C＃中switch語句的替代方法

Question

有人可以建議解決此問題的另一種方法，我不想在代碼中使用SWITCH語句。

類定義：

public class Rootobject
{
    public Must[] must { get; set; }
    public Should[] should { get; set; }        
}  

public class Should 
{
    public Match match { get; set; }
    public Bool _bool { get; set; }
}

public class Must 
{        
    public Match match { get; set; }
    public Bool _bool { get; set; }
}

public class Match
{
    public string pname { get; set; }
}

public class Bool
{
    public string rname { get; set; }
}

功能定義

public root getobject(string op)
        {
            Rootobject root = new Rootobject();

            op ="must";

            switch (op)
            {
                case "should":
                    root.should = new Should[1];
                    Should objShould = new Should();
                    objShould.match = new Match();
                    objShould.match.pname = "hello";
                    root.should[0] = objShould;
                   break;
                case "must":
                    root.must = new Must[1];
                    Must objMust = new Must();
                    objMust.match = new Match();
                    objMust.match.pname = "hello";
                    root.must[0] = objMust;
                    break;
            }                       

       return(root);
        }

Switch語句是一種新類型的開銷，然后我可能需要添加另一個條件。 任何人都可以建議使用switch語句的另一種方法。

Answer 1

根據您的問題下的評論，我發現可以實現@Jon Skeet所說的內容。

您可以在RootObject類中添加一個Initialize方法來創建字典（ 使用ref字典，以避免在RootObject類中設置字典，這可能會更改序列化的結構 ）：

 public void Initialize(ref Dictionary<string, Func<Rootobject>> rootDic)
        {
            Func<Rootobject> shouldFunc = () =>
            {
                Rootobject root = new Rootobject();
                root.should = new Should[1];
                Should objShould = new Should();
                objShould.match = new Match();
                objShould.match.pname = "hello";
                root.should[0] = objShould;

                return root;
            };

            Func<Rootobject> mustFunc = () =>
            {
                Rootobject root = new Rootobject();
                root.must = new Must[1];
                Must objMust = new Must();
                objMust.match = new Match();
                objMust.match.pname = "hello";
                root.must[0] = objMust;

                return root;
            };
            rootDic.Add("should", shouldFunc);
            rootDic.Add("must", mustFunc);
        }

然后像這樣在您的getobject方法中調用它：

 public static Rootobject getobject(string op)
        {
            Dictionary<string, Func<Rootobject>> rootDic = new Dictionary<string,Func<Rootobject>>();
            Rootobject root = new Rootobject();
            root.Initialize(ref rootDic);

            if(rootDic.Count > 0)
            return rootDic[op].Invoke();

            return new Rootobject();
        }

即使將其序列化，您仍將獲得與問題中的解決方案相同的結果。