c - 收到警报信号时，read() 系统调用不会失败

Question

这是我的(部分)服务器端代码

void timeout_handler(int value) {
    printf("Handler\n");
    return;
}

int main (int argc, char **argv) {
    [...]
    signal(SIGALRM, timeout_handler);
    alarm(seconds);
    int result = read(input_socket, buffer, sizeof(buffer));
    if (result == -1 && errno == EINTR) {
        printf("read() failed\n");
    }
    [...]
}

其中 input_socket 是与客户端正确连接的 TCP 套接字(如果我从客户端发送数据，服务器会接收它们)。

作为警报信号的测试，我尝试只打开并连接套接字客户端而不发送任何数据。我希望输出像

Handler
read() failed

但结果只是Handler消息，然后进程依然活跃。

为什么 read() 不会因 errno=EINTR 而失败？

Answer 1

同样在 OSX 上，某些系统调用在被信号中断时默认重新启动。

siginterrupt() 可用于更改此行为。下面一行(在 alarm() 被调用之前的某处)应该让程序按照 OP 的预期运行:

siginterrupt(SIGALRM, 1);

---

来自 OSX 文档(由我强调):

For some system calls, if a signal is caught while the call is executing and the call is prematurely terminated, the call is automatically restarted. Any handler installed with signal(3) will have the SA_RESTART flag set, meaning that any restartable system call will not return on receipt of a signal. The affected system calls include read(2), write(2), sendto(2), recvfrom(2), sendmsg(2), and recvmsg(2) on a communications channel or a low speed device and during a ioctl(2) or wait(2). However, calls that have already committed are not restarted, but instead return a partial success (for example, a short read count). These semantics could be changed with siginterrupt(3).

这在 Linux 上是不同的，例如，在 Linux 上重新启动可能被中断的系统调用需要通过发送的信号明确请求。