference
struct Container<T: Comparable> {
var container = [T?]()
init() {
container.append(0)
}
}
Playground execution failed: error: Algorithms.playground:7:9: error: ambiguous reference to member 'append'
container.append(0)
^~~~~~~~~
I want to understand why this error is occurring?

Answer 1

The issue is that T: Comparable does not necessarily mean Int, it means ANY value that conforms to Comparable, which includes String, Float, Bool, and the thousands of custom structs other programmers have written for their own projects.
Since 0 is not a valid value for all of these Structs, you essentially have one of two issues:
1) Do I actually just want the Container to always use Int - in which case:
struct Container
{
var container: [Int?]
init()
{
container = [nil]
}
}
var a = Container()
a.container.append (0)
2) Why am I using 0, when I mean nil
struct Container<T: Comparable>
{
var container: [T?]
init()
{
container = [nil]
}
}
var a = Container<Int>()
a.container.append (0)

Answer 2

T can by any comparable type (String, a custom type, ...),
and initializing it from 0 is not generally
possible.
You could require that T can be created from an integer literal,
this covers all integer and floating point types:
struct Container<T: Comparable> where T: ExpressibleByIntegerLiteral {
var container = [T?]()
init() {
container.append(0)
}
}
Or provide a separate append() method:
struct Container<T: Comparable> {
var container = [T?]()
init() {
container.append(nil)
}
mutating func append(_ newElement: T) {
container.append(newElement)
}
}
var c = Container<Int>()
c.append(0)