嵌套数组上的 Swift 相等运算符

Question

为什么最后一条语句无法编译并出现错误:二元运算符'=='不能应用于两个'[[Simple]]'操作数，有没有办法修改Simple 结构或扩展 == 运算符以便能够对嵌套数组(或字典)执行相等检查？

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

Answer 1

更新:条件一致性已在 Swift 4.1 中实现。特别是:

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

(来自 Swift CHANGELOG )。

任意嵌套的 Equatable 数组元素是 Equatable现在可以与 == 进行比较。您的代码

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

在 Xcode 9.3 中编译 if Simple是 Equatable .

<小时/>

(旧答案:)原因与Why is Equatable not defined for optional arrays类似。 。数组可以与 == 进行比较如果元素类型是Equatable :

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

这就是为什么

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

编译。

但即使对于可等同类型 T , Array<T> 不符合Equatable协议(protocol)，比较 Why can't I make Array conform to Equatable? 。因此，在

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和y是元素类型 [Simple] 的数组这确实不符合Equatable协议(protocol)，并且没有匹配==运算符。

您可以定义一个通用的 ==简单嵌套数组的运算符为

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

或者更简单(按照@kennytm的建议):

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

这使得x == y编译并按预期工作。目前看来，有无法定义==任意嵌套数组上的运算符。