epoll_wait（）阻止打印到标准输出

Question

使用epoll_wait ，即使我在调用与epoll相关的任何内容之前尝试打印，它似乎epoll_wait “吃掉”所有已写入stdout并延迟打印，直到epoll_wait接收到一个事件为止（甚至可能在我的开始时主要方法，但仍无法打印）。

在epoll_wait收到事件之后epoll_wait显示的打印示例：

printf("This doesn't get printed. ");
fprintf(stdout, "This doesn't get printed either.");
ev.events = EPOLLIN;
ev.data.fd = some_sock_fd; // Same with STDIN_FILENO
if (epoll_ctl(epoll_fd, EPOLL_CTL_ADD, some_sock_fd, &ev) == -1) {
    perror("epoll_ctl");
    exit(EXIT_FAILURE);
}

for (;;) {
    rc = epoll_wait(epoll_fd, &ev, 1, -1);
    // This is where it gets printed

写入stderr可以正常工作，但是如何写入stdout呢？ 如何防止epoll_wait阻止打印到stdout ？

Answer 1

该问题似乎与epoll_wait无关。 以下是违规代码的摘要：

// Since there's no newline, the following stays in the buffer
printf("Some print without newline.");

for (;;) {
    // At this point, the buffer has not been flushed,
    // and epoll_wait blocks the output
    rc = epoll_wait(epoll_fd, &ev, 1, -1);

使用fflush(stdout) 是此代码的解决方案，因为缓冲与epoll_wait无关，但与用户空间缓冲stdout的方式无关：

// Since there's no newline, the following stays in the buffer
printf("Some print without newline.");

// Forces a write of user-space buffered data for stdout
fflush(stdout);

for (;;) {
    // At this point, the buffer has not been flushed,
    // and epoll_wait blocks the output
    rc = epoll_wait(epoll_fd, &ev, 1, -1);

综上所述，这似乎是在错误的地方寻找本应显而易见的问题的情况。