Swift3.0 更新后出现比较运算符方法

在将项目更新到swift3.0之后，在一些controller头部会出现 比较运算符的方法

// FIXME: comparison operators with optionals were removed from the Swift Standard Libary.

// Consider refactoring the code to use the non-optional operators.

fileprivate func < <T : Comparable>(lhs: T?, rhs: T?) -> Bool {

  switch (lhs, rhs) {

  case let (l?, r?):

    return l < r

  case (nil, _?):

    return true

  default:

    return false

  }

}

 

// FIXME: comparison operators with optionals were removed from the Swift Standard Libary.

// Consider refactoring the code to use the non-optional operators.

fileprivate func > <T : Comparable>(lhs: T?, rhs: T?) -> Bool {

  switch (lhs, rhs) {

  case let (l?, r?):

    return l > r

  default:

    return rhs < lhs

  }

}

在swift3.0中移除了这四种运算符。也就是移除了Optional比较运算符

public func < <T : Comparable>(lhs: T?, rhs: T?) -> Bool
public func > <T : Comparable>(lhs: T?, rhs: T?) -> Bool
public func <= <T : Comparable>(lhs: T?, rhs: T?) -> Bool
public func >= <T : Comparable>(lhs: T?, rhs: T?) -> Bool

Swift的进化给出了如下动机说明

let a: Int? = 4
let b: Int = 5
a < b  // b is coerced from "Int" to "Int?" to match the parameter type.

根据上面的代码，如果没有去掉可选类型的比较，b会被强转换为 "Int?"

由此，参照下边的代码可以更好地理解，去掉这个方法的必要，如果没有去掉强转，一些比较就会出现意想不到的结果。

struct Pet {
  let age: Int
}

struct Person {
  let name: String
  let pet: Pet?
}

let peeps = [
  Person(name: "Fred", pet: Pet(age: 5)),
  Person(name: "Jill", pet: .None), // no pet here
  Person(name: "Burt", pet: Pet(age: 10)),
]

let ps = peeps.filter { $0.pet?.age < 6 }

ps == [Fred, Jill] // if you don’t own a pet, your non-existent pet is considered to be younger than any actual pet  🐶

如果你没有宠物，你不应该是查询的结果中的一个，所以Jill不应该在里边。

所以，另一方面，如果强转被移除，为了解决比较的问题，必须对可选值和不可选值的混合体进行明确的处理，这样确保不会出现意外数据。具体代码提现如下

let a: Int? = 4
let b: Int = 5
a < b            // no longer works
a < .some(b)     // works
a < Optional(b)  // works

 

https://github.com/apple/swift-evolution/blob/master/proposals/0121-remove-optional-comparison-operators.md