使用表达式在C＃访问结构特性

Question

我一直在使用下面的代码缓存，以便快速访问该功能属性的getter/setter代表：

class PropertyHelper 
{ 
    public static Func<object, object> BuildGetter(PropertyInfo propertyInfo) 
    { 
     var method = propertyInfo.GetGetMethod(true); 

     var obj = Expression.Parameter(typeof(object), "o"); 
     Expression<Func<object, object>> expr = 
       Expression.Lambda<Func<object, object>>(
         Expression.Convert(
           Expression.Call(
             Expression.Convert(obj, method.DeclaringType), 
             method), 
           typeof(object)), 
         obj); 
     return expr.Compile(); 
    } 

    public static Action<object, object> BuildSetter(PropertyInfo propertyInfo) 
    { 
     var method = propertyInfo.GetSetMethod(true); 

     var obj = Expression.Parameter(typeof(object), "o"); 
     var value = Expression.Parameter(typeof(object)); 

     Expression<Action<object, object>> expr = 
      Expression.Lambda<Action<object, object>>(
       Expression.Call(
        Expression.Convert(obj, method.DeclaringType), 
        method, 
        Expression.Convert(value, method.GetParameters()[0].ParameterType)), 
       obj, 
       value); 

     Action<object, object> action = expr.Compile(); 
     return action; 
    } 
} 

这工作得很好访问类对象的属性时，但是当我将它用于结构对象时它失败了。例如，考虑下面的代码：

public struct LocationStruct 
{ 
    public double X { get; set; } 
    public double Y { get; set; } 
} 

public class LocationClass 
{ 
    public double X { get; set; } 
    public double Y { get; set; } 
} 

public class Tester 
{ 
    public static void TestSetX() 
    { 
     Type locationClassType = typeof(LocationClass); 
     PropertyInfo xProperty = locationClassType.GetProperty("X"); 
     Action<object, object> setter = PropertyHelper.BuildSetter(xProperty); 

     LocationStruct testLocationClass = new LocationClass(); 
     setter(testLocationClass, 10.0); 
     if (testLocationClass.X == 10.0) 
     { 
      MessageBox.Show("Worked for the class!"); 
     } 


     Type locationStructType = typeof(LocationStruct); 
     xProperty = locationStructType.GetProperty("X"); 
     setter = PropertyHelper.BuildSetter(xProperty); 

     LocationStruct testLocationStruct = new LocationStruct(); 
     setter(testLocationStruct, 10.0); 
     if (testLocationStruct.X != 10.0) 
     { 
      MessageBox.Show("Didn't work for the struct!"); 
     } 
    } 
} 

第一部分作品，testLocationClass的X值设置为10。然而，因为LocationStruct是一个结构，该testLocationStruct通过值传递中，该值（内给由委托调用的方法）将其X设置为10，但上述代码块中的testLocationStruct对象保持不变。因此，我需要一种方法来访问与上面类似的结构对象的属性（它只适用于类对象的属性）。我试图使用“通过引用”模式来完成此操作，但我无法使其工作。

任何人都可以提供类似的BuildGetter和BuildSetter方法，可以用来缓存结构属性值的getter/setter委托吗？

Answer 1

你需要照顾在为了两件事情这个工作：

1. 在创建二传手表达式树，你需要使用Expression.Unbox值类型和Expression.Convert引用类型。
2. 当使用值类型调用setter时，需要确保使用指向结构的指针设置值（而不是在结构的副本上工作）。

新的实现看起来像这样（只显示新的setter和测试方法，因为其余的是相同的）：

public static Action<object, object> BuildSetter(PropertyInfo propertyInfo) 
{ 
    // Note that we are testing whether this is a value type 
    bool isValueType = propertyInfo.DeclaringType.IsValueType; 
    var method = propertyInfo.GetSetMethod(true); 
    var obj = Expression.Parameter(typeof (object), "o"); 
    var value = Expression.Parameter(typeof (object)); 

    // Note that we are using Expression.Unbox for value types 
    // and Expression.Convert for reference types 
    Expression<Action<object, object>> expr = 
     Expression.Lambda<Action<object, object>>(
      Expression.Call(
       isValueType ? 
        Expression.Unbox(obj, method.DeclaringType) : 
        Expression.Convert(obj, method.DeclaringType), 
       method, 
       Expression.Convert(value, method.GetParameters()[0].ParameterType)), 
       obj, value); 
    Action<object, object> action = expr.Compile(); 
    return action; 
} 

和代码来调用编译二传手：

... 
Type locationStructType = typeof (LocationStruct); 
xProperty = locationStructType.GetProperty("X"); 
setter = PropertyHelper.BuildSetter(xProperty); 
LocationStruct testLocationStruct = new LocationStruct(); 

// Note the boxing of the struct before calling the setter 
object boxedStruct = testLocationStruct; 
setter(boxedStruct, 10.0); 
testLocationStruct = (LocationStruct)boxedStruct; 
... 

此打印：

Worked for the class! 
Worked for the struct! 

我也有准备一个.net小提琴，这里显示了工作实施：https://dotnetfiddle.net/E6WZmK

看到这个答案为Expression.Unbox步骤的解释：https://stackoverflow.com/a/32158735/521773

Answer 2

结构的参数是按值传递，和REF /出似乎不表现很好地工作，你可以考虑使用新功能的签名，返回一个结构实例，而不是：

static Func<MethodInfo, object, object, object> s1 = (MethodInfo set, object instance, object val) => 
{ 
    set.Invoke(instance, new object[] { val }); 
    return instance; 
}; 

// Non-Generic approach 
static Func<object, object, object> BuildSetter5(PropertyInfo propertyInfo) 
{ 
    var method = propertyInfo.GetSetMethod(true); 

    var obj = Expression.Parameter(typeof(object), "o"); 
    var value = Expression.Parameter(typeof(object)); 

    Expression<Func<object, object, object>> expr = 
     Expression.Lambda<Func<object, object, object>>(
      Expression.Call(
       s1.Method, 
       Expression.Constant(method), 
       obj, 
       Expression.Convert(value, method.GetParameters()[0].ParameterType)), 
      obj, 
      value); 

    Func<object, object, object> action = expr.Compile(); 
    return action; 
} 

// Generic approach 
static Func<T, object, T> BuildSetter6<T>(PropertyInfo propertyInfo) where T : struct 
{ 
    var method = propertyInfo.GetSetMethod(true); 

    var obj = Expression.Parameter(typeof(T), "o"); 
    var value = Expression.Parameter(typeof(object)); 

    Expression<Func<T, object, T>> expr = 
     Expression.Lambda<Func<T, object, T>>(
      Expression.Convert(
       Expression.Call(
        s1.Method, 
        Expression.Constant(method), 
        Expression.Convert(obj, typeof(object)), 
        Expression.Convert(value, method.GetParameters()[0].ParameterType)), 
       typeof(T)), 
      obj, 
      value); 

    Func<T, object, T> action = expr.Compile(); 
    return action; 
}