2

给定两个不包含引用循环的对象，您是否知道一种以“通用”方式（通过反射）测试它们相等性的方法？

我基本上想要与结构等价相同的语义，只在类上。

4

4 回答 4

3

我认为框架中没有这样的方法可用，但它很容易编写。也许不是最短的实现，但似乎可以完成这项工作：

private bool AreEqual(object x, object y)
{
    // if both are null, they are equal
    if (x == null && y == null)
    {
        return true;
    }
    // if one of them are null, they are not equal
    else if (x == null || y == null)
    {
        return false;
    }

    // if they are of different types, they can't be compared
    if (x.GetType() != y.GetType())
    {
        throw new InvalidOperationException("x and y must be of the same type");
    }

    Type type = x.GetType();
    PropertyInfo[] properties = type.GetProperties();

    for (int i = 0; i < properties.Length; i++)
    {
        // compare only properties that requires no parameters
        if (properties[i].GetGetMethod().GetParameters().Length == 0)
        {
            object xValue = properties[i].GetValue(x, null);
            object yValue = properties[i].GetValue(y, null);

            if (properties[i].PropertyType.IsValueType && !xValue.Equals(yValue))
            {
                return false;
            }
            else if (!properties[i].PropertyType.IsValueType)
            {
                if (!AreEqual(xValue, yValue))
                {
                    return false;
                }
            } // if
        } // if
    } // for

    return true;

}
于 2009-05-17T10:41:08.930 回答
0

顺便提一句

 Expression.Lambda<Func<T,T,bool>> Compile()

可以用作动态方法构建器。

在构建表达式时仍然必须使用反射

于 2009-09-22T18:02:58.807 回答
0

如果您想在不对每个调用进行反射的情况下执行此操作，您可能需要考虑DynamicMethod在第一次调用时构建并使用它。（我有一个指向这样做的文章的链接，但我丢失了它 - 抱歉 - 如果有兴趣，请尝试谷歌搜索。）

于 2009-05-17T10:46:38.620 回答
0

这是 Fredrik Mörk 的回答的更新版本，其中考虑了Nullable递归引用：

public static bool AreEqual<T>(T x, T y) =>
    AreEqual(x, y, new HashSet<object>(new IdentityEqualityComparer<object>()));

private static bool AreEqual(object x, object y, ISet<object> visited)
{
    // if both are null, they are equal
    if (x == null && y == null) return true;

    // if one of them are null, they are not equal
    if (x == null || y == null) return false;

    // if they are of different types, they can't be compared
    if (x.GetType() != y.GetType())
    {
        throw new InvalidOperationException("x and y must be of the same type");
    }

    // check for recursive references
    if (visited.Contains(x)) return true;
    if (visited.Contains(y)) return true;
    visited.Add(x);
    visited.Add(y);

    var type = x.GetType();
    var properties = type.GetProperties();

    foreach (var property in properties)
    {
        // compare only properties that requires no parameters
        if (property.GetGetMethod().GetParameters().Length == 0)
        {
            object xValue = property.GetValue(x, null);
            object yValue = property.GetValue(y, null);

            if (property.PropertyType.IsValueType)
            {
                // check for Nullable
                if (xValue == null && yValue == null) continue;
                if (xValue == null || yValue == null) return false;
                if (!xValue.Equals(yValue)) return false;
            }

            if (!property.PropertyType.IsValueType)
            {
                if (!AreEqual(xValue, yValue, visited)) return false;
            }
        }
    }

    return true;
}

private class IdentityEqualityComparer<T> : IEqualityComparer<T> where T : class
{
    public int GetHashCode(T value) => RuntimeHelpers.GetHashCode(value);
    public bool Equals(T left, T right) => left == right;
}
于 2017-01-27T09:39:47.810 回答