对表达式树使用 lambda 返回值答案

【问题标题】：Use lambda return value for expression tree对表达式树使用 lambda 返回值
【发布时间】：2013-09-25 14:41:55
【问题描述】：

我尝试使用表达式树。我有一个带有List<string> 的对象，我想构建一个表达式树，为该属性添加一个值，但我想通过Func 指定要添加的值。目前我正在尝试这个......

public static Action<T> CreateMethodAddObjectToList<T, C>(this Type type, string property, Func<C> ctorFunction)
        {
            PropertyInfo fieldInfo = type.GetProperty(property);

            if (fieldInfo == null)
            {
                return null;
            }

            ParameterExpression targetExp = Expression.Parameter(type, "target");
            MemberExpression fieldExp = Expression.Property(targetExp, property);
            var method = fieldExp.Type.GetMethod("Add", BindingFlags.Public | BindingFlags.Instance);

            Expression<Func<C>> ctorExpression = () => ctorFunction();

// but this doesnt work because I can't use the ctorExpression in this way
            var callExp = Expression.Call(fieldExp, method, ctorExpression);

            var function = Expression.Lambda<Action<T>>(callExp, targetExp).Compile();

            return function;
        }

电话看起来像

var dummyObject = new DummyObject { IntProperty = 5 };

            Action<DummyObject> setter = typeof (DummyObject).CreateMethodAddObjectToList<DummyObject, string>("StringList", () => "Test" );

【问题讨论】：

标签： c# lambda expression-trees

【解决方案1】：

您可以将 ctorFunction 更改为 Expression<Func<C>>，然后在生成的操作中调用它：

public static Action<T> CreateMethodAddObjectToList<T, C>(this Type type, string property, Expression<Func<C>> createExpr)
{
    PropertyInfo fieldInfo = type.GetProperty(property);

    if (fieldInfo == null)
    {
        return null;
    }

    ParameterExpression targetExp = Expression.Parameter(type, "target");
    MemberExpression fieldExp = Expression.Property(targetExp, property);
    var method = fieldExp.Type.GetMethod("Add", BindingFlags.Public | BindingFlags.Instance);

    var valueExpr = Expression.Invoke(createExpr);
    var callExpr = Expression.Call(fieldExp, method, valueExpr);

    var function = Expression.Lambda<Action<T>>(callExpr, targetExp).Compile();

    return function;
}

【讨论】：