这并不是说我知道（更换每一个元素，而不是转换到一个新的数组或序列），但它是非常容易写：

Not that I'm aware of (replacing each element rather than converting to a new array or sequence), but it's incredibly easy to write:

public static void ConvertInPlace<T>(this IList<T> source, Func<T, T> projection)
{
    for (int i = 0; i < source.Count; i++)
    {
        source[i] = projection(source[i]);
    }
}

使用：

int[] values = { 1, 2, 3 };
values.ConvertInPlace(x => x * x);

当然，如果你真的不需要的改变现有的阵列，其他的答案使用选择将更多的功能发布。或现有的 ConvertAll 从.NET 2的方法：

Of course if you don't really need to change the existing array, the other answers posted using Select would be more functional. Or the existing ConvertAll method from .NET 2:

int[] values = { 1, 2, 3 };
values = Array.ConvertAll(values, x => x * x);

此是所有假定一维数组。如果您要包括矩形阵列，它变得棘手，特别是如果你想避免拳击。

This is all assuming a single-dimensional array. If you want to include rectangular arrays, it gets trickier, particularly if you want to avoid boxing.