如何绕过指定泛型参数类型的需要？

Question

我有一个像这样的扩展方法：

      public static void ImplementsAttribute<TX, TY>(this Expression<Func<TY>> expression) 
     where TX : Coupling.PropertiesMergerAttribute
  {
     var memberExpression = expression.Body as MemberExpression;
     var name = MetaHelper.GetPropertyName(expression);
     var property = memberExpression.Expression.Type.GetProperty(name);
     var attributes = property.GetCustomAttributes(true);
     Assert.IsTrue(attributes.Any(a => a is TX));
  }

我可以像这样使用我的代码：

     Expression<Func<String>> nameProperty = () => new ImprovisedExplosiveXML().Name;
     nameProperty.ImplementsAttribute<Coupling.UnresolvablePropertiesMergerAttribute, String>();

但我不需要指定第二个泛型参数类型：

     Expression<Func<String>> nameProperty = () => new ImprovisedExplosiveXML().Name;
     nameProperty.ImplementsAttribute<Coupling.UnresolvablePropertiesMergerAttribute>();

在C# 3.5中有办法做到这一点吗？

Answer 1

C#不支持部分泛型推理。如果编译器无法确定所有类型，则必须自己提供所有类型。

Answer 2

你可以这样做：

public class AttributeTester
{
    public Attribute[] Attributes { get; set; }

    public void ImplementsAttribute<TAttType>()
    {
         Assert.IsTrue(Attributes.Any(x => x is TAttType));
    }
}

public static void ForProperty<TType, TProperty>(this TType obj, Expression<Func<TType, TProperty>> expression)
{
     var memberExpression = expression.Body as MemberExpression;
     var name = MetaHelper.GetPropertyName(expression);
     var property = memberExpression.Expression.Type.GetProperty(name);
     return new AttributeTester { Attributes = property.GetCustomAttributes(true) };
}

然后，您应该能够像这样编写它：

new ImproveisedExplosiveXML().ForProperty(x => x.Name).ImplementsAttribute<SomeAttribute>();