异步方法跳过等待

本文介绍了异步方法跳过等待的处理方法，对大家解决问题具有一定的参考价值，需要的朋友们下面随着小编来一起学习吧！ 问题描述

我有以下代码:

public async Task<string> GetData(Uri source) { if (client.IsBusy == true) client.CancelAsync (); Task<string> tskResult = client.DownloadStringTaskAsync (source); string strResult = await tskResult; return strResult; }

当我从 Task< string> ... 开始逐步执​​行此方法时，调试器会跳过 return strResult; 并且strResult的值为null.

When I step through this method starting with Task<string>... the debugger jumps over return strResult; And the value of strResult is null.

为什么会这样?谢谢.

PS:我正在这样调用此方法:

PS: I am calling this method like this:

StringBuilder strBuild = new StringBuilder();

StringBuilder strBuild = new StringBuilder();

foreach (var image in imageApiUrlLst) { string imageModelPull = await callMgr.GetData(new Uri(image)); ///WHY GETS STUCK! strBuild.AppendLine(imageModelPull); }

推荐答案

如果等待的内容尚未到达，则 async 方法将在到达 await 语句后立即返回.没完.完成后，该方法将在该 await 语句之后继续执行.尝试在return语句上放置一个断点，它应该被击中两次.

An async method returns as soon as an await statement is reached, if the thing awaited hasn't finished. Once it completes, the method continues execution after that await statement. Try putting a break point on the return statement and it should get hit twice.