如果我有两个 Glib::IOChannel 实例,它们会阻塞,直到都写入。 这样做的正确方法是什么?

发布于 2024-07-25 15:38:26 字数 2430 浏览 4 评论 0原文

我修改了此处<的示例/a> 使用两个 io 通道。 在我写入两个通道之前,似乎没有调用任何回调。 之后,在写入 fifo 时会单独调用它们。 我是不是忘记了什么?

  • 在一个 shell 窗口中启动测试程序。
  • 写入 echo "abc" > 第二个 shell 窗口中的 testfifo1。 -> 什么都没发生。
  • 写入 echo "def" > testfifo2 在第三个 shell 窗口中。 -> 现在我将“abc”和“def”
  • 写入fifo之一。 这是立即送达的。

编辑: 正如 Gormley 在下面所暗示的,解决方案是缺乏非阻塞。

read_fd1 = open("testfifo1", O_RDONLY | O_NONBLOCK);
...
read_fd2 = open("testfifo2", O_RDONLY | O_NONBLOCK);

对下面代码的更改使其立即做出响应。

代码:

#include <gtkmm/main.h>
#include <fcntl.h>
#include <iostream>

int read_fd1, read_fd2;
Glib::RefPtr<Glib::IOChannel> iochannel1, iochannel2;

// Usage: "echo "Hello" > testfifo<1|2>", quit with "Q"

bool MyCallback1(Glib::IOCondition io_condition)
{
    Glib::ustring buf;
    iochannel1->read_line(buf);
    std::cout << "io 1: " << buf;

    if (buf == "Q\n")
        Gtk::Main::quit();

    return true;
}

bool MyCallback2(Glib::IOCondition io_condition)
{
    Glib::ustring buf;
    iochannel2->read_line(buf);
    std::cout << "io 2: " << buf;

    if (buf == "Q\n")
        Gtk::Main::quit();

    return true;
}

int main(int argc, char *argv[])
{
    // the usual Gtk::Main object
    Gtk::Main app(argc, argv);

    if (access("testfifo1", F_OK) == -1)
    {
        if (mkfifo("testfifo1", 0666) != 0)
            return -1;
    }

    if (access("testfifo2", F_OK) == -1)
    {
        if (mkfifo("testfifo2", 0666) != 0)
            return -1;
    }

    read_fd1 = open("testfifo1", O_RDONLY);

    if (read_fd1 == -1)
        return -1;

    read_fd2 = open("testfifo2", O_RDONLY);

    if (read_fd2 == -1)
        return -1;

    Glib::signal_io().connect(sigc::ptr_fun(MyCallback1), read_fd1, Glib::IO_IN);
    Glib::signal_io().connect(sigc::ptr_fun(MyCallback2), read_fd2, Glib::IO_IN);

    iochannel1 = Glib::IOChannel::create_from_fd(read_fd1);
    iochannel2 = Glib::IOChannel::create_from_fd(read_fd2);

    app.run();

    if (unlink("testfifo1"))
        std::cerr << "error removing fifo 1" << std::endl;

    if (unlink("testfifo2"))
        std::cerr << "error removing fifo 2" << std::endl;

    return 0;
}

I have modified the example found here to use two io channels. None of the callbacks seem to be called before I have written to both channels. After that they are called individually when writing to the fifos. Am I forgetting something?

  • Start the test program in one shell window.
  • Write echo "abc" > testfifo1 in second shell window. -> nothing happens.
  • Write echo "def" > testfifo2 in a third shell window. -> now I get "abc" and "def"
  • Write to one of the fifos. This is immediately served.

Edit:
The solution, as hinted by Gormley below, was the lack of nonblock.

read_fd1 = open("testfifo1", O_RDONLY | O_NONBLOCK);
...
read_fd2 = open("testfifo2", O_RDONLY | O_NONBLOCK);

This change to the code below made it respond immediately.

The code:

#include <gtkmm/main.h>
#include <fcntl.h>
#include <iostream>

int read_fd1, read_fd2;
Glib::RefPtr<Glib::IOChannel> iochannel1, iochannel2;

// Usage: "echo "Hello" > testfifo<1|2>", quit with "Q"

bool MyCallback1(Glib::IOCondition io_condition)
{
    Glib::ustring buf;
    iochannel1->read_line(buf);
    std::cout << "io 1: " << buf;

    if (buf == "Q\n")
        Gtk::Main::quit();

    return true;
}

bool MyCallback2(Glib::IOCondition io_condition)
{
    Glib::ustring buf;
    iochannel2->read_line(buf);
    std::cout << "io 2: " << buf;

    if (buf == "Q\n")
        Gtk::Main::quit();

    return true;
}

int main(int argc, char *argv[])
{
    // the usual Gtk::Main object
    Gtk::Main app(argc, argv);

    if (access("testfifo1", F_OK) == -1)
    {
        if (mkfifo("testfifo1", 0666) != 0)
            return -1;
    }

    if (access("testfifo2", F_OK) == -1)
    {
        if (mkfifo("testfifo2", 0666) != 0)
            return -1;
    }

    read_fd1 = open("testfifo1", O_RDONLY);

    if (read_fd1 == -1)
        return -1;

    read_fd2 = open("testfifo2", O_RDONLY);

    if (read_fd2 == -1)
        return -1;

    Glib::signal_io().connect(sigc::ptr_fun(MyCallback1), read_fd1, Glib::IO_IN);
    Glib::signal_io().connect(sigc::ptr_fun(MyCallback2), read_fd2, Glib::IO_IN);

    iochannel1 = Glib::IOChannel::create_from_fd(read_fd1);
    iochannel2 = Glib::IOChannel::create_from_fd(read_fd2);

    app.run();

    if (unlink("testfifo1"))
        std::cerr << "error removing fifo 1" << std::endl;

    if (unlink("testfifo2"))
        std::cerr << "error removing fifo 2" << std::endl;

    return 0;
}

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不离久伴 2024-08-01 15:38:26

这两条语句阻止程序进入主循环,直到两个 fifo
开放供写入。 fifos 阻塞,直到双方都连接

iochannel1 = Glib::IOChannel::create_from_fd(read_fd1);
iochannel2 = Glib::IOChannel::create_from_fd(read_fd2);

These two statements block the program from getting into the main loop until both fifos
are open for write. fifos block until both sides are connected

iochannel1 = Glib::IOChannel::create_from_fd(read_fd1);
iochannel2 = Glib::IOChannel::create_from_fd(read_fd2);

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