可能会带来什么麻烦呢?将reversed()组合到一个快速数组中? [英] What trouble could bring assining reversed() to a swift array?

问题描述

我想知道快速数组上的reversed()方法:

 可变项= ["a"，"b"，"c"]items = items.reversed() 

Apple文档中反向方法的签名说，它返回一个

ReversedRandomAccessCollection< Array< Element>>

可以将其分配回项目，而无需执行苹果医生所说的是什么

例如，要获取数组的反向版本，请从reversed()方法的结果中初始化一个新的Array实例.

还是将来会带来问题?(因为编译器没有抱怨)

解决方案

在Swift 3中，数组的3个reversed()重载.

1. 将数组作为RandomAccessCollection处理，
   

   默认情况下， reversed()选择RandomAccessCollection的重载并返回ReversedRandomAccessCollection.但是，当你写

     items = items.reversed() 

   您正在强迫RHS返回可转换为LHS的类型( [String] ).因此，只会选择返回数组的第三个重载.

   该重载将复制整个序列(因此为O(n))，因此覆盖原始数组没有问题.

   ---

   代替 items = items.reversed()，它创建数组的副本，将其反转并复制回去，您可以使用变异函数

   the signature of the reversed method from the Apple doc says that it returns a

   ReversedRandomAccessCollection<Array<Element>>

   could that be assigned back to items without doing what the apple doc say which is

   For example, to get the reversed version of an array, initialize a new Array instance from the result of this reversed() method.

   or would it give problem in the future? (since the compiler doesn't complain)

   解决方案

   There are 3 overloads of reversed() for an Array in Swift 3:

   1. Treating the Array as a RandomAccessCollection,
      func reversed() -> ReversedRandomAccessCollection<Self> (O(1))
   2. Treating the Array as a BidirectionalCollection,
      func reversed() -> ReversedCollection<Self> (O(1))
   3. Treating the Array as a Sequence,
      func reversed() -> [Self.Iterator.Element] (O(n))

   By default, reversed() pick the RandomAccessCollection's overload and return a ReversedRandomAccessCollection. However, when you write

   items = items.reversed()
   

   you are forcing the RHS to return a type convertible to the LHS ([String]). Thus, only the 3rd overload that returns an array will be chosen.

   That overload will copy the whole sequence (thus O(n)), so there is no problem overwriting the original array.

   ---

   Instead of items = items.reversed(), which creates a copy of the array, reverse that and copy it back, you could reach the same effect using the mutating function items.reverse(), which does the reversion in-place without copying the array twice.

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