distance(from:to :)'不可用:任何String视图索引转换都可能在Swift 4中失败; 请打开可选索引

Question

我试图将我的应用程序迁移到Swift 4,Xcode 9.我收到此错误.它来自第三方框架.

distance(from:to :)'不可用:任何String视图索引转换都可能在Swift 4中失败; 请打开可选索引

func nsRange(from range: Range<String.Index>) -> NSRange {
    let utf16view = self.utf16
    let from = range.lowerBound.samePosition(in: utf16view)
    let to = range.upperBound.samePosition(in: utf16view)
    return NSMakeRange(utf16view.distance(from: utf16view.startIndex, to: from), // Error: distance(from:to:)' is unavailable: Any String view index conversion can fail in Swift 4; please unwrap the optional indices
                       utf16view.distance(from: from, to: to))// Error: distance(from:to:)' is unavailable: Any String view index conversion can fail in Swift 4; please unwrap the optional indices
}

Answer 1

你可以简单地打开像这样的可选索引:

func nsRange(from range: Range<String.Index>) -> NSRange? {
    let utf16view = self.utf16
    if let from = range.lowerBound.samePosition(in: utf16view), let to = range.upperBound.samePosition(in: utf16view) {
       return NSMakeRange(utf16view.distance(from: utf16view.startIndex, to: from), utf16view.distance(from: from, to: to))
    }
    return nil
}

Answer 2

该错误表明您生成的距离是可选的，需要展开。尝试这个：

func nsRange(from range: Range<String.Index>) -> NSRange {
    let utf16view = self.utf16
    guard let lowerBound = utf16view.distance(from: utf16view.startIndex, to: from), let upperBound = utf16view.distance(from: from, to: to) else { return NSMakeRange(0, 0) }
    return NSMakeRange(lowerBound, upperBound)
}

然而，在声明中可以更好地处理退货guard。我建议设置函数的返回类型NSRange?，并在调用函数的任何地方检查 nil 以避免返回不准确的值。