Question

因此，我有一些来自某些硬件的格式良好的数据流。该流由一堆8位数据组成，其中一些数据块形成32位整数。这一切都很好。数据一直在移动，现在我想把序列包起来。

数据实际上是一个连续字节块，其中的段被映射到有用的数据。因此，例如，第一个字节是确认代码，以下四个字节表示某些特定于应用程序的含义的UInt32，后跟两个字节表示UInt16，依此类推几十个字节。

我找到了两种不同的方法来做到这一点，这两种方式看起来都有点过时了。这可能就是当你接近金属时会发生什么。

但是 - 这两个代码习惯用语通常是人们应该做的吗？或者我错过了更紧凑的东西？

  // data : Data exists before this code, and has what we're transforming into UInt32

  // One Way to get 4  bytes from Data into a UInt32
  var y : [UInt8] = [UInt8](repeating:UInt8(0x0), count: 4)
  data.copyBytes(to: &y, from: Range(uncheckedBounds: (2,6)))
  let u32result = UnsafePointer(y).withMemoryRebound(to: UInt32.self, capacity: 1, {
       $0.pointee
  })
  // u32result contains the 4 bytes from data

  // Another Way to get 4 bytes from Data into a UInt32 via NSData
  var result : UInt32 = 0
  let resultAsNSData : NSData = data.subdata(in: Range(uncheckedBounds: (2,6))) as NSData
  resultAsNSData.getBytes(&result, range: NSRange(location: 0, length: 4))
  // result contains the 4 bytes from data

Answer 1

从格式良好的数据对象创建UInt32数组。

Swift 3

// Create sample data
let data = "foo".data(using: .utf8)!

// Using pointers style constructor
let array = data.withUnsafeBytes {
    [UInt32](UnsafeBufferPointer(start: $0, count: data.count))
}

Swift 2

// Create sample data
let data = "foo".dataUsingEncoding(NSUTF8StringEncoding)!

// Using pointers style constructor
let array = Array(UnsafeBufferPointer(start: UnsafePointer<UInt32>(data.bytes), count: data.length))

Answer 2

我发现另外两种方法可以让我相信有很多方法可以做到这一点，我认为这很好。在Ray Wenderlich

上以某种方式描述了另外两种方式

此代码放入您的Xcode游乐场将揭示这两个其他成语。

do {
let count = 1 // number of UInt32s
let stride = MemoryLayout<UInt32>.stride
let alignment = MemoryLayout<UInt32>.alignment
let byteCount = count * stride

var bytes : [UInt8] = [0x0D, 0x0C, 0x0B, 0x0A] // little-endian LSB -> MSB
var data : Data = Data.init(bytes: bytes) // In my situtation, I actually start with an instance of Data, so the [UInt8] above is a conceit.

print("---------------- 1 ------------------")

let placeholder = UnsafeMutableRawPointer.allocate(bytes: byteCount, alignedTo:alignment)

withUnsafeBytes(of: &data, { (bytes) in
    for (index, byte) in data.enumerated() {
        print("byte[\(index)]->\(String(format: "0x%02x",byte)) data[\(index)]->\(String(format: "0x%02x", data[index])) addr: \(bytes.baseAddress!+index)")

        placeholder.storeBytes(of: byte, toByteOffset: index, as: UInt8.self)

    }
})
let typedPointer1 = placeholder.bindMemory(to: UInt32.self, capacity: count)
print("u32: \(String(format: "0x%08x", typedPointer1.pointee))")



print("---------------- 2 ------------------")

for (index, byte) in bytes.enumerated() {
    placeholder.storeBytes(of: byte, toByteOffset: index, as: UInt8.self)
   // print("byte \(index): \(byte)")
    print("byte[\(index)]->\(String(format: "0x%02x",byte))")

}
let typedPointer = placeholder.bindMemory(to: UInt32.self, capacity: count)
print(typedPointer.pointee)
let result : UInt32 = typedPointer.pointee
print("u32: \(String(format: "0x%08x", typedPointer.pointee))")
}

输出：

---------------- 1 ------------------
byte[0]->0x0d data[0]->0x0d addr: 0x00007fff57243f68
byte[1]->0x0c data[1]->0x0c addr: 0x00007fff57243f69
byte[2]->0x0b data[2]->0x0b addr: 0x00007fff57243f6a
byte[3]->0x0a data[3]->0x0a addr: 0x00007fff57243f6b
u32: 0x0a0b0c0d
---------------- 2 ------------------
byte[0]->0x0d
byte[1]->0x0c
byte[2]->0x0b
byte[3]->0x0a
168496141
u32: 0x0a0b0c0d

这里是Gist。

Answer 3

let a = [ 0x00, 0x00, 0x00, 0x0e ]
let b = a[0] << 24 + a[1] << 16 + a[2] << 8 + a[3]
print(b) // will print 14.

我应该描述这个操作吗？

两种方法将（Swift3）数据4字节输入UInt32

3 个答案:

Swift 3

Swift 2