友善三星2440的buttons_test程序，里面是这样的
for (;;) {
        char current_buttons[3];
        int count_of_changed_key;
        int i;
        
        if (read(buttons_fd, current_buttons, sizeof current_buttons) != sizeof current_buttons) {
            perror("read buttons:");
            exit(1);
        }

        for (i = 0, count_of_changed_key = 0; i < sizeof buttons / sizeof buttons[0]; i++) {
            if (buttons != current_buttons) {
                buttons = current_buttons;
                printf("%skey %d is %s", count_of_changed_key? ", ": "", i+1, buttons == '0' ? "up" : "down");
                count_of_changed_key++;
            }
        }
        if (count_of_changed_key) {
            printf("\n");
        }
        printf("ok\n");    //这是我添加的，为什么不是一直打印OK？？？？？
        
    }

只打印了一次，然后按键才打印一次？？？不是死循环来的吗？

>read(buttons_fd, current_buttons, sizeof current_buttons)
这个读取按键的接口可能会阻塞等待按键输入？（猜测）

就是那里的read问题，等待中断发生。

这是阻塞型IO，参考《Linux Device Driver》第三版 6.2节"阻塞I/O"的描述