C# Winform递归循环-方法被反复调用

请随意创建windows窗体应用程序。要再现错误，请禁用网络连接并运行代码。它每1秒尝试重新连接一次。在4-5次尝试后，启用网络连接，并在调试模式下，您将注意到，即使获取产品，Reconnect（）方法也会被调用4-5次。获取产品后，为什么它会一次又一次地调用Reconnect（）方法

        string apiUrl = "https://api.gdax.com/products";
        string json;

        private void Form1_Load(object sender, EventArgs e)
        {            
            try
            {                
                if (FillProducts()) // product need first
                {
                }
            }
            catch (WebException ex)
            {
                ReconnectOnError(ex);
            }
        }
        private bool FillProducts()
        {
            bool isDone = false;
            try
            {
                json = GetGDXJSONData(apiUrl);
                JsonSerializer serializer = new JsonSerializer();
                DataTable dt = (System.Data.DataTable)Newtonsoft.Json.JsonConvert.DeserializeObject(json, (typeof(System.Data.DataTable)));

                count = dt.Rows.Count;

                if (count > 0)
                    isDone = true;
            }
            catch (Exception ex)
            {               
                isDone = false;
                ReconnectOnError(ex);
            }
            return isDone;
        }

        int count = 0;
        private void ReconnectOnError(Exception errorMessage)
        {
            try
            {
                Thread.Sleep(1000);
                if (count < 1)
                {
                    FillProducts();     // it comes on this point again and again even the count is greater than 1
                    Reconnect();
                }
                else
                {
                    Reconnect();
                }
            }
            catch (WebException ex)
            {
                ReconnectOnError(ex);
            }
        }

        private void Reconnect()
        {
            // why this is called the number of times the attempt was made to fill the products?
        }      

        private string GetGDXJSONData(string apiUrl)
        {
            HttpWebRequest request = (HttpWebRequest)WebRequest.Create(apiUrl);

            request.Method = "GET";
            request.ContentType = "application/json";
            request.UserAgent = "gdax-node-client";
            request.Accept = "application/json";

            HttpWebResponse response = (HttpWebResponse)request.GetResponse();
            string responseString = new StreamReader(response.GetResponseStream()).ReadToEnd();

            return responseString;
        }
    }

---

这是因为如果没有网络连接，那么调用API url将无法连接，并且在

catch

块中，您正试图重新连接

        catch (Exception ex)
        {               
            isDone = false;
            ReconnectOnError(ex);
        }

---

没有internet连接时。计数变量始终为零。

这里有一个意外循环：

如果

FillProducts

失败，它将调用自己

迭代：FP失败

迭代：FP调用ReconnectOnEx调用FP，再次失败

迭代：FP调用重新连接调用FP调用重新连接

n。迭代：。。。。调用成功的FP并返回

现在，随着每次迭代调用

重新连接

，整个堆栈将展开

private bool FillProducts()
    {
        bool isDone = false;
        try
        {
            /* ... Fails if no connection ... */
        }
        catch (Exception ex)
        {               
            isDone = false;
            ReconnectOnError(ex); // ==> BLOCKS !!
        }
        return isDone;
    }

    int count = 0;
    private void ReconnectOnError(Exception errorMessage)
    {
        try
        {
            Thread.Sleep(1000);
            if (count < 1)
            {
                FillProducts();     // <== Will result in another call to this method. Returns on 1st Succeeding call to FillProducts.
                Reconnect();        // <== Will be called as soon as FillProducts returns.
            }
            else
            {
                Reconnect();
            }
        }
        catch (WebException ex)
        {
            ReconnectOnError(ex);
        }
    }

需要考虑的其他几点：

您不应该在GUI线程上执行网络I/O。您的GUI可能会变得无响应。考虑使用异步/等待（任务异步模式）< /P>

捕获异常可能不是最好的主意。您应该捕获可能的最特定异常，并让调用方处理其余异常

---

这是由于

中的else部分，如果（count<0）

你再次调用它，我必须这样做。。不过，它不会出现在这条线上。如果您运行代码，您将看到它调用FillProducts（）；重新连接（）；一次又一次，即当没有Internet时，它第一次尝试重新连接的次数您有一个间接递归调用-

FillProducts（）

调用

重新连接错误（ex）如果出现调用
FillProducts（）
的异常。请运行代码并在不使用internet的情况下进行调试，然后在尝试4-5次重新连接后启动internet。请不要停止我正在尝试重新连接的应用程序，但我的问题是，一旦internet恢复，为什么它仍然调用4-5次reconnect（）方法？请单击“在此处输入图像描述”链接
private bool FillProducts()
    {
        bool isDone = false;
        try
        {
            /* ... Fails if no connection ... */
        }
        catch (Exception ex)
        {               
            isDone = false;
            ReconnectOnError(ex); // ==> BLOCKS !!
        }
        return isDone;
    }

    int count = 0;
    private void ReconnectOnError(Exception errorMessage)
    {
        try
        {
            Thread.Sleep(1000);
            if (count < 1)
            {
                FillProducts();     // <== Will result in another call to this method. Returns on 1st Succeeding call to FillProducts.
                Reconnect();        // <== Will be called as soon as FillProducts returns.
            }
            else
            {
                Reconnect();
            }
        }
        catch (WebException ex)
        {
            ReconnectOnError(ex);
        }
    }

private bool FillProducts()
    {
        // To prevent waiting forever ...
        int retryCount = 10;

        bool isDone = false;
        while ( !isDone && (retryCount-- > 0) )
        {
            try
            {
                /* ... Fails if no connection ... */

                // OnSuccess=>
                isDone = true; // will break the loop.
            }
            catch (Exception ex) // You should actually catch a more specific Exception here
                                 // ... but that's a different question.
            {               
                isDone = false;
                // Just omit this! >>> ReconnectOnError(ex); // ==> BLOCKS !!
                // If you want, you can add a little delay here ...
                Thread.Sleep(1000);
                // From your code, I guess this has to be called on failure ...
                Reconnect();
            }
        }
        return isDone;
    }