作为辅助方法扩展

public static bool Implements<I>(this Type type, I @interface) where I : class
{
    if(((@interface as Type)==null) || !(@interface as Type).IsInterface)
        throw new ArgumentException("Only interfaces can be 'implemented'.");

    return (@interface as Type).IsAssignableFrom(type);
}

使用示例:

var testObject = new Dictionary<int, object>();
result = testObject.GetType().Implements(typeof(IDictionary<int, object>)); // true!

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。

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));

    }

您必须检查泛型接口的构造类型。

你必须这样做:

foo is IBar<String>

因为IBar<String>表示构造的类型。你必须这样做的原因是，如果T在检查中是未定义的，编译器不知道你是指IBar<Int32>还是IBar<SomethingElse>。

试试下面的扩展。

public static bool Implements(this Type @this, Type @interface)
{
    if (@this == null || @interface == null) return false;
    return @interface.GenericTypeArguments.Length>0
        ? @interface.IsAssignableFrom(@this)
        : @this.GetInterfaces().Any(c => c.Name == @interface.Name);
}

为了测试它。创建

public interface IFoo { }
public interface IFoo<T> : IFoo { }
public interface IFoo<T, M> : IFoo<T> { }
public class Foo : IFoo { }
public class Foo<T> : IFoo { }
public class Foo<T, M> : IFoo<T> { }
public class FooInt : IFoo<int> { }
public class FooStringInt : IFoo<string, int> { }
public class Foo2 : Foo { }

以及测试方法

public void Test()
{
    Console.WriteLine(typeof(Foo).Implements(typeof(IFoo)));
    Console.WriteLine(typeof(FooInt).Implements(typeof(IFoo)));
    Console.WriteLine(typeof(FooInt).Implements(typeof(IFoo<>)));
    Console.WriteLine(typeof(FooInt).Implements(typeof(IFoo<int>)));
    Console.WriteLine(typeof(FooInt).Implements(typeof(IFoo<string>)));
    Console.WriteLine(typeof(FooInt).Implements(typeof(IFoo<,>)));
    Console.WriteLine(typeof(FooStringInt).Implements(typeof(IFoo<,>)));
    Console.WriteLine(typeof(FooStringInt).Implements(typeof(IFoo<string,int>)));
    Console.WriteLine(typeof(Foo<int,string>).Implements(typeof(IFoo<string>)));
 }