在Swift String中查找字符的索引

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.countcountcountElements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.startIndexString.endIndexString.Index实现BidirectionalIndexType)创建任何索引。可以使用successorpredecessor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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