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));
  }

我实际上可以像这样使用我的代码：

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

但我不想指定第二个通用参数类型：

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

有没有办法在C＃3.5中执行此操作？

Answer 1

C＃不支持部分通用推理。如果编译器无法确定您必须自己提供的所有类型。

Answer 2

您可以这样做：

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，你应该能够这样写：

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

如何绕过指定通用参数类型的需要？

2 个答案: