因此,在Swift 3中,我使用此函数将代表Hexa数据的String转换为Data类型:
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)
? 最佳答案
您必须像这样使用:
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
}
}