Swift 2 学习笔记 22.类型检查和类型转换
2018-12-03 本文已影响0人
Maserati丶
课程来自慕课网liuyubobobo老师
类型检查和类型转换
- 类型检查
let library: [MediaItem] = [
Movie(name: "Zootopia", genre: "Animation"),
Music(name: "Hello", artistName: "Adele"),
Game(name: "LIMBO", developer: "Playdead"),
Music(name: "See you agian", artistName: "Wiz Khalifa"),
Game(name: "The Bridge", developer: "The Quantum Astrophysicists Guild")
]
for mediaItem in library{
if mediaItem is Movie{
movieCount += 1
}
else if mediaItem is Music{
musicCount += 1
}
else if mediaItem is Game{
gameCount += 1
}
}
- 向下类型转换
//let item0 = library[0] as? Movie
//let item0 = library[0] as? Music
//let item0 = library[0] as! Movie
//let item0 = library[0] as! Music
for mediaItem in library{
if let movie = mediaItem as? Movie{
print("Movie:", movie.name, "Genre:", movie.genre)
}
else if let music = mediaItem as? Music{
print("Music:", music.name, "Artist:", music.artist)
}
else if let game = mediaItem as? Game{
print("Game:", game.name, "Developer(s):", game.developer)
}
- 检查协议遵守
protocol Shape{
var name: String{get}
}
protocol HasArea{
func area() -> Double
}
struct Point: Shape{
let name: String = "point"
var x: Double
var y: Double
}
struct Rectangle: Shape, HasArea{
let name: String = "rectangle"
var origin: Point
var width: Double
var height: Double
func area() -> Double{
return width * height
}
}
struct Circle: Shape, HasArea{
let name = "circle"
var center: Point
var radius: Double
func area() -> Double{
return Double.pi * radius * radius
}
}
struct Line: Shape{
let name = "line"
var a: Point
var b: Point
}
let shapes:[Shape] = [
Rectangle(origin: Point(x:0.0,y:0.0), width: 3.0, height: 4.0),
Point(x: 0.0, y: 0.0),
Circle(center: Point(x:0.0,y:0.0), radius: 1.0),
Line(a: Point(x:1.0,y:1.0), b: Point(x:5.0,y:5.0))
]
for shape in shapes{
if shape is HasArea{
print("\(shape.name) has area.")
}
else{
print("\(shape.name) has no area.")
}
}
print("==========")
for shape in shapes{
if let areaShape = shape as? HasArea{
print("The area of \(shape.name) is \(areaShape.area()).")
}
else{
print("\(shape.name) has no area.")
}
}