.NET PropertyInfos的平等性

时间:2012-11-28 22:58:11

标签: c# .net reflection propertyinfo

我有一些代码可以将2个PropertyInfos与Equals()进行比较。虽然这通常似乎有效,但我遇到了一个奇怪的情况,即同一底层属性的两个反射属性信息对象不相等:

PropertyInfo prop1, prop2; // both are public and not static
Console.WriteLine(prop1 == prop2); // false ???
Console.WriteLine(Equals(prop1, prop2)); // false ???
Console.WriteLine(prop1.DeclaringType == prop2.DeclaringType); // true
Console.WriteLine(prop1.ReturnType == prop2.ReturnType); // true
Console.WriteLine(prop1.Name == prop2.Name); // true
Console.WriteLine(prop1.DeclaringType.GetProperties().Contains(prop1)); // true
Console.WriteLine(prop2.DeclaringType.GetProperties().Contains(prop2)); // false ???

看起来PropertyInfo实际上并没有实现Equals(),但我认为.NET缓存会反映成员,因此总是会返回相同的实例。你当然总是看到a.GetType()== b.GetType()。 PropertyInfos不是这样吗?

其他一些说明: - 在.NET 4,VS2012,x86构建目标中运行NUnit测试时发生了这种奇怪现象 - 这种情况甚至不会发生在我们用这种方式比较的所有属性上,但它在一个属性上一致地失败。

任何人都可以解释这种行为吗?

编辑:如果有人有兴趣,这里是我写的比较MemberInfos的EqualityComparison函数:

public class MemberEqualityComparer : EqualityComparer<MemberInfo> {
    public override bool Equals(MemberInfo @this, MemberInfo that) {
        if (@this == that) { return true; }
        if (@this == null || that == null) { return false; }

                        // handles everything except for generics
                    if (@this.MetadataToken != that.MetadataToken
                        || !Equals(@this.Module, that.Module)
                        || this.Equals(@this.DeclaringType, that.DeclaringType))
                    {
                        return false;
                    }

                    bool areEqual;
                    switch (@this.MemberType)
                    {
                        // constructors and methods can be generic independent of their types,
                        // so they are equal if they're generic arguments are equal
                        case MemberTypes.Constructor:
                        case MemberTypes.Method:
                            var thisMethod = @this as MethodBase;
                            var thatMethod = that as MethodBase;
                                                areEqual = thisMethod.GetGenericArguments().SequenceEqual(thatMethod.GetGenericArguments(), 
this);
                            break;
                        // properties, events, and fields cannot be generic independent of their types,
                        // so if we've reached this point without bailing out we just return true.
                        case MemberTypes.Property:
                        case MemberTypes.Event:
                        case MemberTypes.Field:
                            areEqual = true;
                            break;
                        // the system guarantees reference equality for types, so if we've reached this point
                        // without returning true the two are not equal
                        case MemberTypes.TypeInfo:
                        case MemberTypes.NestedType:
                            areEqual = false;
                            break;
                        default:
                            throw new NotImplementedException(@this.MemberType.ToString());
    }

    public override int GetHashCode(MemberInfo memberInfo) {
        if (memberInfo == null) { return 0; }

    var hash = @this.MetadataToken 
        ^ @this.Module.GetHashCode() 
        ^ this.GetHashCode(@this.DeclaringType);
    return hash;
    }
}

2 个答案:

答案 0 :(得分:12)

对象标识仅承诺用于Type类,而不是其他反射类。比较相等性的可能声音方法是检查属性是否具有相同的元数据标记并来自同一模块。所以试试这个:

bool equal = prop1.MetadataToken == prop2.MetadataToken &&
             prop1.Module.Equals(prop2.Module);

只要ecma 335适用,这是有道理的。我没有对你的代码进行测试,因为你没有发布它。所以试试吧。

答案 1 :(得分:6)

我猜他们有不同的ReflectedType。例如,继承:

class A {
   public int Foo {get;set;}
}
class B : A {}

现在查看typeof(A).GetProperty("Foo")typeof(B).GetProperty("Foo")