看起来您找到了适合您的问题的答案,但是由于标题要求提供多维数组(我读为 2 或更多),这是我得到的第一个搜索结果搜索时,我将添加我的解决方案:
public static class MultidimensionalArrayExtensions
{
/// <summary>
/// Projects each element of a sequence into a new form by incorporating the element's index.
/// </summary>
/// <typeparam name="T">The type of the elements of the array.</typeparam>
/// <param name="array">A sequence of values to invoke the action on.</param>
/// <param name="action">An action to apply to each source element; the second parameter of the function represents the index of the source element.</param>
public static void ForEach<T>(this Array array, Action<T, int[]> action)
{
var dimensionSizes = Enumerable.Range(0, array.Rank).Select(i => array.GetLength(i)).ToArray();
ArrayForEach(dimensionSizes, action, new int[] { }, array);
}
private static void ArrayForEach<T>(int[] dimensionSizes, Action<T, int[]> action, int[] externalCoordinates, Array masterArray)
{
if (dimensionSizes.Length == 1)
for (int i = 0; i < dimensionSizes[0]; i++)
{
var globalCoordinates = externalCoordinates.Concat(new[] { i }).ToArray();
var value = (T)masterArray.GetValue(globalCoordinates);
action(value, globalCoordinates);
}
else
for (int i = 0; i < dimensionSizes[0]; i++)
ArrayForEach(dimensionSizes.Skip(1).ToArray(), action, externalCoordinates.Concat(new[] { i }).ToArray(), masterArray);
}
public static void PopulateArray<T>(this Array array, Func<int[], T> calculateElement)
{
array.ForEach<T>((element, indexArray) => array.SetValue(calculateElement(indexArray), indexArray));
}
}
使用示例:
var foo = new string[,] { { "a", "b" }, { "c", "d" } };
foo.ForEach<string>((value, coords) => Console.WriteLine("(" + String.Join(", ", coords) + $")={value}"));
// outputs:
// (0, 0)=a
// (0, 1)=b
// (1, 0)=c
// (1, 1)=d
// Gives a 10d array where each element equals the sum of its coordinates:
var bar = new int[4, 4, 4, 5, 6, 5, 4, 4, 4, 5];
bar.PopulateArray(coords => coords.Sum());
一般的想法是通过维度向下递归。我确信这些函数不会赢得效率奖,但它可以作为我的格的一次性初始化程序,并带有一个足够好的 ForEach 来公开值和索引。我还没有解决的主要缺点是让它自动从数组中识别 T,所以在类型安全方面需要小心。