Swift equality operator on nested arrays

Question

Why does the last statement fail to compile with the error: Binary operator '==' cannot be applied to two '[[Simple]]’ operands, and is there a way way to modify the Simple struct or extend the == operator to be able to perform equality checks on nested arrays (or dictionaries)?

var i1: [Int] = [1]
var i2: [Int] = [1]
i1 == i2 // -> true


var i3: [[Int]] = [[1], [2]]
var i4: [[Int]] = [[1], [2]]
i3 == i4 // -> true


struct Simple: Equatable, Hashable {
    let message: String

    var hashValue: Int {
        return message.hashValue
    }
}
func ==(lhs: Simple, rhs: Simple) -> Bool {
    return lhs.message == rhs.message
}

var a: [Simple] = [Simple(message: "a")]
var b: [Simple] = [Simple(message: "a")]
a == b // -> true

var x: [[Simple]] = [[Simple(message: "a")], [Simple(message: "b")]]
var y: [[Simple]] = [[Simple(message: "a")], [Simple(message: "b")]]
x == y // -> ERROR! Binary operator '==' cannot be applied to two '[[Simple]]’ operands

Your problem has been solved in Swift 4.1.

Martin R
– Martin R

2018-04-27 17:49:41 +00:00
Commented Apr 27, 2018 at 17:49 — Martin R
– Martin R, Commented Apr 27, 2018 at 17:49

Martin R · Accepted Answer · 2018-04-27 17:49:26Z

15

Update: Conditional conformance has been implemented in Swift 4.1. In particular:

The standard library types Optional, Array, and Dictionary now conform to the Equatable protocol when their element types conform to Equatable. ...

(from the Swift CHANGELOG).

Arbitrarily nested arrays of Equatable elements are Equatable now and can be compared with ==. Your code

var x: [[Simple]] = [[Simple(message: "a")], [Simple(message: "b")]]
var y: [[Simple]] = [[Simple(message: "a")], [Simple(message: "b")]]
x == y

compiles in Xcode 9.3 if Simple is Equatable.

(Old answer:) The reason is similar as in Why is Equatable not defined for optional arrays. Arrays can be compared with == if the element type is Equatable:

/// Returns true if these arrays contain the same elements.
public func ==<Element : Equatable>(lhs: [Element], rhs: [Element]) -> Bool

That's why

var a: [Simple] = [Simple(message: "a")]
var b: [Simple] = [Simple(message: "a")]
a == b // -> true

compiles.

But even for equatable types T, Array<T> does not conform to the Equatable protocol, compare Why can't I make Array conform to Equatable?. Therefore, in

var x: [[Simple]] = [[Simple(message: "a")], [Simple(message: "b")]]
var y: [[Simple]] = [[Simple(message: "a")], [Simple(message: "b")]]
x == y // -> ERROR! Binary operator '==' cannot be applied to two '[[Simple]]’ operands

x and y are arrays with the element type [Simple] which does not conform to the Equatable protocol, and there is no matching == operator.

You could define a generic == operator for simply nested arrays as

func ==<Element : Equatable> (lhs: [[Element]], rhs: [[Element]]) -> Bool {
    return lhs.count == rhs.count && !zip(lhs, rhs).contains {$0 != $1 }
}

or more simply (as suggested by @kennytm):

func ==<Element : Equatable> (lhs: [[Element]], rhs: [[Element]]) -> Bool {
    return lhs.elementsEqual(rhs, by: ==)
}

This makes x == y compile and work as expected. At present, there seems to be no way to define a == operator on arbitrarily nested arrays.

edited Apr 27, 2018 at 17:49

answered Oct 27, 2015 at 20:45

Martin R

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3 Comments

kennytm Over a year ago

Perhaps update the implementation to lhs.elementsEqual(rhs, by: ==)?

Petro Korienev Over a year ago

Suggestion: since github.com/apple/swift-evolution/blob/master/proposals/… was not included in Swift 4, update the answer

Martin R Over a year ago

@PetroKorienev: Done! – SE-0143 has not been implemented yet, but the special case for arrays of equatable elements has, and that is actually sufficient for this problem.

Adam · Accepted Answer · 2015-10-27 20:33:35Z

1

You can do it by implementing a == function for it, like following:

func == (lhs: [[Simple]], rhs: [[Simple]]) -> Bool {
    //your code
}

answered Oct 27, 2015 at 20:33

Adam

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4 Comments

Jeff Over a year ago

Sorry for the down vote, but this question is way more complicated than what your answer implies.

Adam Over a year ago

@Jeff You are entitled to your opinion but I always prefer simple solutions over complicated ones. YAGNI applies here.

Jeff Over a year ago

My comment was not about preferring an unnecessarily complex answer. Your answer is not a solution to this question. Your answer is incomplete. The question is more complex than you seem to comprehend.

Adam Over a year ago

My answer is a solution and is complete. It is the simplest way of solving the problem that was described. You seem to make things more complicated than they really are.

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