c# - 从 int[] 中选择每 2 个 int 并使用 Lamda 表达式将它们组合成一个 uint64[] 数组

Question

我使用的第三方 API 返回一个类型为 int[] 的数组。我们将其称为 readWords:

int[] readWords

然而，这个重新调整的数组应该是一个 uint[]。所以我正在进行以下转换:

Uint Ulongs = readWords(item => (ulong)item).ToArray();

现在的问题是，在某些情况下，我需要将每 2 个接收到的 int 字组合成一个 Uint64[]。
使用 lambda 表达式执行此操作的优雅方式是什么？

我正在考虑使用类似的东西，但我不确定如何:

Uint64[] longs = readWords
                 .Select(every 2 int into 1 Uint64)
                 .GroupBy(x => ...);

请注意，每 2 个 int 必须按照以下方式组合成一个 Uint64:

Uint64 word0 = (readWords[0] & 0xFFFFFF) | ((Uint64)(readWords[1] & 0xFFFFFF) << 24);

例如，如果我们有:

int[] = new int[]{0x00123456, 0x00456789}

生成的组合 64 位字必须是:

0x0000456789123456

Answer 1

两个项目的每个组合都是笛卡尔连接(readWords 与自身):

int[] readWords = ...

var result = readWords
  .SelectMany(left => readWords
     .Select(right => /*TODO: put left and right combination here*/))
  .ToArray();

为了从 2 int 创建 ulong，让我们使用您提供的位操作:

(lelf, right) => unchecked(((ulong)left & 0xFFFFFF) | ((ulong)(right & 0xFFFFFF) << 24)))

最后

int[] readWords = ...

var result = readWords
  .SelectMany(left => readWords
     .Select(right => unchecked(((ulong)left & 0xFFFFFF) | 
                                ((ulong)(right & 0xFFFFFF) << 24))))
  .ToArray(); // materialization, if required

让我们看看:

  int[] readWords = new int[] { 1, 2, 3 };

  var result = readWords
    .SelectMany(left => readWords
      .Select(right => unchecked(((ulong)left & 0xFFFFFF) | 
                                 ((ulong)(right & 0xFFFFFF) << 24))))
    .ToArray();

  string report = string.Join(Environment.NewLine, 
    result.Select(item => item.ToString("x16")));

  Console.WriteLine(report); 

结果:

0000000001000001
0000000002000001
0000000003000001
0000000001000002
0000000002000002
0000000003000002
0000000001000003
0000000002000003
0000000003000003

编辑:同样的想法(Cartesian Join)如果你想组合，比如说，每个偶数项与每个奇数 一个:

  int[] readWords = new int[] { 0x00123456, 0x00456789 };

  var result = readWords
    .Where((value, index) => index % 2 == 0)    // even indexes only
    .SelectMany(left => readWords
      .Where((value, index) => index % 2 != 0)  // odd indexes only
      .Select(right => unchecked(((ulong)left & 0xFFFFFF) | 
                                 ((ulong)(right & 0xFFFFFF) << 24))))
    .ToArray();

  string report = string.Join(Environment.NewLine, 
    result.Select(item => item.ToString("x16")));

  Console.Write(report);

结果:

  0000456789123456 

编辑 2: 但是，如果您想合并，这里没有 Cartesian Join

 item_0 and item_1,
 item_2 and item_3,
 item_4 and item_5,
...
item_2N and item_2N+1,
...

(请参阅下面的评论；请注意，我们不会合并每个 项目)只是一个简单的选择:

var result = Enumerable
  .Range(0, readWords.Length / 2)
  .Select(index => new {
     left = readWords[2 * index],
     right = readWords[2 * index + 1]
   })
  .Select(pair => unchecked(((ulong)(pair.left) & 0xFFFFFF) | 
                            ((ulong)(pair.right & 0xFFFFFF) << 24)))
  .ToArray();