Question

我知道如何在java中做到这一点（参见here），但我找不到java的ByteBuffer的快速等价物，因此找不到它的.putDouble（double value）方法。

基本上，我正在寻找这样的函数：

func doubleToByteArray(value: Double) -> [UInt8]? {
    . . .
}
doubleToByteArray(1729.1729) // should return [64, 155, 4, 177, 12, 178, 149, 234]

Answer 1

typealias Byte = UInt8

func toByteArray<T>(var value: T) -> [Byte] {
    return withUnsafePointer(&value) {
        Array(UnsafeBufferPointer(start: UnsafePointer<Byte>($0), count: sizeof(T)))
    }
}

toByteArray(1729.1729)
toByteArray(1729.1729 as Float)
toByteArray(1729)
toByteArray(-1729)

但结果与您的预期相反（因为有字节顺序）：

[234, 149, 178, 12, 177, 4, 155, 64]
[136, 37, 216, 68]
[193, 6, 0, 0, 0, 0, 0, 0]
[63, 249, 255, 255, 255, 255, 255, 255]

添加了：

func fromByteArray<T>(value: [Byte], _: T.Type) -> T {
    return value.withUnsafeBufferPointer {
        return UnsafePointer<T>($0.baseAddress).memory
    }
}

let a: Double = 1729.1729
let b = toByteArray(a) // -> [234, 149, 178, 12, 177, 4, 155, 64]
let c = fromByteArray(b, Double.self) // -> 1729.1729

对于Xcode8 / Swift3.0：

func toByteArray<T>(_ value: T) -> [UInt8] {
    var value = value
    return withUnsafePointer(to: &value) {
        $0.withMemoryRebound(to: UInt8.self, capacity: MemoryLayout<T>.size) {
            Array(UnsafeBufferPointer(start: $0, count: MemoryLayout<T>.size))
        }
    }
}

func fromByteArray<T>(_ value: [UInt8], _: T.Type) -> T {
    return value.withUnsafeBufferPointer {
        $0.baseAddress!.withMemoryRebound(to: T.self, capacity: 1) {
            $0.pointee
        }
    }
}

对于Xcode8.1 / Swift3.0.1

func toByteArray<T>(_ value: T) -> [UInt8] {
    var value = value
    return withUnsafeBytes(of: &value) { Array($0) }
}

func fromByteArray<T>(_ value: [UInt8], _: T.Type) -> T {
    return value.withUnsafeBytes {
        $0.baseAddress!.load(as: T.self)
    }
}

Answer 2

嗯，这并不容易，但现在是：

func doubleToByteArray(value: Double) -> [UInt8] {
    let count = sizeof(Double)
    var doubles: [Double] = [value]
    let data = NSData(bytes: doubles, length: count)
    var result = [UInt8](count: count, repeatedValue: 0)
    data.getBytes(&result, length: count)
    return result
}

谨慎使用。

Answer 3

这是我对原始解决方案的更新版本。

/// input: array of bytes 
/// -> get pointer to byte array (UnsafeBufferPointer<[Byte]>)
/// -> access its base address
/// -> rebind memory to target type T (UnsafeMutablePointer<T>)
/// -> extract and return the value of target type
func binarytotype <T> (_ value: [Byte], _: T.Type) -> T
{
    return value.withUnsafeBufferPointer {
        $0.baseAddress!
          .withMemoryRebound(to: T.self, capacity: 1) {
            $0.pointee
        }
    }
}

/// input type: value of type T
/// -> get pointer to value of T
/// -> rebind memory to the target type, which is a byte array
/// -> create array with a buffer pointer initialized with the     source pointer
/// -> return the resulted array
func typetobinary <T> (_ value: T) -> [Byte]
{
    var mv : T = value
    let s : Int = MemoryLayout<T>.size
    return withUnsafePointer(to: &mv) {
        $0.withMemoryRebound(to: Byte.self, capacity: s) {
            Array(UnsafeBufferPointer(start: $0, count: s))
        }
    }
}

PS：别忘了用UInt8替换Byte。

Answer 4

swift 3中的解决方案：

public func toByteArray<T>(_ value: T) -> [Byte] {
  let totalBytes = MemoryLayout<T>.size
  var value = value
  return withUnsafePointer(to: &value) { valuePtr in
    return valuePtr.withMemoryRebound(to: Byte.self, capacity: totalBytes) { reboundPtr in
      return Array(UnsafeBufferPointer(start: reboundPtr, count: totalBytes))
    }
  }
}

Answer 5

由于MemoryLayout为您提供静态类型T的大小，因此接受的答案很危险！

要解决此问题，您应该创建自定义协议并在其中询问Self：

protocol ByteConvertible {}

extension ByteConvertible {

    func toBytes() -> [UInt8] {

        let capacity = MemoryLayout<Self>.size
        var mutableValue = self
        return withUnsafePointer(to: &mutableValue) {

            return $0.withMemoryRebound(to: UInt8.self, capacity: capacity) {

                return Array(UnsafeBufferPointer(start: $0, count: capacity))
            }
        }
    }
}

我之前提到，接受的答案很危险，这是一个例子：

let num = UInt8(42)
MemoryLayout.size(ofValue: num) //=> 1 byte as expected
let any: Any = num
MemoryLayout.size(ofValue: any) //=> 32 bytes which is what will happen in the generic functions from the all the answers

Swift 3.0

Answer 6

上面的方法可以使用Swift 2，但是，我发现了一种更简单，更快速的方法来进行这种转换，反之亦然：

func binarytotype <T> (value: [UInt8], _: T.Type) -> T
{
    return value.withUnsafeBufferPointer
    {
        return UnsafePointer<T>($0.baseAddress).memory
    }
}

func typetobinary <T> (var value: T) -> [UInt8]
{
    return withUnsafePointer(&value)
    {
        Array(UnsafeBufferPointer(start: UnsafePointer<UInt8>($0), count: sizeof(T)))
    }
}

let a: Double = 0.25
let b: [UInt8] = typetobinary(a) // -> [0, 0, 0, 0, 0, 0, 208, 63]
let c = binarytotype(b, Double.self) // -> 0.25

我在操场上用Xcode 7.2测试了它。

Answer 7

func byteArray<T>(_ value: T) -> [UInt8] {
    var value = value
    var initialArray = withUnsafeBytes(of: &value) { Array($0) }

    initialArray.reverse()
    var count = initialArray.count
    while initialArray.first == 0 && count > 1 {
        initialArray[0...count - 2] = initialArray[1...count - 1]
        count -= 1
    }
    if initialArray[0] >= 128 {
        var newArray = [UInt8](repeating: 0, count: count + 1)
        newArray[0] = UInt8(0)
        newArray[1...count] = initialArray[0...count - 1]
        return newArray
    } else {
        return Array(initialArray[0...count - 1])
    }
}

如何在Flash中将double转换为字节数组？

7 个答案: