问题描述
这种比较在 Swift 2 中有效，但在 Swift 3 中不再适用:
This comparison worked in Swift 2 but doesn't anymore in Swift 3:
让 myStringContainsOnlyOneCharacter = mySting.rangeOfComposedCharacterSequence(at: myString.startIndex) == mySting.characters.indices
如何比较 Range 和 DefaultBidirectionalIndices?
How do I compare Range and DefaultBidirectionalIndices?
推荐答案
来自 SE-0065 - 集合和索引的新模型
在 Swift 2 中，collection.indices 返回了一个 Range，但是因为范围是一对简单的索引，并且索引不能再在它们的拥有，Range 不再是可迭代的.
In Swift 2,
collection.indicesreturned aRange<Index>, but because a range is a simple pair of indices and indices can no longer be advanced on their own,Range<Index>is no longer iterable.
为了保持上述代码正常工作，Collection 获取了一个始终可迭代的关联 Indices 类型，...
In order to keep code like the above working, Collection has acquired an associated Indices type that is always iterable, ...
由于 rangeOfComposedCharacterSequence 返回一个 range字符索引，解决方案是不使用indices，但是startIndex..
Since rangeOfComposedCharacterSequence returns a range of
character indices, the solution is not to use indices, but
startIndex..<endIndex:
myString.rangeOfComposedCharacterSequence(at: myString.startIndex)
== myString.startIndex..<myString.endIndex
这篇关于如何比较 Range<String.Index>和 DefaultBidirectionalIndices<String.CharacterView>?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编程学习网！