Mark Byers的回答非常棒，但我想补充以下内容:

OutputDataReceived和ErrorDataReceived委托需要在outputWaitHandle和errorWaitHandle被释放之前被移除。如果进程在超过超时时间后继续输出数据，然后终止，则outputWaitHandle和errorWaitHandle变量将在被释放后被访问。

(仅供参考，我不得不加上这个警告作为回答，因为我不能评论他的帖子。)

流程的文档。StandardOutput说在你等待之前读取，否则你会死锁，代码片段复制如下:

 // Start the child process.
 Process p = new Process();
 // Redirect the output stream of the child process.
 p.StartInfo.UseShellExecute = false;
 p.StartInfo.RedirectStandardOutput = true;
 p.StartInfo.FileName = "Write500Lines.exe";
 p.Start();
 // Do not wait for the child process to exit before
 // reading to the end of its redirected stream.
 // p.WaitForExit();
 // Read the output stream first and then wait.
 string output = p.StandardOutput.ReadToEnd();
 p.WaitForExit();

未处理的ObjectDisposedException问题发生在进程超时时。在这种情况下，条件的其他部分:

if (process.WaitForExit(timeout) 
    && outputWaitHandle.WaitOne(timeout) 
    && errorWaitHandle.WaitOne(timeout))

不执行。我用以下方法解决了这个问题:

using (AutoResetEvent outputWaitHandle = new AutoResetEvent(false))
using (AutoResetEvent errorWaitHandle = new AutoResetEvent(false))
{
    using (Process process = new Process())
    {
        // preparing ProcessStartInfo

        try
        {
            process.OutputDataReceived += (sender, e) =>
                {
                    if (e.Data == null)
                    {
                        outputWaitHandle.Set();
                    }
                    else
                    {
                        outputBuilder.AppendLine(e.Data);
                    }
                };
            process.ErrorDataReceived += (sender, e) =>
                {
                    if (e.Data == null)
                    {
                        errorWaitHandle.Set();
                    }
                    else
                    {
                        errorBuilder.AppendLine(e.Data);
                    }
                };

            process.Start();

            process.BeginOutputReadLine();
            process.BeginErrorReadLine();

            if (process.WaitForExit(timeout))
            {
                exitCode = process.ExitCode;
            }
            else
            {
                // timed out
            }

            output = outputBuilder.ToString();
        }
        finally
        {
            outputWaitHandle.WaitOne(timeout);
            errorWaitHandle.WaitOne(timeout);
        }
    }
}

我试图通过考虑Mark Byers, Rob, stevejay的回答，创建一个类来解决你使用异步流读取的问题。这样做，我意识到有一个与异步流程输出流读取相关的错误。

我在微软报告了这个漏洞:https://connect.microsoft.com/VisualStudio/feedback/details/3119134

简介:

You can't do that: process.BeginOutputReadLine(); process.Start(); You will receive System.InvalidOperationException : StandardOut has not been redirected or the process hasn't started yet. ============================================================================================================================ Then you have to start asynchronous output read after the process is started: process.Start(); process.BeginOutputReadLine(); Doing so, make a race condition because the output stream can receive data before you set it to asynchronous:

process.Start(); 
// Here the operating system could give the cpu to another thread.  
// For example, the newly created thread (Process) and it could start writing to the output
// immediately before next line would execute. 
// That create a race condition.
process.BeginOutputReadLine();

============================================================================================================================ Then some people could say that you just have to read the stream before you set it to asynchronous. But the same problem occurs. There will be a race condition between the synchronous read and set the stream into asynchronous mode. ============================================================================================================================ There is no way to acheive safe asynchronous read of an output stream of a process in the actual way "Process" and "ProcessStartInfo" has been designed.

对于您的情况，您最好使用其他用户建议的异步读取。但是您应该意识到，由于竞态条件，您可能会错过一些信息。

我最终使用的解决方案来避免所有的复杂性:

var outputFile = Path.GetTempFileName();
info = new System.Diagnostics.ProcessStartInfo("TheProgram.exe", String.Join(" ", args) + " > " + outputFile + " 2>&1");
info.CreateNoWindow = true;
info.WindowStyle = System.Diagnostics.ProcessWindowStyle.Hidden;
info.UseShellExecute = false;
System.Diagnostics.Process p = System.Diagnostics.Process.Start(info);
p.WaitForExit();
Console.WriteLine(File.ReadAllText(outputFile)); //need the StandardOutput contents

所以我创建了一个临时文件，通过使用> outputfile > 2>&1将输出和错误重定向到它，然后在进程完成后读取文件。

其他解决方案适用于希望对输出执行其他操作的场景，但对于简单的操作，这可以避免很多复杂性。

我也有同样的问题，但原因不同。但是在Windows 8下会发生，而在Windows 7下不会。下面这行似乎是导致这个问题的原因。

pProcess.StartInfo.UseShellExecute = False

解决方案是不禁用UseShellExecute。我现在收到一个Shell弹出窗口，这是不需要的，但比程序等待什么特别的事情发生要好得多。所以我添加了如下的解决方法:

pProcess.StartInfo.WindowStyle = ProcessWindowStyle.Hidden

现在唯一让我困扰的是为什么在Windows 8下会出现这种情况。