為任何方法創建Func或Action(使用c#中的反射)

c# expression-trees lambda reflection

我的應用程序可以根據數據庫中的設置(文件,類和方法名稱)動態加載dll。為了方便,加快和減少反射的使用,我希望有一個緩存....

遵循使用的想法:

 MethodInfo.Invoke

沒有任何表演( 反射性能 - 創建代表(屬性C#) )我想翻譯任何方法調用。我想到了一些像這樣的東西:

public static T Create<T>(Type type, string methodName) // or
public static T Create<T>(MethodInfo info) // to use like this:
var action = Create<Action<object>>(typeof(Foo), "AnySetValue");

一個要求是所有參數都可以是對象。

我正在嘗試處理表達式,到目前為止我有這樣的事情:

    private void Sample()
    {
        var assembly = Assembly.GetAssembly(typeof(Foo));

        Type customType = assembly.GetType("Foo");

        var actionMethodInfo = customType.GetMethod("AnyMethod");
        var funcMethodInfo = customType.GetMethod("AnyGetString");
        var otherActionMethod = customType.GetMethod("AnySetValue");
        var otherFuncMethodInfo = customType.GetMethod("OtherGetString");

        var foo = Activator.CreateInstance(customType);
        var actionAccessor = (Action<object>)BuildSimpleAction(actionMethodInfo);
        actionAccessor(foo);

        var otherAction = (Action<object, object>)BuildOtherAction(otherActionMethod);
        otherAction(foo, string.Empty);

        var otherFuncAccessor = (Func<object, object>)BuildFuncAccessor(funcMethodInfo);
        otherFuncAccessor(foo);

        var funcAccessor = (Func<object,object,object>)BuildOtherFuncAccessor(otherFuncMethodInfo);
        funcAccessor(foo, string.Empty);
    }

    static Action<object> BuildSimpleAction(MethodInfo method)
    {
        var obj = Expression.Parameter(typeof(object), "o");

        Expression<Action<object>> expr =
            Expression.Lambda<Action<object>>(
                Expression.Call(
                    Expression.Convert(obj, method.DeclaringType),
                    method), obj);

        return expr.Compile();
    }

    static Func<object, object> BuildFuncAccessor(MethodInfo method)
    {
        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();

    }

    static Func<object, object, object> BuildOtherFuncAccessor(MethodInfo method)
    {
        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(
                        Expression.Convert(obj, method.DeclaringType),
                        method,
                        Expression.Convert(value, method.GetParameters()[0].ParameterType)), 
                        obj, value);

        return expr.Compile();

    }

    static Action<object, object> BuildOtherAction(MethodInfo method)
    {
        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);

        return expr.Compile();
    }

public class Foo
{
    public void AnyMethod() {}

    public void AnySetValue(string value) {}

    public string AnyGetString()
    {            return string.Empty;        }

    public string OtherGetString(string value)
    {            return string.Empty;        }
}

有沒有辦法簡化這段代碼? (我相信有可能只使用泛型創建一個方法..)當你有3,4,5,任何參數和我一樣?


我在想,如果有這樣的事情怎麼辦:

https://codereview.stackexchange.com/questions/1070/generic-advanced-delegate-createdelegate-using-expression-trees

但是我會有更多的參數(在動作或函數中),這個參數(第一個參數)是一個要執行的對象。這可能嗎?

一般承認的答案

我製作了一個滿足您所有要求的示例程序(我想!)

class Program
{
    class MyType
    {
        public MyType(int i) { this.Value = i; }

        public void SetValue(int i) { this.Value = i; }

        public void SetSumValue(int a, int b) { this.Value = a + b; }

        public int Value { get; set; }
    }

    public static void Main()
    {
        Type type = typeof(MyType);

        var mi = type.GetMethod("SetValue");

        var obj1 = new MyType(1);
        var obj2 = new MyType(2);

        var action = DelegateBuilder.BuildDelegate<Action<object, int>>(mi);

        action(obj1, 3);
        action(obj2, 4);

