Swift2.2 failable initializer容許提早返回nil以及和Java的不一樣

發現swift和java有一個徹底不同的地方

 在swift中, 子類必須先初始化子類的全部屬性, 而後才能調用父類的構造器. 而在java中.super調用必須出如今構造函數的第一行.

java代碼

public class Dog {
	String name;
	Dog(String name){
		this.name = name;
	}
}


class NoisyDog extends Dog {
	int age;
	
	NoisyDog(String name) {
        // 交換如下兩行的順序會報錯: Constructor call must be the first statement in a constructor
		super(name);
        this.age = 5;
	}
	
}

對應的swift代碼:

class Dog {
    var name: String;
    init(name: String){
      self.name = name;
    }
}


class NoisyDog: Dog {
    var age: Int
    
    override init(name: String) {
        //交換如下兩行的順序會報錯error: property 'self.age' not initialized at super.init call
        self.age = 5;
        super.init(name: name);
       
    }
    
}

 

書中關於failable initializer描述有錯誤

如下代碼在swift2.1及以前會編譯錯誤, 在swift2.2中修正了這個bug
swift2.2: 子類failable designated 構造器在返回nil前沒必要初始化子類的屬性也沒必要調用父類的designated initializer, 也就是說, 在子類的failable initilizer中容許提早返回nil

//: Playground - noun: a place where people can play
import Foundation

class Dog{
    var name: String
    init(name: String){
        self.name = name
    }
    
}

class NoisyDog : Dog {
    var age: Int
    
    override init(name: String){
        self.age = 5
        super.init(name: name)
    }
    
    init?(name: String, age: Int){
        // as of swift2.2: 子類failable designated 構造器在返回nil前沒必要初始化子類的屬性
        // 也沒必要調用父類的designated initializer
        if age < 0 {
            return nil
        }
        
        self.age = age;
        super.init(name: name)
    }
    
    
}

見:
http://stackoverflow.com/questions/26495586/best-practice-to-implement-a-failable-initializer-in-swift/26497229#26497229