信号警报过早失效

时间:2018-10-02 03:58:53

标签: c udp signals go-back-n

我有一个signal,在其中设置了回调处理程序,然后在函数中调用alarm(1),这样我的程序将在1秒钟后超时。超时后,我希望它重试相同的阻塞调用recvfrom(),直到设置为MAX_ATTEMPTS的5。

问题:

它仅重试2次,然后退出程序。知道可能出什么问题了吗?

/* declate signal for setting alarm */
signal(SIGALRM, timeout_hdler);
int attempts = 0;

while(1) {
    if(attempts > MAX_ATTEMPTS) {
        printf("Connection is not working, EXITING program");
        s.curr_state = CLOSED;
        /* we assume connection is failing so we shut down */
        exit(-1);
    }

    if (s.curr_state == CLOSED) {

        printf("Sending SYN_packet with seqnum: %d...\n", SYN_packet->seqnum);

        if (sendto(sockfd, SYN_packet, sizeof(SYN_packet), 0, server, socklen) == -1) {
            perror("Sendto error");
            exit(-1);
        }

        s.curr_state = SYN_SENT;

        printf("Current state SYN_SENT: %d\n", s.curr_state);
    }

    if (s.curr_state == SYN_SENT) {



        alarm(1);
        attempts++;
        printf("\nAttempt number: %d\n", attempts);
        printf("Waiting for SYNACK_packet...\n");

        if (recvfrom(
                sockfd, SYNACK_packet, sizeof(*SYNACK_packet), 0, (struct sockaddr *) &server, &socklen) == -1)
        {
            if (errno != EINTR) {
                perror("Recvfrom SYNACK_packet error\n");
                s.curr_state = CLOSED;
                exit(-1);
            }
        }

        if ((SYNACK_packet->type == SYNACK) && (validate_packet(SYNACK_packet) == 1)) {

            printf("SYNACK_packet received\n");

            s.address = *(struct sockaddr *) &server;
            s.sock_len = socklen;
            s.curr_state = ESTABLISHED;
            s.seq_num = SYNACK_packet->seqnum;
            printf("Current state ESTABLISHED: %d\n", s.curr_state);
            return sockfd;
        }
    }
}

处理程序(除了打印外什么也不做):

void timeout_hdler(int signum) {
    printf("TIMEOUT has occured with signum: %d", signum);
}

这是我的控制台中的输出(来自printf语句):

In Connect() with socket: 4, server: 2, socklen: 16
Sending SYN_packet with seqnum: 67...
Current state SYN_SENT: 1

Attempt number: 1
Waiting for SYNACK_packet...
TIMEOUT has occured with signum: 14
Attempt number: 2
Waiting for SYNACK_packet...
Alarm clock

为什么仅2次尝试就退出程序?理想情况下,我希望它在关闭连接之前重试5次(这是使用UDP的Go Back N实现)

更新

我通过在处理程序中重新安装signalsignal(SIGALRM, timeout_hdler);解决了该问题。但是为什么呢?我做错了吗?

void timeout_hdler(int signum) {
    printf("TIMEOUT has occured with signum: %d", signum);
    signal(SIGALRM, timeout_hdler);
}

1 个答案:

答案 0 :(得分:1)

signal的语义取决于操作系统和libc。但是根据您的特定情况下的输出,在首次调用处理程序函数后,信号处理程序处理程序将重置为默认值。如果再次触发信号(即Alarm clock),则默认值将导致程序退出,并在输出中看到退出。

来自the documentation of signal in Linux

  

signal()的唯一可移植用法是将信号的处置设置为          SIG_DFL或SIG_IGN。 ....
  在原始UNIX系统中,当建立处理程序时          使用signal()通过传递信号来调用,信号的处置将重置为SIG_DFL ,而系统没有          阻止其他信号实例的传递。 ...

换句话说:不要使用signal。相反,您应该使用sigaction

  

POSIX.1通过指定sigaction(2)解决了可移植性问题,          当信号处理程序处于以下状态时,提供对语义的显式控制          援引;使用该接口而不是signal()。