Convert Int16 to String Swift 4 - swift

Question

I am trying to convert an Int16 value to String but it gives value with Optional, won't allow me to forced unwrap it.
String(describing:intValue)
Result : Optional(intValue)

Answer 1

Unwrap intValue first, then pass it to the string initializer: String(unwrappedIntValue)
Here are some ways of handling the optional. I've added explicit string# variables with type annotations, to make it clear what types are involved
let optionalInt: Int? = 1
// Example 1
// some case: print the value within `optionalInt` as a String
// nil case: "optionalInt was nil"
if let int = optionalInt {
let string1: String = String(int)
print(string1)
}
else {
print("optionalInt was nil")
}
// Example 2, use the nil-coalescing operator (??) to provide a default value
// some case: print the value within `optionalInt` as a String
// nil case: print the default value, 123
let string2: String = String(optionalInt ?? 123)
print(string2)
// Example 3, use Optional.map to convert optionalInt to a String only when there is a value
// some case: print the value within `optionalInt` as a String
// nil case: print `nil`
let string3: String? = optionalInt.map(String.init)
print(string3 as Any)
// Optionally, combine it with the nil-coalescing operator (??) to provide a default string value
// for when the map function encounters nil:
// some case: print the value within `optionalInt` as a String
// nil case: print the default string value "optionalInt was nil"
let string4: String = optionalInt.map(String.init) ?? "optionalInt was nil"
print(string4)

Answer 2

You can convert a number to a String with string interpolation:
let stringValue = "\(intValue)"
Or you can use a standard String initializer:
let stringValue = String(intValue)
If the number is an Optional, just unwrap it first:
let optionalNumber: Int? = 15
if let unwrappedNumber = optionalNumber {
let stringValue = "\(unwrappedNumber)"
}
Or
if let unwrappedNumber = optionalNumber {
let stringValue = String(unwrappedNumber)
}