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
最佳答案
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();
关于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,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/58215783/