How to get a NSString substring using an NSRange? Returning NSString, not String-CodePudding

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

How can I do this, but returning an NSString instead of a String. I'm not looking for a solution that uses Range<String.Index>, I'm aware how to do it that way, but for what i'm doing the speed difference is noticeable. I'd like to keep things in the NSString world.

EDIT: Why does a question like this get downvoted? It's a very clear question. I've explained my goal, and given a code example.

Benchmark code:

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

let string = String(repeating: "This is it. ", count: 10000)
let s1 = 19000
let s2 = 19020

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

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

CodePudding user response：

Can't you cast NSString to String and vice versa?

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

CodePudding user response：

These are being done 100 times, Right?

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

How about doing this instead:

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

tc {
    string[start..<end]
}

Which one is more optimized now?