Question

我收到警告＆＃34;这种异步方法缺乏等待＆＃39;等等。运营商并将同步运行。考虑使用＆＃39; await＆＃39;运营商等待非阻塞API调用，或等待Task.Run（...）＆＃39;在后台线程上进行CPU绑定的工作。虽然听起来很明显，但我认为我被覆盖了，因为我运行了所有进程，并在我的processFiles方法中有一个Await Task.WhenAll（）。我需要以不同的方式做这件事吗？任何想法都会受到赞赏。

    using System;
    using System.Collections.Generic;
    using System.ComponentModel;
    using System.Data;
    using System.Data.SqlClient;
    using System.Drawing;
    using System.Linq;
    using System.Text;
    using System.Threading.Tasks;
    using System.Windows.Forms;
    using System.IO;


    namespace ProcessFiles
    {
        public partial class ProcessFiles : Form 
        {
            public ProcessFiles(int userLevel, int userID)
            {
            }

            private void btnProcess_Click(object sender, EventArgs e)
            {
                processFiles();

            }

            async Task<int> processFiles()
            {
                var processingTasks = new List<Task>();
                foreach (string fileName in listBox1.Items)
                {
                    processingTasks.Add(process012(fileName));
                    processingTasks.Add(process123(fileName));
                    processingTasks.Add(process234(fileName));

                }
                await Task.WhenAll(processingTasks);
                return (1);
            }


            async Task<int> process173(string fileName)
            {
                return (retVal);

            }

            async Task<int> process032(string fileName)
            {
                return (retVal);
            }

            async Task<int> process018(string fileName)
            {
                return (retVal);
            }

Answer 1

不需要在processXXX方法中使用async Task只返回Task＆lt; int＆gt;代替。您将等待主processFiles（）方法中的任务：

        private async void btnProcess_Click(object sender, EventArgs e)
        {
            await processFiles();
        }

        async Task<int> processFiles()
        {
            var processingTasks = new List<Task>();
            foreach (string fileName in listBox1.Items)
            {
                processingTasks.Add(process012(fileName));
                processingTasks.Add(process123(fileName));
                processingTasks.Add(process234(fileName));

            }
            await Task.WhenAll(processingTasks);
            return (1);
        }


        Task<int> process173(string fileName)
        {
            return Task.FromResult(retVal)
        }

        Task<int> process032(string fileName)
        {
            return Task.FromResult(retVal)
        }

        Task<int> process018(string fileName)
        {
           return Task.FromResult(retVal)
        }

异步等待持续警告

1 个答案: