Swift: shortcut unwrapping of array of optionals

Since it is an array of optionals, it is possible some of the entries are nil. Instead of force unwrapping with !, use the nil coalescing operator to turns nils into empty strings.

let arrayOfOptionals: [String?] = ["This", "array", nil, "has", "some", "nils", nil]

let array:[String] = arrayOfOptionals.map{ $0 ?? "" }
// array is now ["This", "array", "", "has", "some", "nils", ""]

This solution will get you a new array with all values unwrapped and all nil's filtered away.

Swift 4.1:

let arrayOfOptionals: [String?] = ["Seems", "like", "an", nil, "of", "optionals"]
let arrayWithNoOptionals = arrayOfOptionals.compactMap { $0 }

Swift 2.0:

let arrayOfOptionals: [String?] = ["Seems", "like", "an", nil, "of", "optionals"]
let arrayWithNoOptionals = arrayOfOptionals.flatMap { $0 }

Swift 1.0:

let arrayOfOptionals: [String?] = ["Seems", "like", "an", nil, "of", "optionals"]
let arrayWithNoOptionals = arrayOfOptionals.filter { $0 != nil }.map { $0! }