        Console.WriteLine(obj1.Value);
        Console.WriteLine(obj2.Value);

        // Sample passing a default value for the 2nd param of SetSumValue.
        var mi2 = type.GetMethod("SetSumValue");

        var action2 = DelegateBuilder.BuildDelegate<Action<object, int>>(mi2, 10);

        action2(obj1, 3);
        action2(obj2, 4);

        Console.WriteLine(obj1.Value);
        Console.WriteLine(obj2.Value);

        // Sample without passing a default value for the 2nd param of SetSumValue.
        // It will just use the default int value that is 0.
        var action3 = DelegateBuilder.BuildDelegate<Action<object, int>>(mi2);

        action3(obj1, 3);
        action3(obj2, 4);

        Console.WriteLine(obj1.Value);
        Console.WriteLine(obj2.Value);
    }
}

DelegateBuilder類:

public class DelegateBuilder
{
    public static T BuildDelegate<T>(MethodInfo method, params object[] missingParamValues)
    {
        var queueMissingParams = new Queue<object>(missingParamValues);

        var dgtMi = typeof(T).GetMethod("Invoke");
        var dgtRet = dgtMi.ReturnType;
        var dgtParams = dgtMi.GetParameters();

        var paramsOfDelegate = dgtParams
            .Select(tp => Expression.Parameter(tp.ParameterType, tp.Name))
            .ToArray();

        var methodParams = method.GetParameters();

        if (method.IsStatic)
        {
            var paramsToPass = methodParams
                .Select((p, i) => CreateParam(paramsOfDelegate, i, p, queueMissingParams))
                .ToArray();

            var expr = Expression.Lambda<T>(
                Expression.Call(method, paramsToPass),
                paramsOfDelegate);

            return expr.Compile();
        }
        else
        {
            var paramThis = Expression.Convert(paramsOfDelegate[0], method.DeclaringType);

            var paramsToPass = methodParams
                .Select((p, i) => CreateParam(paramsOfDelegate, i + 1, p, queueMissingParams))
                .ToArray();

            var expr = Expression.Lambda<T>(
                Expression.Call(paramThis, method, paramsToPass),
                paramsOfDelegate);

            return expr.Compile();
        }
    }

    private static Expression CreateParam(ParameterExpression[] paramsOfDelegate, int i, ParameterInfo callParamType, Queue<object> queueMissingParams)
    {
        if (i < paramsOfDelegate.Length)
            return Expression.Convert(paramsOfDelegate[i], callParamType.ParameterType);

        if (queueMissingParams.Count > 0)
            return Expression.Constant(queueMissingParams.Dequeue());

        if (callParamType.ParameterType.IsValueType)
            return Expression.Constant(Activator.CreateInstance(callParamType.ParameterType));

        return Expression.Constant(null);
    }
}

怎麼運行的

核心是BuildDelegate方法:

static T BuildDelegate<T>(MethodInfo method)

  • T是您要創建的委託類型。
  • method是您希望由生成的委託調用的方法的MethodInfo。

示例調用: var action = BuildDelegate<Action<object, int>>(mi);

參數規則:

  • 如果傳遞的方法是實例方法,則生成的委託的第一個參數將接受包含方法本身的對象的實例。所有其他參數將傳遞給該方法。

  • 如果傳遞的方法是靜態方法,則生成的委託的所有參數都將傳遞給方法。

  • 缺少參數將傳遞默認值。

  • 額外的參數將被丟棄。

熱門答案

Delegate.CreateDelegate比構建表達式樹簡單得多。

    var assembly = Assembly.GetAssembly(typeof(Foo));

    Type customType = assembly.GetType("Foo");

    var actionMethodInfo = customType.GetMethod("AnyMethod");

    var foo = Activator.CreateInstance(customType);

    Action action = (Action)Delegate.CreateDelegate(typeof(Action), foo, actionMethodInfo);


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許可下: CC-BY-SA with attribution
不隸屬於 Stack Overflow