Extending typed Arrays (of primitive types like Bool) in Swift 3?

Just extend the Collection or the Sequence

extension Collection where Element == Bool { 
    var allTrue: Bool { return !contains(false) }
}

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edit/update:

Xcode 10 • Swift 4 or later

You can use Swift 4 or later collection method allSatisfy

let alltrue = [true, true,true, true,true, true].allSatisfy{$0}  // true

let allfalse = [false, false,false, false,false, false].allSatisfy{!$0} // true

---

extension Collection where Element == Bool {
    var allTrue: Bool { return allSatisfy{ $0 } }
    var allFalse: Bool { return allSatisfy{ !$0 } }
}

---

Testing Playground:

[true, true, true, true, true, true].allTrue // true
[false, false, false, false, false, false].allFalse // true

As of Swift 3.1 (included in Xcode 8.3), you can now extend a type with a concrete constraint:

extension Array where Element == Bool {
    var allTrue: Bool {
        return !contains(false)
    }
}

You can also extend Collection instead of Array, but you'll need to constrain Iterator.Element, not just Element.

Apple replaced _ArrayType with _ArrayProtocol in Swift 3.0 (see Apple's Swift source code on GitHub) so you can do the same thing you did in Swift 2.2 by doing the following:

extension _ArrayProtocol where Iterator.Element == Bool {
    var allTrue : Bool { return !self.contains(false) }
}