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

Question

Previously in Swift 2.2 I'm able to do:

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

which extends [Bool] with .allTrue. E.g.

[true, true, false].allTrue == false

But in Swift 3.0 I'm getting this error:

undeclared type _ArrayType

So I tried switching it to Array and using the new keyword Iterator

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

But I got a different error complaining that I'm forcing element to be non-generic

Same-type requirement makes generic parameter 'Element' non-generic

I've also tried the solutions in this 2 years old post but to no avail.

So how does one extend arrays of primitive types like Bool in Swift 3?

related: http://stackoverflow.com/questions/31806324/extend-array-types-using-where-clause-in-swift — Archy Will He 何魏奇, Sep 22 '16 at 01:09

Leo Dabus · Accepted Answer · 2019-03-10T19:33:54.493

Just extend the Collection or the Sequence

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

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

score 8 · Answer 2 · edited Sep 24 '16 at 07:58

8

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) }
}

edited Sep 24 '16 at 07:58

Archy Will He 何魏奇

9,589
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answered Sep 22 '16 at 14:23

Felipe Mulinari Rocha Campos

91
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8

3

This is almost correct, If you extend _ArrayProtocol you would need to constraint it `where Iterator.Element == Bool` – Leo Dabus Sep 23 '16 at 00:50
Thanks for the info. Very interesting. I almost forgot it was open-sourced. – Archy Will He 何魏奇 Sep 24 '16 at 07:57
No problem, I was having a similar issue (in addition to several others) migrating from Swift 2 to 3 and figured looking through the source code couldn't hurt. @LeoDabus any chance you could explain why you need to constrain `_ArrayProtocol` with `where Iterator.Element == Bool` as opposed to just `where Element == Bool`? – Felipe Mulinari Rocha Campos Sep 24 '16 at 15:14
This doesn't work for me. `Use of undeclared type _ArrayProtocol` – Declan McKenna Oct 19 '16 at 12:49
Yes. Extending the Collection didn't work either but extending the Sequence did. – Declan McKenna Oct 20 '16 at 14:31

score 1 · Answer 3 · answered Mar 27 '17 at 22:23

1

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.

answered Mar 27 '17 at 22:23

erikstrottmann

11
3

You could generalize it even further, to an extension on `Sequence` – Alexander Jul 28 '17 at 00:33

score 0 · Answer 4 · answered Oct 20 '16 at 14:33

0

Extending _ArrayProtocol or Collection didn't work for me but Sequence did.

public extension Sequence where Iterator.Element == String
{
    var allTrue: Bool { return !contains(false)
}

answered Oct 20 '16 at 14:33

Declan McKenna

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

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