异步异常未被捕获或被吞没

Question

从未来的更新: TL; DR赶上你要异步方法表达await,Task.WaitAll或.Result.

我创建了一个有点复杂的异步方法,只运行其他异步方法.您可以忽略大部分内容,因为只有行var mSpekTask ...是有意义的,我也不关心逻辑,我只想知道我的异常去了哪里.我的主要问题是ex.ToString()永远不会被击中,即使在mSpecTask内部确实发生了异常.

public async Task LoadAsync(IEnumerable<ProductRequest> feed, int? customerId,
      IProgress<int> mSpecProgress, Action<Task> mSpecCompletionHandler)
  {
        var ids = feed.Select(x => x.ProductId.ToString()).Distinct().ToList();

        try
        {
           var mSpecTask = this.LoadMSpecAsync(mSpecProgress, ids);
        }
        catch (Exception ex)
        {
           ex.ToString();
        }
  }

这是LoadMSpecAsync的代码

public Task<ResultSet> LoadMSpecAsync(IProgress<int> prg,
     IEnumerable<string> ids)
  {
     return this.LoadAsync(prg, ids, Selector.M, SPMS, x => x.Order);
  }

下面是LoadAsync的代码,等待db.ExecuteTVP(progress,spName,id,parameters)生成异常.

      private async Task<Dictionary<Pair, dynamic>> LoadAsync(IProgress<int> progress,
     IEnumerable<string> ids, Selector s, string spName, Func<dynamic, int> k,
      Func<dynamic, dynamic> f = null, object parameters = null)
  {
     parameters = new ExpandoObject().CopyFromSafe(parameters);
     if (spName != SPMAP) ((dynamic)parameters).lang = this.languageCode;

     using (var db = new SqlConnection(this.connectionString))
     {
        await db.OpenAsync();

        var results = await db.ExecuteTVP(progress, spName, ids, parameters);

        db.Close();
     }

     return this.data[s];
  }

Answer 1

当async方法抛出异常时,该异常将放在返回的异常上Task.它不是直接向调用者提出的.这是设计的.

所以,你必须要么await在Task从返回LoadMSpecAsync或者让你mSpecCompletionHandler检查它的Task例外参数.它会出现在那里.