Question

所以在Swift 3中，我有这个函数将表示Hexa数据的String转换为Data type：

extension String {

public func fromHexStringtoData() -> Data? {
    func convertToUInt8(u: UInt16) -> UInt8? {
        switch(u) {
        case 0x30 ... 0x39:
            return UInt8(u - 0x30)
        case 0x41 ... 0x46:
            return UInt8(u - 0x41 + 10)
        case 0x61 ... 0x66:
            return UInt8(u - 0x61 + 10)
        default:
            return nil
        }
    }

    let utf16 = self.utf16
    guard let data = NSMutableData(capacity: utf16.count/2) else { return nil }
    var i = utf16.startIndex
    while i != utf16.endIndex {
        guard let hi = convertToUInt8(u: utf16[i]) else { return nil }

        //Need to convert following line to Swift 4
        guard let lo = convertToUInt8(u: utf16[i.advanced(by: 1)]) else { return nil }

        var value = hi << 4 + lo
        data.append(&value, length: 1)

        //Need to convert following line to Swift 4
        i = i.advanced(by: 2)
    }
    return data as Data
}

}

如何以最佳方式转换advanced(by: n)？

Answer 1

你必须这样使用：

extension String {
    public func fromHexStringtoData() -> Data? {
        func convertToUInt8(u: UInt16) -> UInt8? {
            switch(u) {
            case 0x30 ... 0x39:
                return UInt8(u - 0x30)
            case 0x41 ... 0x46:
                return UInt8(u - 0x41 + 10)
            case 0x61 ... 0x66:
                return UInt8(u - 0x61 + 10)
            default:
                return nil
            }
        }

        let utf16 = self.utf16
        guard let data = NSMutableData(capacity: utf16.count/2) else { return nil }
        var i = utf16.startIndex
        while i != utf16.endIndex {
            guard let hi = convertToUInt8(u: utf16[i]) else { return nil }

            //changed to Swift 4
            guard let lo = convertToUInt8(u: utf16[utf16.index(i, offsetBy: 1)]) else { return nil }

            var value = hi << 4 + lo
            data.append(&value, length: 1)

            //changed to Swift 4
            i = utf16.index(i, offsetBy: 2)
        }
        return data as Data
    }
}

Answer 2

这不是您问题的直接答案，但您的代码会从头到尾扫描输入字符串。在这种情况下，您可以在不使用索引的情况下重写代码：

extension String {
    public func fromHexStringToData() -> Data? {
        func convertToUInt8(u: UInt16) -> UInt8? {
            //...
        }

        var utf16Iterator = utf16.makeIterator()
        var data = Data(capacity: utf16.count/2)
        while let hiChar = utf16Iterator.next() {
            guard
                let hi = convertToUInt8(u: hiChar),
                let loChar = utf16Iterator.next(),
                let lo = convertToUInt8(u: loChar)
                else { return nil }

            let value = hi << 4 + lo
            data.append(value)
        }
        return data
    }
}

适用于Swift 3和Swift 4。

在Swift 4中按功能转换UTF16View.Index

2 个答案: