在Swift String中查找字符的索引

  • Z时代
  • 2024-01-10
  • 分类：问答

现在该承认失败了…

在Objective-C中，我可以使用类似以下内容的东西：

NSString* str = @"abcdefghi";
[str rangeOfString:@"c"].location; // 2

在Swift中，我看到了类似的内容：

var str = "abcdefghi"
str.rangeOfString("c").startIndex

…但是这给了我一个String.Index，可以用来将其下标回到原始字符串，但不能从中提取位置。

FWIW，其中String.Index有一个私有ivar _position，其中具有正确的值。我只是看不到它的暴露方式。

我知道自己可以轻松地将其添加到String中。我对这个新API中缺少的内容感到好奇。

回答：

您不是唯一找不到解决方案的人。

String没有实现RandomAccessIndexType。可能是因为它们启用了具有不同字节长度的字符。这就是为什么我们必须使用string.characters.count（count或countElements在Swift

1.x中）获取字符数的原因。这也适用于职位。的_position可能是一个索引字节的原始阵列，他们不希望公开这一点。本String.Index是为了保护我们从人物的中间访问字节。

这意味着您必须从String.startIndex或String.endIndex（String.Index实现BidirectionalIndexType）创建任何索引。可以使用successor或predecessor方法创建任何其他索引。

现在可以使用索引来帮助我们，它提供了一组方法（Swift 1.x中的函数）：

斯威夫特4.x

let text = "abc"
let index2 = text.index(text.startIndex, offsetBy: 2) //will call succ 2 times
let lastChar: Character = text[index2] //now we can index!
let characterIndex2 = text.index(text.startIndex, offsetBy: 2)
let lastChar2 = text[characterIndex2] //will do the same as above
let range: Range<String.Index> = text.range(of: "b")!
let index: Int = text.distance(from: text.startIndex, to: range.lowerBound)

斯威夫特3.0

let text = "abc"
let index2 = text.index(text.startIndex, offsetBy: 2) //will call succ 2 times
let lastChar: Character = text[index2] //now we can index!
let characterIndex2 = text.characters.index(text.characters.startIndex, offsetBy: 2)
let lastChar2 = text.characters[characterIndex2] //will do the same as above
let range: Range<String.Index> = text.range(of: "b")!
let index: Int = text.distance(from: text.startIndex, to: range.lowerBound)

斯威夫特2.x

let text = "abc"
let index2 = text.startIndex.advancedBy(2) //will call succ 2 times
let lastChar: Character = text[index2] //now we can index!
let lastChar2 = text.characters[index2] //will do the same as above
let range: Range<String.Index> = text.rangeOfString("b")!
let index: Int = text.startIndex.distanceTo(range.startIndex) //will call successor/predecessor several times until the indices match

斯威夫特1.x

let text = "abc"
let index2 = advance(text.startIndex, 2) //will call succ 2 times
let lastChar: Character = text[index2] //now we can index!
let range = text.rangeOfString("b")
let index: Int = distance(text.startIndex, range.startIndex) //will call succ/pred several times

使用String.Index它很麻烦，但是使用包装器按整数索引很危险，因为它掩盖了实际索引的效率低下。

请注意，Swift索引实现存在一个问题，即为 ，例如：

斯威夫特2.x

let text: String = "abc"
let text2: String = "????????????"
let range = text.rangeOfString("b")!
//can randomly return a bad substring or throw an exception
let substring: String = text2[range]
//the correct solution
let intIndex: Int = text.startIndex.distanceTo(range.startIndex)
let startIndex2 = text2.startIndex.advancedBy(intIndex)
let range2 = startIndex2...startIndex2
let substring: String = text2[range2]

斯威夫特1.x

let text: String = "abc"
let text2: String = "????????????"
let range = text.rangeOfString("b")
//can randomly return nil or a bad substring 
let substring: String = text2[range]
//the correct solution
let intIndex: Int = distance(text.startIndex, range.startIndex)    
let startIndex2 = advance(text2.startIndex, intIndex)
let range2 = startIndex2...startIndex2
let substring: String = text2[range2]

以上是 在Swift String中查找字符的索引 的全部内容， 来源链接： utcz.com/qa/423248.html

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