Swift: shortcut unwrapping of array of optionals

Question

Assume we have an array of optionals defined:

var arrayOfOptionals: [String?] = [\"Seems\", \"like\", \"an\", nil, \"of\", \"optionals\"]

I

Answer 1

Although you can use flatMap { $0 } to remove nils, flatMap is actually a much more powerful function, and has an overloaded version which does something completely different (e.g. flatten [[Int]] to [Int]). If you're not careful, you may accidentally invoke the wrong function.

I would recommend using an extension on SequenceType to remove nils. If you use removeNils(), you'll be able to essentially do the following:

[1, nil, 2].removeNils() == [1, 2]

It works by making Optional conform to an OptionalType protocol which allows extending SequenceTypes that contain Optional values.

For more information see the original answer I posted.