I have an application that creates two threads. (thread_1 for a Qt GUI and thread_2 for an app that runs a TCL interpreter).
我有一个创建两个线程的应用程序。 （用于Qt GUI的thread_1和用于运行TCL解释器的应用程序的thread_2）。
I want thread_1 (Qt GUI) to create a command and send it to thread_2 (TCL interpreter).
我想要thread_1（Qt GUI）创建命令并将其发送到thread_2（TCL解释器）。
I'm thinking of connecting thread_1's stdout to thread_2's stdin, and I don't know how to do it ? if you know how to do it or can suggest different way of work, I'd appreciate your help.
我正在考虑将thread_1的stdout连接到thread_2的stdin，我不知道该怎么做？如果你知道如何做或者可以提出不同的工作方式，我将非常感谢你的帮助。
2 个解决方案
#1
1
The solution I propose requires to set up 2 std::queue<> or std::list to let each thread pass a message to the other one and vice versa. The simplest way is to have each thread setup its own incoming message queue, and let other threads get a pointer to it. First you need a synchronized version of the queue datatype: as I gave in the comment, there's an implementation there.
我建议的解决方案需要设置2 std :: queue <>或std :: list让每个线程将消息传递给另一个，反之亦然。最简单的方法是让每个线程都设置自己的传入消息队列，让其他线程获得指向它的指针。首先，您需要队列数据类型的同步版本：正如我在评论中给出的，那里有一个实现。
Then you only need to upgrade your thread class (or runnable class, or whatever you're using as an abstraction of a task) with one such queue available internally, and a send method publicly accessible so that other tasks may post a message to it. Your task will then have to periodically check that queue for incoming message, and eventually process it.
然后你只需要升级你的线程类（或者可运行的类，或者你用作任务的抽象的任何东西），内部有一个这样的队列，一个可公开访问的发送方法，以便其他任务可以向它发送消息。然后，您的任务必须定期检查该队列是否有传入消息，并最终处理它。
NB: I got that page from stack overflow itself, since the blog owner is a member of this community. See that page talking about queue synchronization issue.
注意：我从堆栈溢出本身获得了该页面，因为博客所有者是该社区的成员。请参阅该页面，讨论队列同步问题。
#2
0
I am not sure why you would like to go through standard input and output here, but I think the issue might be much simpler than you think. I would just personally use the qt signal-slot mechanism as follows:
我不确定你为什么要在这里通过标准输入和输出，但我认为问题可能比你想象的要简单得多。我个人会使用qt信号槽机制，如下所示：
connect(guiThreadSender, SIGNAL(sendCommand(const QByteArray&)),
tclThreadReceiver, SLOT(handleCommand(const QByteArray&)));
#1
1
The solution I propose requires to set up 2 std::queue<> or std::list to let each thread pass a message to the other one and vice versa. The simplest way is to have each thread setup its own incoming message queue, and let other threads get a pointer to it. First you need a synchronized version of the queue datatype: as I gave in the comment, there's an implementation there.
我建议的解决方案需要设置2 std :: queue <>或std :: list让每个线程将消息传递给另一个，反之亦然。最简单的方法是让每个线程都设置自己的传入消息队列，让其他线程获得指向它的指针。首先，您需要队列数据类型的同步版本：正如我在评论中给出的，那里有一个实现。
Then you only need to upgrade your thread class (or runnable class, or whatever you're using as an abstraction of a task) with one such queue available internally, and a send method publicly accessible so that other tasks may post a message to it. Your task will then have to periodically check that queue for incoming message, and eventually process it.
然后你只需要升级你的线程类（或者可运行的类，或者你用作任务的抽象的任何东西），内部有一个这样的队列，一个可公开访问的发送方法，以便其他任务可以向它发送消息。然后，您的任务必须定期检查该队列是否有传入消息，并最终处理它。
NB: I got that page from stack overflow itself, since the blog owner is a member of this community. See that page talking about queue synchronization issue.
注意：我从堆栈溢出本身获得了该页面，因为博客所有者是该社区的成员。请参阅该页面，讨论队列同步问题。
#2
0
I am not sure why you would like to go through standard input and output here, but I think the issue might be much simpler than you think. I would just personally use the qt signal-slot mechanism as follows:
我不确定你为什么要在这里通过标准输入和输出，但我认为问题可能比你想象的要简单得多。我个人会使用qt信号槽机制，如下所示：
connect(guiThreadSender, SIGNAL(sendCommand(const QByteArray&)),
tclThreadReceiver, SLOT(handleCommand(const QByteArray&)));