使用對像初始值設定項創建實例的表達式

Question

使用對像初始值設定項創建實例的表達式

c# dynamic expression expression-trees lambda

題

有沒有辦法用對像初始化器和表達式樹創建對象的實例？我的意思是創建一個表達式樹來構建這個lambda：

// my class
public class MyObject {
    public bool DisplayValue { get; set; }
}

// my lambda:
var lambda = (Func<bool, MyObject>)
             (displayValue => new MyObject { DisplayValue = displayValue });

如何使用表達式樹創建此lambda？

更新：

我試著自己編寫以下代碼：

    public static Func<bool, dynamic> Creator;

    static void BuildLambda() {
        var expectedType = typeof(MyObject);
        var displayValueParam = Expression.Parameter(typeof(bool), "displayValue");
        var ctor = Expression.New(expectedType);
        var local = Expression.Parameter(expectedType, "obj");
        var displayValueProperty = Expression.Property(ctor, "DisplayValue");

        var returnTarget = Expression.Label(expectedType);
        var returnExpression = Expression.Return(returnTarget,local, expectedType);
        var returnLabel = Expression.Label(returnTarget, Expression.Default(expectedType));

        var block = Expression.Block(
            new[] { local },
            Expression.Assign(local, ctor),
            Expression.Assign(displayValueProperty, displayValueParam),
            Expression.Return(Expression.Label(expectedType), local, expectedType),
            returnExpression,
            returnLabel
            );
        Creator =
            Expression.Lambda<Func<bool, dynamic>>(block, displayValueParam)
                .Compile();
    }

但它會引發以下錯誤：

無法跳轉到未定義的標籤''。

大家能幫幫我嗎？

agent47

熱門答案

最後我找到了答案：

public static Func<bool, dynamic> Creator; 

static void BuildLambda() { 
    var expectedType = typeof(MyObject); 
    var displayValueParam = Expression.Parameter(typeof(bool), "displayValue"); 
    var ctor = Expression.New(expectedType); 
    var local = Expression.Parameter(expectedType, "obj"); 
    var displayValueProperty = Expression.Property(local, "DisplayValue"); 

    var returnTarget = Expression.Label(expectedType); 
    var returnExpression = Expression.Return(returnTarget,local, expectedType); 
    var returnLabel = Expression.Label(returnTarget, Expression.Default(expectedType)); 

    var block = Expression.Block( 
        new[] { local }, 
        Expression.Assign(local, ctor), 
        Expression.Assign(displayValueProperty, displayValueParam), 
        /* I forgot to remove this line:
         * Expression.Return(Expression.Label(expectedType), local, expectedType), 
         * and now it works.
         * */
        returnExpression, 
        returnLabel 
        ); 
    Creator = 
        Expression.Lambda<Func<bool, dynamic>>(block, displayValueParam) 
            .Compile(); 
}

更新：

雖然它工作正常，但@svick在他的回答中提供了一個更好更短的方式，我正在尋找： MemberInit 。請參閱@ svick的回答。

agent47

在 Stack Overflow 上查看更多

Accepted Answer

要在Expression中表示對像初始值設定項，您應該使用Expression.MemberInit() ：

Expression<Func<bool, MyObject>> BuildLambda() { 
    var createdType = typeof(MyObject);
    var displayValueParam = Expression.Parameter(typeof(bool), "displayValue"); 
    var ctor = Expression.New(createdType);
    var displayValueProperty = createdType.GetProperty("DisplayValue");
    var displayValueAssignment = Expression.Bind(
        displayValueProperty, displayValueParam);
    var memberInit = Expression.MemberInit(ctor, displayValueAssignment);

    return
        Expression.Lambda<Func<bool, MyObject>>(memberInit, displayValueParam);
}

要驗證這實際上是否符合您的要求，您可以在創建的表達式上調用ToString() 。在這種情況下，輸出符合預期：

displayValue => new MyObject() {DisplayValue = displayValue}

使用對像初始值設定項創建實例的表達式

題

Fastest Entity Framework Extensions

一般承認的答案

熱門答案

Performance

Expression Evaluator