如何从表达式树中设置属性值？

Question

我想设置表达式树中引用的属性值。

using System;
using System.Diagnostics;
using System.Linq.Expressions;
using System.Reflection;

namespace ConsoleApp8
{
    class TestObject
    {
        public double X { get; set; }
    }

    class Program
    {
        static Action<double> GetSetterForX(Expression<Func<double>> expression)
        {
            var body = expression.Body;
            var operand = body as MemberExpression;
            var propertyInfo = (PropertyInfo) (operand.Member);
            var setter = propertyInfo.GetSetMethod(true);

            // At this point I have the setter. But how do I get access to the testObject?

            return null;
        }

        static void Main(string[] args)
        {
            var testObject = new TestObject();
            var setter = GetSetterForX(() => testObject.X);
            setter.Invoke(5);
            Debug.Assert(testObject.X == 5);
        }
    }
}

我可以获得 setter 但无法找到访问实例 (testObject) 的方法。有办法吗？

请注意，这是一个简化的示例。我将使用更复杂的表达式，其中包含许多属性引用，我希望能够（单独）设置所有这些引用。

更新

澄清一下。我想要返回一个 setter，它只需要一个双精度值，并且它分配给 testObject 的 X 属性。这应该是可能的，而无需在 setter 中显式传递对 testObject 的引用。我可以做同样的事情来获得 getter 但不是 setter。这是 getter 代码：

static Func<double> GetGetterForX(Expression<Func<double>> expression)
{
    var body = expression.Body;
    var operand = body as MemberExpression;
    var result = new Func<double>(() => (double) GetValue(operand));

    return result;
}

private static object GetValue(MemberExpression member)
{
    var objectMember = Expression.Convert(member, typeof(object));
    var getterLambda = Expression.Lambda<Func<object>>(objectMember);
    var getter = getterLambda.Compile();
    return getter();
}

返回的 getter 始终在 testObject 实例上工作。不用再传入testObject了

Answer 1

您需要 return MethodInfo 而不是 Action<double>，Invoke(object obj, object[] params) 也采用原始对象和参数：

static MethodInfo GetSetterForX(Expression<Func<double>> expression)
{
    var body = expression.Body;
    var operand = body as MemberExpression;
    var propertyInfo = (PropertyInfo)(operand.Member);
    var setter = propertyInfo.GetSetMethod(true);
    return setter;
}

public static void Main()
{
    var testObject = new TestObject();
    var setter = GetSetterForX(() => testObject.X);
    setter.Invoke(testObject, new object[]{5});
    Debug.Assert(testObject.X == 5);
}

Fiddle : https://dotnetfiddle.net/CHJGbk

Answer 2

只要输入的lambda表达式表示成员访问器，就可以使用Expression.Assign传递输入的lambda表达式主体和表示值的参数，例如

static Action<double> GetSetterForX(Expression<Func<double>> expression)
{
    var parameter = Expression.Parameter(typeof(double), "value");
    var body = Expression.Assign(expression.Body, parameter);
    var lambda = Expression.Lambda<Action<double>>(body, parameter);
    return lambda.Compile();
}

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