Question

如何在循环中构建Action操作？解释（对不起，这太长了）

我有以下内容：

public interface ISomeInterface {
    void MethodOne();
    void MethodTwo(string folder);
}

public class SomeFinder : ISomeInterface 
{ // elided 
}

以及使用上述内容的类：

public Map Builder.BuildMap(Action<ISomeInterface> action, 
                            string usedByISomeInterfaceMethods) 
{
    var finder = new SomeFinder();
    action(finder);
}

我可以用其中任何一种来调用它，效果很好：

var builder = new Builder();

var map = builder.BuildMap(z => z.MethodOne(), "IAnInterfaceName");
var map2 = builder(z =>
                   {
                     z.MethodOne();
                     z.MethodTwo("relativeFolderName");
                   }, "IAnotherInterfaceName");

如何以编程方式构建第二个实现？即，

List<string> folders = new { "folder1", "folder2", "folder3" };
folders.ForEach(folder =>
               {
                 /* do something here to add current folder to an expression
                  so that at the end I end up with a single object that would
                  look like:
                  builder.BuildMap(z => {
                                   z.MethodTwo("folder1");
                                   z.MethodTwo("folder2");
                                   z.MethodTwo("folder3");
                                   }, "IYetAnotherInterfaceName");
                */
                });

我一直以为我需要一个

Expression<Action<ISomeInterface>> x

或类似的东西，但对于我的生活，我没有看到如何构建我想要的东西。任何想法都将不胜感激！

Answer 1

这很简单，因为代表已经是多播：

Action<ISomeInterface> action1 = z => z.MethodOne();
Action<ISomeInterface> action2 = z => z.MethodTwo("relativeFolderName");
builder.BuildMap(action1 + action2, "IAnotherInterfaceName");

或者，如果你出于某种原因收集了它们：

IEnumerable<Action<ISomeInterface>> actions = GetActions();
Action<ISomeInterface> action = null;
foreach (Action<ISomeInterface> singleAction in actions)
{
    action += singleAction;
}

甚至：

IEnumerable<Action<ISomeInterface>> actions = GetActions();
Action<ISomeInterface> action = (Action<ISomeInterface>)
    Delegate.Combine(actions.ToArray());

如何将多个Action <t>组合成C＃中的单个Action <t>？</t> </t>

1 个答案: