Question

有人可以解释为什么半开放和关闭范围在Swift 3中的字符串上不再相同吗？

此代码有效：

var hello = "hello"
let start = hello.index(hello.startIndex, offsetBy: 1)
let end = hello.index(hello.startIndex, offsetBy: 4)
let range = start..<end   // <-- Half Open Range Operator still works
let ell = hello.substring(with: range)

但这不是：

var hello = "hello"
let start = hello.index(hello.startIndex, offsetBy: 1)
let end = hello.index(hello.startIndex, offsetBy: 4)
let range = start...end   // <-- Closed Range Operator does NOT work
let ello = hello.substring(with: range)   // ERROR

导致如下错误：

Cannot convert value of type 'ClosedRange<String.Index>' (aka 'ClosedRange<String.CharacterView.Index>') to expected argument type 'Range<String.Index>' (aka 'Range<String.CharacterView.Index>')

Answer 1

要执行您要执行的操作，请不要致电substring(with:)。直接下标：

var hello = "hello"
let start = hello.index(hello.startIndex, offsetBy: 1)
let end = hello.index(hello.startIndex, offsetBy: 4)
let ello = hello[start...end] // "ello"

Answer 2

为什么let range = start..<end有效？

如果您想使用NSRange（described here），

substring(with:)就是您所需要的。

以下是Range的{{3}}：

相比类型的半开区间，从下限到，但不包括上限。

创建Range：

您可以使用半开放范围运算符创建Range实例（..≤）。

所以波纹管代码完全正确（它会起作用）：

var hello = "hello"
let start = hello.index(hello.startIndex, offsetBy: 1)
let end = hello.index(hello.startIndex, offsetBy: 4)
let range = start..<end   // <-- Half Open Range Operator still works
let ell = hello.substring(with: range)

为什么let range = start...end不起作用：

如果吼叫，你强迫ClosedRange成为Range：

var hello = "hello"
let start = hello.index(hello.startIndex, offsetBy: 1)
let end = hello.index(hello.startIndex, offsetBy: 4)
let range = start...end   // <-- Closed Range Operator does NOT work
let ello = hello.substring(with: range)   // ERROR

如何将ClosedRange转换为Range？

the definition

Swift Strings上的范围运算符（..＆lt;和...）

2 个答案: