epoll_wait（）阻止打印到標准輸出

Question

使用epoll_wait ，即使我在調用與epoll相關的任何內容之前嘗試打印，它似乎epoll_wait “吃掉”所有已寫入stdout並延遲打印，直到epoll_wait接收到一個事件為止（甚至可能在我的開始時主要方法，但仍無法打印）。

在epoll_wait收到事件之后epoll_wait顯示的打印示例：

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

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

Answer 1

該問題似乎與epoll_wait無關。 以下是違規代碼的摘要：

// 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) 是此代碼的解決方案，因為緩沖與epoll_wait無關，但與用戶空間緩沖stdout的方式無關：

// 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);

綜上所述，這似乎是在錯誤的地方尋找本應顯而易見的問題的情況。