为什么不能Swift初始化器调用他们的超类的方便初始化器？ [英] Why can't Swift initializers call convenience initializers on their superclass?

问题描述

考虑两个类：

class A {
    var x: Int

    init(x: Int) {
        self.x = x
    }

    convenience init() {
        self.init(x: 0)
    }
}

class B: A {
    init() {
        super.init() // Error: Must call a designated initializer of the superclass 'A'
    }
}

我不明白为什么'允许。最终，每个类的指定初始化器被调用任何他们需要的值，所以为什么我需要重复自己在 B 的 init x 的默认值，当 A中的方便 init

推荐答案

这是指定的初始化链接规则的规则1在Swift编程指南中，它写为：

This is Rule 1 of the "Initializer Chaining" rules as specified in the Swift Programming Guide, which reads:

规则1：指定初始化函数必须调用指定

Rule 1: Designated initializers must call a designated initializer from their immediate superclass.

https://developer.apple.com/library/prerelease/ios/documentation/swift/conceptual/swift_programming_language /Initialization.html

强调我的。指定的初始化程序不能调用方便初始化程序。

Emphasis mine. Designated initializers cannot call convenience initializers.

有一个图表，随规则一起显示允许初始化程序路线：

There is a diagram that goes along with the rules to demonstrate what initializer "directions" are allowed:

这篇关于为什么不能Swift初始化器调用他们的超类的方便初始化器？的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！