README.md

Tips and tricks for iOS developers

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• Clearly explain the given problem & solution
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Optionals

Use Optional.map and Optional.flatMap to avoid unnecessary unwrapping

This:

let foo: Int? = 6
let bar: Double?

if let foo = foo {
    bar = Double(foo)
} else {
    bar = nil
}

becomes this:

let foo: Int? = 6
let bar = foo.map(Double.init)

This:

let foo: String? = "4"
let bar: Int?

if let foo = foo {
    bar = Int(foo)
} else {
    bar = nil
}

becomes this:

let foo: String? = "4"
let bar = foo.flatMap(Int.init)

The difference between Optional.map and Optional.flatMap lies in their return types. These are their respective function signatures: public func map<U>(_ transform: (Wrapped) throws -> U) rethrows -> U? public func flatMap<U>(_ transform: (Wrapped) throws -> U?) rethrows -> U?

Using Optional.map on a function which returns an Optional (e.g. Double.init(_ text: StringProtocol)) will make it a double Optional. For example:

The type of bar here will be Double??

let foo: String? = "4"
let bar = foo.map(Double.init)

The type of bar here will be Double?

let foo: String? = "4"
let bar = foo.flatMap(Double.init)

Use Collection.compactMap to transform a collection of Optional<T> to a collection of T

// foo has a type of Array<Optional<Int>>, i.e. [Int?]
let foo = [5, 4, nil, 2, nil, nil]

// this works, but it's inefficient — O(n^2) — not to mention ugly
// bar has a type of [Int]
let bar = foo.filter { $0 != nil }.map { $0! }

// clean, efficient
// baz has a type of [Int]
let baz = foo.compactMap { $0 }