对于某些应用可能更好:

string.Join(",", arr1) == string.Join(",", arr2)

对于单元测试，可以使用CollectionAssert。AreEqual而不是assert。AreEqual。

这可能是最简单的方法。

可列举的使用。LINQ中的SequenceEqual。

int[] arr1 = new int[] { 1,2,3};
int[] arr2 = new int[] { 3,2,1 };

Console.WriteLine(arr1.SequenceEqual(arr2)); // false
Console.WriteLine(arr1.Reverse().SequenceEqual(arr2)); // true

Elementwise比较?是什么

public void Linq78a()
{
 int[] numbers1 = { 5, 4, 1, 3, 9, 8, 6, 7, 2, 0 };
 int[] numbers = { 5, 4, 1, 3, 9, 8, 6, 7, 2, 0 };
 bool bb = numbers.Zip(numbers1, (a, b) => (a == b)).Any(p => !p);
 if (!bb) Console.WriteLine("Lists are equal (bb)");
   else Console.WriteLine("Lists are not equal (bb)");
}

将(a==b)条件替换为您想在a和b中进行比较的任何内容。

(这结合了来自MSDN开发者Linq的两个例子)

如果您不想比较顺序，但确实想比较每一项的计数，包括处理空值，那么我为此编写了一个扩展方法。

例如，它给出了以下结果:

new int?[]{  }.IgnoreOrderComparison(new int?{ });                            // true
new int?[]{ 1 }.IgnoreOrderComparison(new int?{ });                           // false
new int?[]{ }.IgnoreOrderComparison(new int?{ 1 });                           // false
new int?[]{ 1 }.IgnoreOrderComparison(new int?{ 1 });                         // true
new int?[]{ 1, 2 }.IgnoreOrderComparison(new int?{ 2, 1 });                   // true
new int?[]{ 1, 2, null }.IgnoreOrderComparison(new int?{ 2, 1 });             // false
new int?[]{ 1, 2, null }.IgnoreOrderComparison(new int?{ null, 2, 1 });       // true
new int?[]{ 1, 2, null, null }.IgnoreOrderComparison(new int?{ null, 2, 1 }); // false
new int?[]{ 2 }.IgnoreOrderComparison(new int?{ 2, 2 });                      // false
new int?[]{ 2, 2 }.IgnoreOrderComparison(new int?{ 2, 2 });                   // true

代码如下:

public static class ArrayComparisonExtensions
{
    public static bool IgnoreOrderComparison<TSource>(this IEnumerable<TSource> first, IEnumerable<TSource> second) =>
        IgnoreOrderComparison(first, second, EqualityComparer<TSource>.Default);

    public static bool IgnoreOrderComparison<TSource>(this IEnumerable<TSource> first, IEnumerable<TSource> second, IEqualityComparer<TSource> comparer)
    {
        var a = ToDictionary(first, out var firstNullCount);
        var b = ToDictionary(second, out var secondNullCount);

        if (a.Count != b.Count)
            return false;

        if (firstNullCount != secondNullCount)
            return false;

        foreach (var item in a)
        {
            if (b.TryGetValue(item.Key, out var count) && item.Value == count)
                continue;
            return false;
        }


        return true;

        Dictionary<TSource, int> ToDictionary(IEnumerable<TSource> items, out int nullCount)
        {
            nullCount = 0;
            var result = new Dictionary<TSource, int>(comparer);
            foreach (var item in items)
            {
                if (item is null)
                    nullCount++;
                else if (result.TryGetValue(item, out var count))
                    result[item] = count + 1;
                else
                    result[item] = 1;
            }

            return result;
        }
    }
}

它只枚举每个枚举对象一次，但它确实为每个枚举对象创建了一个字典，并迭代这些字典一次。我对如何改善这一点很感兴趣。

我想确定两个集合是否有相同的内容，以任何顺序。这意味着，对于集合A中的每个元素，在两个集合中都有相同数量的具有该值的元素。我想要解释重复(所以{1,2,2,3}和{1,2,3,3}不应该被认为是“相同的”)。

这是我想出的(注意，IsNullOrEmpty是另一个静态扩展方法，如果枚举对象为空或有0个元素，则返回true):

    public static bool HasSameContentsAs<T>(this IEnumerable<T> source, IEnumerable<T> target) 
        where T : IComparable
    {
        //If our source is null or empty, then it's just a matter of whether or not the target is too
        if (source.IsNullOrEmpty())
            return target.IsNullOrEmpty();

        //Otherwise, if the target is null/emtpy, they can't be equal
        if (target.IsNullOrEmpty())
            return false;

        //Neither is null or empty, so we'll compare contents.  To account for multiples of 
        //a given value (ex. 1,2,2,3 and 1,1,2,3 are not equal) we'll group the first set
        foreach (var group in source.GroupBy(s => s))
        {
            //If there are a different number of elements in the target set, they don't match
            if (target.Count(t => t.Equals(group.Key)) != group.Count())
                return false;
        }

        //If we got this far, they have the same contents
        return true;
    }