为什么 String.subscript(_:) 在不涉及 `Int` 的情况下要求 `String.Index` 和 `Int` 类型相等? [英] Why does String.subscript(_:) require the types `String.Index` and `Int` to be equal when there is no `Int` involved?

问题描述

我无法理解 Xcode 在这一行中遇到的问题:

I fail to understand the problem Xcode is confronting me with in this line:

iteration.template = template[iterationSubstring.endIndex...substring.startIndex]

template 是 String 和 iterationSubstring 和 substring 是 Substring<代码>模板.Xcode 使用以下消息突​​出显示左方括号:

template is a String and iterationSubstring and substring are Substrings of template. Xcode highlights the opening square bracket with the following message:

下标 'subscript(_:)' 要求类型 'Substring.Index' 和 'Int' 是等价的

Subscript 'subscript(_:)' requires the types 'Substring.Index' and 'Int' be equivalent

错误信息对我来说没有任何意义.我尝试通过使用 [template.startIndex...template.endIndex] 下标创建 Range 来获取 Substring.这与 Int 有什么关系?为什么同样的模式在其他地方也能用?

The error message does not make any sense to me. I try to obtain a Substring by creating a Range<String.Index> with the [template.startIndex...template.endIndex] subscript. How is this related to Int? And why does the same pattern work elsewhere?

重现问题的Xcode Playground 代码:

Xcode playground code reproducing the problem:

import Foundation
let template = "This is an ordinary string literal."

let firstSubstringStart = template.index(template.startIndex, offsetBy: 5)
let firstSubstringEnd = template.index(template.startIndex, offsetBy: 7)
let firstSubstring = template[firstSubstringStart...firstSubstringEnd]

let secondSubstringStart = template.index(template.startIndex, offsetBy: 10)
let secondSubstringEnd = template.index(template.startIndex, offsetBy: 12)
let secondSubstring = template[secondSubstringStart...secondSubstringEnd]

let part: String = template[firstSubstring.endIndex...secondSubstring.startIndex]

毕竟我有一个模板字符串和它的两个子字符串.我想得到一个 String，范围从第一个 Substring 的结尾到第二个 Substring 的开头.

After all I have a template string and two substrings of it. I want to get a String ranging from the end of the first Substring to the start of the second Substring.

推荐答案

当前版本的 Swift 与 Substring 结构体，它是一个切片的 String.

The current version of Swift works with the Substring struct which is a sliced String.

该错误似乎具有误导性，如果您要将(范围下标的)Substring 分配给 String 变量，则会发生该错误.

The error seems to be misleading and occurs if you are going to assign a (range-subscripted) Substring to a String variable.

要修复错误，请从 Substring

iteration.template = String(template[iterationSubstring.endIndex...substring.startIndex])

尽管如此，强烈建议您不要使用来自不同字符串(iterationSubstring 和 substring)的索引创建范围.对主字符串进行切片，保留索引.

Nevertheless you are strongly discouraged from creating ranges with indices from different strings (iterationSubstring and substring). Slice the main string, the indices are preserved.

第二个(同时删除)示例中的崩溃发生是因为字符串的最后一个字符在endIndex之前的索引处，它是

The crash in the second (meanwhile deleted) example occurred because the last character of a string is at index before endIndex, it's

template[template.startIndex..<template.endIndex] 

或更短

template[template.startIndex...]

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