如果你想要一个支持泛型基类型和接口的扩展方法，我扩展了sduplooy的答案:

    public static bool InheritsFrom(this Type t1, Type t2)
    {
        if (null == t1 || null == t2)
            return false;

        if (null != t1.BaseType &&
            t1.BaseType.IsGenericType &&
            t1.BaseType.GetGenericTypeDefinition() == t2)
        {
            return true;
        }

        if (InheritsFrom(t1.BaseType, t2))
            return true;

        return
            (t2.IsAssignableFrom(t1) && t1 != t2)
            ||
            t1.GetInterfaces().Any(x =>
              x.IsGenericType &&
              x.GetGenericTypeDefinition() == t2);
    }

为了完全解决类型系统，我认为你需要处理递归，例如，IList<T>: ICollection<T>: IEnumerable<T>，否则你不会知道IList<int>最终实现了IEnumerable<>。

    /// <summary>Determines whether a type, like IList&lt;int&gt;, implements an open generic interface, like
    /// IEnumerable&lt;&gt;. Note that this only checks against *interfaces*.</summary>
    /// <param name="candidateType">The type to check.</param>
    /// <param name="openGenericInterfaceType">The open generic type which it may impelement</param>
    /// <returns>Whether the candidate type implements the open interface.</returns>
    public static bool ImplementsOpenGenericInterface(this Type candidateType, Type openGenericInterfaceType)
    {
        Contract.Requires(candidateType != null);
        Contract.Requires(openGenericInterfaceType != null);

        return
            candidateType.Equals(openGenericInterfaceType) ||
            (candidateType.IsGenericType && candidateType.GetGenericTypeDefinition().Equals(openGenericInterfaceType)) ||
            candidateType.GetInterfaces().Any(i => i.IsGenericType && i.ImplementsOpenGenericInterface(openGenericInterfaceType));

    }

我使用一个稍微简单的版本的@GenericProgrammers扩展方法:

public static bool Implements<TInterface>(this Type type) where TInterface : class {
    var interfaceType = typeof(TInterface);

    if (!interfaceType.IsInterface)
        throw new InvalidOperationException("Only interfaces can be implemented.");

    return (interfaceType.IsAssignableFrom(type));
}

用法:

    if (!featureType.Implements<IFeature>())
        throw new InvalidCastException();

您必须遍历继承树并找到树中每个类的所有接口，并将typeof(IBar<>)与调用Type的结果进行比较。GetGenericTypeDefinition如果接口是泛型的。当然，这一切都有点痛苦。

更多信息和代码请参见这个答案和这些答案。

如果你想要一个支持泛型基类型和接口的扩展方法，我扩展了sduplooy的答案:

    public static bool InheritsFrom(this Type t1, Type t2)
    {
        if (null == t1 || null == t2)
            return false;

        if (null != t1.BaseType &&
            t1.BaseType.IsGenericType &&
            t1.BaseType.GetGenericTypeDefinition() == t2)
        {
            return true;
        }

        if (InheritsFrom(t1.BaseType, t2))
            return true;

        return
            (t2.IsAssignableFrom(t1) && t1 != t2)
            ||
            t1.GetInterfaces().Any(x =>
              x.IsGenericType &&
              x.GetGenericTypeDefinition() == t2);
    }

以下内容应该没有任何问题:

bool implementsGeneric = (anObject.Implements("IBar`1") != null);

如果您想为IBar查询提供一个特定的泛型类型参数，您可以捕获AmbiguousMatchException。