Convert Int16 to String Swift 4

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) 

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) }