如何繞過指定通用參數類型的需要？

Question

我有一個擴展方法，像這樣的：

 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)); 
    } 

其實我可以用我的代碼是這樣的：

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

，但我想並不需要指定第二個泛型參數類型：

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

有沒有辦法在C＃3.5中做到這一點？

Answer 1

C＃不支持部分泛型推理。如果編譯器無法確定您必須自己提供所有類型的所有類型。

Answer 2

你可以做這樣的事情：

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) }; 
} 

然後，你應該能夠只寫它像這樣：

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