如何让孩子从表达式中声明类型？

Question

我有一个父/子类层次结构中的父抽象声明一个字符串属性和子类实现它：

abstract class Parent 
{ 
    public abstract string Value { get; } 
} 

class Child : Parent 
{ 
    public override string Value { get { return null; } } 
} 

当我使用显式（或隐式）使用儿童类的表情，我期待表达的的MemberInfo的DeclaringType是“孩子”，而是它是父：

Child child = new Child(); 
Expression<Func<string>> expression = (() => child.Value); 
MemberInfo memberInfo = expression.GetMemberInfo(); 
Assert.AreEqual(typeof(Child), memberInfo.DeclaringType); // FAILS! 

因为DeclaringType是家长断言失败。

有什么我可以做声明我的表达或消耗它来揭示儿童类型的实际使用？

注：GetMemberInfo（）上面的扩展方法：

public static class TypeExtensions 
{ 
    /// <summary> 
    /// Gets the member info represented by an expression. 
    /// </summary> 
    /// <param name="expression">The member expression.</param> 
    /// <returns>The member info represeted by the expression.</returns> 
    public static MemberInfo GetMemberInfo(this Expression expression) 
    { 
     var lambda = (LambdaExpression)expression; 

     MemberExpression memberExpression; 
     if (lambda.Body is UnaryExpression) 
     { 
      var unaryExpression = (UnaryExpression)lambda.Body; 
      memberExpression = (MemberExpression)unaryExpression.Operand; 
     } 
     else memberExpression = (MemberExpression)lambda.Body; 

     return memberExpression.Member; 
    } 
} 

Answer 1

否 - 这是什么得到由C＃编译器发出准确的表示（我甚至忘了我们这样写的！）。当寻找成员时，覆盖被忽略 - 编译器只关心最初声明为成员的类型。您可以通过编译代码然后查看IL来看到这一点。这种方法：

static void Main() 
{ 
    Child c = new Child(); 
    string x = c.Value; 
} 

被编译成这个IL：

IL_0000: nop 
IL_0001: newobj  instance void Child::.ctor() 
IL_0006: stloc.0 
IL_0007: ldloc.0 
IL_0008: callvirt instance string Parent::get_Value() 
IL_000d: stloc.1 
IL_000e: ret 

一个琐事点：VB编译不相同的方式工作，所以这种方法：

Public Shared Sub Main(Args As String()) 
    Dim x As Child = New Child() 
    Dim y As String = x.Value 
End Sub 

编译为：

IL_0000: newobj  instance void [lib]Child::.ctor() 
IL_0005: stloc.0 
IL_0006: ldloc.0 
IL_0007: callvirt instance string [lib]Child::get_Value() 
IL_000c: stloc.1 
IL_000d: ret 

Answer 2

如果你不想使用静态类型的方法，而是使用最新的重写，那么这是可能的。我没有测试，但是类似于以下的东西应该做的工作：

public bool FindOverride(MethodInfo baseMethod, Type type) 
{ 
    if(baseMethod==null) 
     throw new ArgumentNullException("baseMethod"); 
    if(type==null) 
     throw new ArgumentNullException("type"); 
    if(!type.IsSubclassOf(baseMethod.ReflectedType)) 
     throw new ArgumentException(string.Format("Type must be subtype of {0}",baseMethod.DeclaringType)); 
    while(true) 
    { 
     var methods=type.GetMethods(BindingFlags.Instance| 
            BindingFlags.DeclaredOnly| 
            BindingFlags.Public| 
            BindingFlags.NonPublic); 
     var method=methods.FirstOrDefault(m=>m.GetBaseDefinition()==baseMethod)) 
     if(method!=null) 
      return method; 
     type=type.BaseType; 
    } 
} 

你在哪里传递MemberInfo作为第一个参数，以及对象作为第二的运行时类型。请注意，这可能很慢，因此您可能需要添加一些缓存。

Answer 3

我的解决方案的基础上，从@JonSkeet和@CodeInChaos信息是不是在表达的纯粹的PropertyInfo看，也是MemberExpression的成员组成的类型：

/// <summary> 
/// Extracts the PropertyInfo for the propertybeing accessed in the given expression. 
/// </summary> 
/// <remarks> 
/// If possible, the actual owning type of the property is used, rather than the declaring class (so if "x" in "() => x.Foo" is a subclass overriding "Foo", then x's PropertyInfo for "Foo" is returned rather than the declaring base class's PropertyInfo for "Foo"). 
/// </remarks> 
/// <typeparam name="T"></typeparam> 
/// <param name="propertyExpression"></param> 
/// <returns></returns> 
internal static PropertyInfo ExtractPropertyInfo<T>(Expression<Func<T>> propertyExpression) 
{ 
    if (propertyExpression == null) 
    { 
     throw new ArgumentNullException("propertyExpression"); 
    } 

    var memberExpression = propertyExpression.Body as MemberExpression; 
    if (memberExpression == null) 
    { 
     throw new ArgumentException(string.Format("Expression not a MemberExpresssion: {0}", propertyExpression), "propertyExpression"); 
    } 

    var property = memberExpression.Member as PropertyInfo; 
    if (property == null) 
    { 
     throw new ArgumentException(string.Format("Expression not a Property: {0}", propertyExpression), "propertyExpression"); 
    } 

    var getMethod = property.GetGetMethod(true); 
    if (getMethod.IsStatic) 
    { 
     throw new ArgumentException(string.Format("Expression cannot be static: {0}", propertyExpression), "propertyExpression"); 
    } 

    Type realType = memberExpression.Expression.Type; 
    if(realType == null) throw new ArgumentException(string.Format("Expression has no DeclaringType: {0}", propertyExpression), "propertyExpression"); 

    return realType.GetProperty(property.Name); 
}