产卵过程，但只有5在一个时间 [英] Spawn processes, but only 5 at a time

问题描述

我工作了与文件名的队列。每个文件必须由一外部二进制处理。这工作得很好，但它只能处理一个文件的时候。难道是可能的双平行产卵多个进程？

i'm working off a queue with filenames. Each file has to be processed by a external binary. This works fine, but it only processes one file at a time. Is it possible two spawn a number of processes parallel?

Queue<string> queue = new Queue<string>();
queue.Enqueue("1.mp3");
queue.Enqueue("2.mp3");
queue.Enqueue("3.mp3");
...
queue.Enqueue("10000.mp3");

while (queue.Count > 0)
{
    string file = queue.Dequeue();

    Process p = new Process();    
    p.StartInfo.FileName = @"binary.exe";
    p.StartInfo.Arguments = file;
    p.StartInfo.UseShellExecute = false;
    p.StartInfo.CreateNoWindow = true;
    p.StartInfo.RedirectStandardOutput = true;
    p.Start();
    p.WaitForExit();
}

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更新：我喜欢从亚历克斯LE（每次产卵的过程，但只有5）的解决方案，但有可能等待子进程要退出由本·福格特建议

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Update: I like the solution from Alex LE (Spawn processes, but only 5 at a time), but is it possible to wait for the child processes to exit as suggested by Ben Voigt?

编辑1：我需要检查p.Exit code == 0，使一些数据库更新

Edit 1: i need to check for p.ExitCode == 0 to make some database updates.

推荐答案

下面就是应该是可能的，如果与进程相关联的等待句柄被打上公众，而不是内部的，因为它是目前<一个href=\"https://connect.microsoft.com/VisualStudio/feedback/details/628663/expose-waithandle-from-system-diagnostics-process\"相对=nofollow>（投票在这里问微软更改）：

Here's what should have been possible, if the wait handle associated with the process was marked public instead of internal as it currently is (vote here to ask Microsoft to change that):

void BatchProcess()
{
    Queue<string> queue = new Queue<string>();
    queue.Enqueue("1.mp3");
    queue.Enqueue("2.mp3");
    queue.Enqueue("3.mp3");
    ...
    queue.Enqueue("10000.mp3");

    WaitHandle[] subprocesses = new WaitHandle[Math.Min(queue.Count, 5)];
    for( int i = 0; i < subprocesses.Length; i++ ) {
        subprocesses[i] = Spawn(queue.Dequeue());
    }

    while (queue.Count > 0) {
        int j = WaitHandle.WaitAny(subprocesses);
        subprocesses[j].Dispose();
        subprocesses[j] = Spawn(queue.Dequeue());
    }

    WaitHandle.WaitAll(subprocesses);
    foreach (var wh in subprocesses) {
        wh.Dispose();
    }
}

ProcessWaitHandle Spawn(string args)
{
    using (Process p = new Process()) {
        p.StartInfo.FileName = @"binary.exe";
        p.StartInfo.Arguments = args;
        p.StartInfo.UseShellExecute = false;
        p.StartInfo.CreateNoWindow = true;
        p.Start();
        return p.WaitHandle;
    }
}

这将是最有效的解决方案成为可能，因为不需要同步对象除了Win32的过程本身。有在C＃code不需要额外的线程，没有异步方法调用，因此任何不需要的锁定或其他同步。

This would be the most efficient solution possible, because no synchronization objects are needed besides the Win32 processes themselves. There are no extra threads needed in the C# code and no asynchronous method invocations, therefore no locking or other synchronization is needed whatsoever.

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