继承(Inheritance)
值类型(枚举、结构体)不支持继承,只有类支持继承
没有父类的类,称为:基类,Swift并没有像OC、Java那样的规定:任何类最终都要继承自某个基类。
子类可以重写父类的下标、方法、属性,重写必须加上override关键字
创建三个类,继承关系: ErHa 继承于 Dog 继承于 Animal
class Animal {var age = 0}class Dog: Animal {var weight = 0}class ErHa: Dog {var iq = 0}
分别创建Animal、Dog、ErHa实例对象,观察内存信息:
let a = Animal()a.age = 10print(Mems.size(ofRef: a)) // 32print(Mems.memStr(ofRef: a))// 一个Animal的实例对象占用32个字节/*0x000000010000c3d0 8个字节:指向类型信息的地址0x0000000000000003 8个字节:引用计数相关0x000000000000000a 8个字节:保存age属性0x0000000105a35830 8个字节:内存对齐*/let d = Dog()d.age = 10d.weight = 20print(Mems.size(ofRef: d)) // 32print(Mems.memStr(ofRef: d))// 一个Dog的实例对象占用32个字节/*0x000000010000c480 8个字节:指向类型信息的地址0x0000000000000003 8个字节:引用计数相关0x000000000000000a 8个字节:保存age属性0x0000000000000014 8个字节:保存weight属性*/let e = ErHa()e.age = 10e.weight = 20e.iq = 30print(Mems.size(ofRef: e)) // 48print(Mems.memStr(ofRef: e))// 一个ErHa的实例对象占用48个字节/*0x000000010000c550 8个字节:指向类型信息的地址0x0000000000000003 8个字节:引用计数相关0x000000000000000a 8个字节:保存age属性0x0000000000000014 8个字节:保存weight属性0x000000000000001e 8个字节:保存iq属性0x0003000000000078 8个字节:内存对齐*/
重写实例方法、下标
class Animal {func speak() {print("Animal speak")}subscript(index: Int) -> Int {return index}}var a = Animal()a.speak()print(a[6])class Dog: Animal {/// 重写实例方法override func speak() {super.speak()print("Dog speak")}/// 重写下标override subscript(index: Int) -> Int {return super[index] + 1}}var d = Dog()d.speak()print(d[6])
打印结果:
Animal speak6Animal speakDog speak7
重写类型方法、下标
被class修饰的类型方法、下标,允许被子类重写
class Animal {class func speak() {print("Animal speak")}class subscript(index: Int) -> Int {return index}}Animal.speak()print(Animal[6])class Dog: Animal {override class func speak() {super.speak()print("Dog speak")}override class subscript(index: Int) -> Int {return super[index] + 1}}Dog.speak()print(Dog[6])
打印结果:
Animal speak6Animal speakDog speak7
重写属性
1、子类可以将父类的属性(存储、计算)重写为计算属性
2、子类不可以将父类属性重写为存储属性
3、只能重写var属性,不能重写let属性
4、重写时,属性名、类型要一致
5、子类重写后的属性权限,不能小于父类的属性权限
- 如果如类属性是只读的,那么子类重写后的属性可以是只读的、也可以是可读写的
如果父类属性是可读写的,那么子类重写后的属性也必须是可读写的
重写实例属性
```swift class Point { /// 存储属性 var x: Int = 0
/// 计算属性 var y: Int {
set {print("Point set y")x = newValue / 2}get {print("Point get y")return x * 2}
} }
class SubPoint: Point {
/// 重写x为计算属性override var x: Int {set {print("SubPoint set x")super.x = newValue}get {print("SubPoint get x")return super.x}}/// 重写y为计算属性override var y: Int {set {print("SubPoint set y")super.y = newValue}get {print("SubPoint get y")return super.y}}
}
let sp = SubPoint() print(sp.y)
打印结果:
SubPoint get y Point get y SubPoint get x 0
分析:<br />sp.y调用的是子类y的get方法,所以打印SubPoint get y<br />子类y的get方法中调用了super.y,所以打印了Point get y<br />父类y的get方法调用了x * 2,但是当前调用对象是SubPoint,会走到子类x的get方法,所以打印了SubPoint get x<br />从父类继承过来的存储属性,都会分配存储空间,无论是否被重写为计算属性。<a name="UoYV3"></a>## 重写类型属性被class修饰的计算类型属性,可以被子类重写<br />被static修饰的类型属性(存储、计算),不可以被子类重写<br /><a name="o3pJj"></a># 属性观察器可以在子类中为父类属性(除了只读计算属性、let属性)增加属性观察器```swiftclass Point {var x: Int = 0}class SubPoint: Point {override var x: Int {willSet {print("SubPoint willSet x", newValue)}didSet {print("SubPoint didSet x", x, oldValue)}}}var p = SubPoint()p.x = 10
打印结果:
SubPoint willSet x 10SubPoint didSet x 10 0
如果父类有自己的属性观察器:
class Point {var x: Int = 0 {willSet {print("Point willSet x", newValue)}didSet {print("Point didSet x", x, oldValue)}}}class SubPoint: Point {override var x: Int {willSet {print("SubPoint willSet x", newValue)}didSet {print("SubPoint didSet x", x, oldValue)}}}var p = SubPoint()p.x = 10
打印结果:
SubPoint willSet x 10Point willSet x 10Point didSet x 10 0SubPoint didSet x 10 0
如果父类有一个计算属性
class Point {var x: Int {set {print("Point set x", newValue)}get {print("Point get x")return 0}}}class SubPoint: Point {override var x: Int {willSet {print("SubPoint willSet x", newValue)}didSet {print("SubPoint didSet x", x, oldValue)}}}var p = SubPoint()p.x = 10
打印结果:
Point get xSubPoint willSet x 10Point set x 10Point get xSubPoint didSet x 0 0
final
被final修饰的方法、下标、属性,禁止被重写
被final修饰的类,禁止被继承
多态
结构体
struct Animal {func speak() {print("Animal speak")}func eat() {print("Animal eat")}func sleep() {print("Animal sleep")}}var animal = Animal()animal.speak()animal.eat()animal.sleep()
定一个结构体,执行三个方法,观察汇编代码
调用方法时,直接调用了函数地址,原因是struct不存在继承,所以编译完函数的地址就是确定的。
类
class Animal {func speak() {print("Animal speak")}func eat() {print("Animal eat")}func sleep() {print("Animal sleep")}}var animal = Animal()animal.speak()animal.eat()animal.sleep()
观察汇编代码:
函数的地址是不确定的,是通过计算得出的,在编译时是不确定的。
实现原理探究
class Animal {func speak() {print("Animal speak")}func eat() {print("Animal eat")}func sleep() {print("Animal sleep")}}class Dog: Animal {override func speak() {print("Dog speak")}override func eat() {print("Dog eat")}func run() {print("Dog run")}}var animal = Animal()animal.speak()animal.eat()animal.sleep()// 父类类型指针,指向子类类型animal = Dog()animal.speak()animal.eat()animal.sleep()
打印结果:
Animal speakAnimal eatAnimal sleepDog speakDog eatAnimal sleep
以dog变量调用sleep方法为例,流程是通过dog指针找到堆空间的dog对象,找到dog对象中的前8个字节
这8个字节是一个指针,指向一块内存区域,这块内存前面保存着dog的类型信息,后面存放了Dog和父类Animal的方法信息。
通过打印这块内存的地址,可以发现是保存在内存的全局区。
