创建范围< String.Index>从常量Int [英] Creating Range<String.Index> from constant Ints

问题描述

这段代码构造了一个范围，然后应该在调用substringWithRange中使用时，这是怎么回事?

What is wrong with this piece of code for constructing a range that should then serve in a call to substringWithRange?

let range = Range<String.Index>(start: 0, end: 3)

Swift编译器(在Xcode 7.1.1中)使用以下错误消息对其进行标记:

The Swift compiler (in Xcode 7.1.1) marks it with this error message:

无法为类型"Range< Index>"调用初始化程序参数类型为'(start:Int，end:Int)'

Cannot invoke initializer for type 'Range<Index>' with an argument list of type '(start: Int, end: Int)'

推荐答案

您需要引用特定字符串的startIndex，然后前进:

You need to reference the startIndex of a specific string, then advance:

let longString = "Supercalifragilistic"
let startIndex = longString.startIndex
let range = Range(start: startIndex, end: startIndex.advancedBy(3))

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