Swift2.2 failable initializer容许提早返回nil以及和Java的不一样

发现swift和java有一个彻底不同的地方

 在swift中, 子类必须先初始化子类的全部属性, 而后才能调用父类的构造器. 而在java中.super调用必须出如今构造函数的第一行.

java代码

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


class NoisyDog extends Dog {
	int age;
	
	NoisyDog(String name) {
        // 交换如下两行的顺序会报错: Constructor call must be the first statement in a constructor
		super(name);
        this.age = 5;
	}
	
}

对应的swift代码:

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


class NoisyDog: Dog {
    var age: Int
    
    override init(name: String) {
        //交换如下两行的顺序会报错error: property 'self.age' not initialized at super.init call
        self.age = 5;
        super.init(name: name);
       
    }
    
}

 

书中关于failable initializer描述有错误

如下代码在swift2.1及以前会编译错误, 在swift2.2中修正了这个bug
swift2.2: 子类failable designated 构造器在返回nil前没必要初始化子类的属性也没必要调用父类的designated initializer, 也就是说, 在子类的failable initilizer中容许提早返回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: 子类failable designated 构造器在返回nil前没必要初始化子类的属性
        // 也没必要调用父类的designated initializer
        if age < 0 {
            return nil
        }
        
        self.age = age;
        super.init(name: name)
    }
    
    
}

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