[Swift5.1] 16-泛型

泛型（Generics）

泛型函数

泛型可以将类型参数化，提高代码复用率，减少代码量

• T代表 不确定类型参数

func swapValues<T>(_ a: inout T, _ b: inout T) {
    (a, b) = (b, a)
}


var i1 = 10  var i2 = 20
swapValues(&i1, &i2)

var d1 = 10.0  var d2 = 20.0
swapValues(&d1, &d2)

struct Date {
    var year = 0, month = 0, day = 0
}
var dd1 = Date(year: 2011, month: 9, day: 10)
var dd2 = Date(year: 2012, month: 10, day: 11)
swapValues(&dd1, &dd2)

泛型函数赋值给变量

func test<T1, T2>(_ t1: T1, _ t2: T2) {}
var fn: (Int, Double) -> () = test

泛型类型

class Stack<E> {
    var elements = [E]()
    func push(_ element: E) { elements.append(element) }
    func pop() -> E { elements.removeLast() }
    func top() -> E { elements.last! }
    func size() -> Int { elements.count }
}

var stack = Stack<Int>()
stack.push(11)
stack.push(22)
stack.push(33)
print(stack.top()) // 33
print(stack.pop()) // 33
print(stack.pop()) // 22
print(stack.pop()) // 11
print(stack.size()) // 0

class SubStack<E> : Stack<E> {}

注意: 修改结构体数组元素, 需要添加mutating

struct Stack<E> {
    var elements = [E]()
    mutating func push(_ element: E) { elements.append(element) }
    mutating func pop() -> E { elements.removeLast() }
    func top() -> E { elements.last! }
    func size() -> Int { elements.count }
}

enum Score<T> {
    case point(T)
    case grade(String)
}
let score0 = Score<Int>.point(100)
let score1 = Score.point(99)
let score2 = Score.point(99.5)
let score3 = Score<Int>.grade("A")

关联类型（Associated Type）

• 关联类型的作用：给协议中用到的类型定义一个占位名称.
• 协议中可以拥有多个关联类型.(协议中使用泛型 只能用关联类型)

protocol Stackable {
    associatedtype Element // 关联类型
    mutating func push(_ element: Element)
    mutating func pop() -> Element
    func top() -> Element
    func size() -> Int
}

• 类遵守协议, 给关联类型设定真实类型

class StringStack : Stackable {
    // 给关联类型设定真实类型, 也可以省略
    // typealias Element = String
    var elements = [String]()
    func push(_ element: String) { elements.append(element) }
    func pop() -> String { elements.removeLast() }
    func top() -> String { elements.last! }
    func size() -> Int { elements.count }
}
var ss = StringStack()
ss.push("Jack")
ss.push("Rose")

• 类中泛型赋值给协议关联类型

class Stack<E> : Stackable {
    // typealias Element = E
    var elements = [E]()
    func push(_ element: E) {
        elements.append(element)
    }
    func pop() -> E { elements.removeLast() }
    func top() -> E { elements.last! }
    func size() -> Int { elements.count }
}

类型约束

下面类型约束: 要求泛型必须是Person子类型和遵守Runnable协议

protocol Runnable { }
class Person { }
func swapValues<T : Person & Runnable>(_ a: inout T, _ b: inout T) {
    (a, b) = (b, a)
}

• 协议中关联类型也可以类型约束

protocol Stackable {
    associatedtype Element: Equatable
}
class Stack<E : Equatable> : Stackable { typealias Element = E }


func equal<S1: Stackable, S2: Stackable>(_ s1: S1, _ s2: S2) -> Bool  where S1.Element == S2.Element, S1.Element : Hashable {
    return false
}

var stack1 = Stack<Int>()
var stack2 = Stack<String>()
// error: requires the types 'Int' and 'String' be equivalent
equal(stack1, stack2)

协议类型的注意点

protocol Runnable {}
class Person : Runnable {}
class Car : Runnable {}

func get(_ type: Int) -> Runnable {
    if type == 0 {
        return Person()
    }
    return Car()
}
var r1 = get(0)
var r2 = get(1)

• 如果协议中有associatedtype, 以下写法会报错.

protocol Runnable {
    associatedtype Speed
    var speed: Speed { get }
}
class Person : Runnable {
    var speed: Double { 0.0 }
}
class Car : Runnable {
    var speed: Int { 0 }
}

func get(_ type: Int) -> Runnable {
    if type == 0 {
        return Person()
    }
    return Car()
}
var r1 = get(0)
var r2 = get(1)

• 因为编译后, 不确定关联类型是什么类型.

1)泛型解决

解决方案①：使用泛型

func get<T : Runnable>(_ type: Int) -> T {
    if type == 0 {
        return Person() as! T
    }
    return Car() as! T
}
var r1: Person = get(0)
var r2: Car = get(1)

2)不透明类型（Opaque Type）

解决方案②：使用some关键字声明一个不透明类型

func get(_ type: Int) -> some Runnable { Car() }
var r1 = get(0)
var r2 = get(1)

some限制只能返回一种类型

some

应用: 想返回遵守某个协议对象, 不希望外部知道对象类型, 外部使用对象时只想暴露协议接口调用, 可以用不透明类型.

• some除了用在返回值类型上，一般还可以用在属性类型上

protocol Runnable { associatedtype Speed }
class Dog : Runnable { typealias Speed = Double }
class Person {
    var pet: some Runnable {
        return Dog()
    }
}

可选项的本质

• 可选项的本质是enum类型

1)可选项在.h中定义

public enum Optional<Wrapped> : ExpressibleByNilLiteral {
    case none
    case some(Wrapped)
    public init(_ some: Wrapped)
}

2)以下可选项赋值等效

var age: Int? = 10
var age0: Optional<Int> = Optional<Int>.some(10)
var age1: Optional = .some(10)
var age2 = Optional.some(10)
var age3 = Optional(10)
age = nil
age3 = .none

var age: Int? = nil
var age0 = Optional<Int>.none
var age1: Optional<Int> = .none

3)可选项混写

var age: Int? = .none
age = 10
age = .some
age = nil

4)可选类型可用于swich

switch age {
case let v?:   //如果可选类型age 有值, 会解包赋值给v
    print("some", v)
case nil:    //如果可选类型age为 nil 
    print("none")
}

switch age {
case let .some(v):
    print("some", v)
case .none:
    print("none")
}

等价于

if let v = age {
    print("some", v)
} else {
    print("none")
}

多重可选项的本质

var age_: Int? = 10
var age: Int?? = age_
age = nil

等价于

var age0 = Optional.some(Optional.some(10))
age0 = .none
var age1: Optional<Optional> = .some(.some(10))
age1 = .none

var age: Int?? = 10
var age0: Optional<Optional> = 10