swift - 如何在 Swift 3 中将 UInt16 转换为 UInt8？

Question

我想将 UInt16 数组转换为 UInt8 数组，但收到以下错误消息:

'init' is unavailable: use 'withMemoryRebound(to:capacity:_)' to temporarily view memory as another layout-compatible type.

代码:

    let statusByte: UInt8 = UInt8(status)
    let lenghtByte: UInt16 = UInt16(passwordBytes.count)

    var bigEndian = lenghtByte.bigEndian

    let bytePtr = withUnsafePointer(to: &bigEndian) {
        UnsafeBufferPointer<UInt8>(start: UnsafePointer($0), count: MemoryLayout.size(ofValue: bigEndian))
    }

Answer 1

如错误消息所示，您必须使用 withMemoryRebound()将指向 UInt16 的指针重新解释为指向 UInt8 的指针:

let bytes = withUnsafePointer(to: &bigEndian) {
    $0.withMemoryRebound(to: UInt8.self, capacity: MemoryLayout.size(ofValue: bigEndian)) {
        Array(UnsafeBufferPointer(start: $0, count: MemoryLayout.size(ofValue: bigEndian)))
    }
}

闭包是用指针 ($0) 调用的，这些指针只有效在闭包的生命周期内，不得传递给外部供以后使用。这就是创建 Array 并将其用作返回值的原因。

但是有一个更简单的解决方案:

let bytes = withUnsafeBytes(of: &bigEndian) { Array($0) }

解释: withUnsafeBytes 调用带有 UnsafeRawBufferPointer 的闭包到 bigEndian 变量的存储。由于 UnsafeRawBufferPointer 是 UInt8 的 Sequence，因此数组可以用 Array($0) 创建。