将 UnsafeMutablePointer<Int16> 转换为 UInt8

Question

我正在尝试将此Int16 可变指针转换为UInt8 以写入输出流。我尝试使用函数.withMemoryRebound，但我不知道如何正确使用。我想用这个功能来做，我试过一次但没有成功。我可以使用下面的代码得到一些东西，但我认为它不正确。

unwrappedOutputStream.open()

let buffer: UnsafeMutablePointer<Int16> = avAudioPCMBuffer.int16ChannelData![0]
let size = MemoryLayout<UInt8>.size

let bound: UnsafeMutablePointer<UInt16> = UnsafeMutablePointer.allocate(capacity: 1)
bound.pointee = UInt16(bitPattern: buffer.pointee)

let bytePointer: UnsafeMutablePointer<UInt8> = UnsafeMutablePointer.allocate(capacity: 1)

bytePointer.pointee = UInt8(bound.pointee >> 0x8)
unwrappedOutputStream.write(bytePointer, maxLength: size)

bytePointer.pointee = UInt8(bound.pointee & 0xff)
unwrappedOutputStream.write(bytePointer, maxLength: size)

bound.deallocate(capacity: 1)
bytePointer.deallocate(capacity: 1)

unwrappedOutputStream.close()

我目前正在使用 Swift 4，有什么可以做的吗？ 谢谢你，感谢你的耐心。

Answer 1

将Unsafe(Mutable)Pointer<Int16> 转换为UnsafePointer<Int8> 只是：

let buffer: UnsafeMutablePointer<Int16> = ...
let count: Int = ... // # of Int16 values

let result = buffer.withMemoryRebound(to: UInt8.self, capacity: 2 * count) {
    outputStream.write($0, maxLength: 2 * count)
}