如何使用 NS 范围获取 NS 字符串子字符串？返回 NS 字符串，而不是字符串

Question

let nsString = NSString("Some string")
let nsRange = NSRange(5...10)
type(of: nsString.substring(with: nsRange))
// => String.Type

如何做到这一点，但返回NSString而不是String。我不是在寻找使用Range的解决方案

基准代码：

let string = String(repeating: "This is it. ", count: 10000)
let i1 = 19000
let i2 = 19020

// time computation
func tc(computation: (Int, Int) -> Void) {
    let startTime = DispatchTime.now()
    for i in 0..<100 {
        computation(i1 + i, i2 + i)
    }
    let endTime = DispatchTime.now()
    let ns = (endTime.uptimeNanoseconds - startTime.uptimeNanoseconds)
    print("Time: \(ns)")
}



tc { (s1, s2) in
  let start = string.index(string.startIndex, offsetBy: s1)
  let end = string.index(start, offsetBy: s2 - s1)
  string[start..<end]
}

tc { (s1, s2) in
  (string as NSString).substring(with: NSRange(s1..<s2))
}

< code >字符串块:< code > 55 _ 394 _ 353 ns

NS 字符串块：1_389_647ns - 快 39.86 倍

Answer 1

这些已经做了100次了，对吗？

let start = string.index(string.startIndex, offsetBy: s1)
let end = string.index(string.startIndex, offsetBy: s2)

如何执行此操作：

let start = string.index(string.startIndex, offsetBy: s1)
let end = string.index(string.startIndex, offsetBy: s2)

tc {
    string[start..<end]
}

现在哪款更优化

Answer 2

难道不能将 NS 字符串转换为字符串，反之亦然吗？

let string = NSString("nsString") as String
let nsString = String("string") as NSString