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我很难理解多进程系统中的 IPC。我有这个系统，其中有三个子进程向它们的进程组发送两种类型的信号。有四种类型的信号处理过程负责特定类型的信号。

有一个监视过程等待两个信号，然后进行相应的处理。当我运行这个程序一段时间后，监控进程似乎并没有像信号处理进程一样接收到信号。我可以在日志中看到信号只是在生成但根本没有处理。

我的代码如下

#include <cstdlib>
#include <iostream>
#include <iomanip>
#include <unistd.h>
#include <sys/types.h>
#include <sys/wait.h>
#include <sys/time.h>
#include <signal.h>
#include <unistd.h>
#include <fcntl.h>
#include <cstdio>
#include <stdlib.h>
#include <stdio.h>
#include <pthread.h>

using namespace std;

double timestamp() {
  struct timeval tp;
  gettimeofday(&tp, NULL);
  return (double)tp.tv_sec + tp.tv_usec / 1000000.;
}

double getinterval() { 
  srand(time(NULL));
  int r = rand()%10 + 1;
  double s = (double)r/100;
}

int count;
int count_1;
int count_2;
double time_1[10];
double time_2[10];


pid_t senders[1];
pid_t handlers[4];
pid_t reporter;

void catcher(int sig) {
  printf("Signal catcher called for %d",sig);
}

int main(int argc, char *argv[]) {

  void signal_catcher_int(int);

  pid_t pid,w;
  int status;

  if(signal(SIGUSR1, SIG_IGN) == SIG_ERR) {
    perror("1");
    return 1;
  }

  if(signal(SIGUSR2 ,SIG_IGN) == SIG_ERR) {
    perror("2");
    return 2;
  }

  if(signal(SIGINT,signal_catcher_int) == SIG_ERR) {
    perror("3");
    return 2;
  }

  //Registering the signal handler
  for(int i=0; i<4; i++) {
    if((pid = fork()) == 0) {
      cout << i << endl;
      //struct sigaction sigact;
      sigset_t sigset;
      int sig;
      int result = 0;

      sigemptyset(&sigset);

      if(i%2 == 0) {

        if(signal(SIGUSR2, SIG_IGN) == SIG_ERR) {
          perror("2");
          return 2;
        }

        sigaddset(&sigset, SIGUSR1);
        sigprocmask(SIG_BLOCK, &sigset, NULL);
      } else {
            if(signal(SIGUSR1, SIG_IGN) == SIG_ERR) {
          perror("2");
          return 2;
        }

        sigaddset(&sigset, SIGUSR2);
        sigprocmask(SIG_BLOCK, &sigset, NULL);
      }

      while(true) {
        int result = sigwait(&sigset, &sig);
        if(result == 0) {
          cout << "The caught signal is " << sig << endl;
        }
      }
      exit(0);

    } else {
      cout << "Registerd the handler " << pid << endl;
      handlers[i] = pid;
    }
  }

  //Registering the monitoring process
  if((pid = fork()) == 0) {
    sigset_t sigset;
    int sig;
    int result = 0;

    sigemptyset(&sigset);
    sigaddset(&sigset, SIGUSR1);
    sigaddset(&sigset, SIGUSR2);

    sigprocmask(SIG_BLOCK, &sigset, NULL);

    while(true) {
      int result = sigwait(&sigset, &sig);
      if(result == 0) {
        cout << "The monitored signal is " << sig << endl;
      } else {
        cout << "error" << endl;
      } 
    }

  } else {
    reporter = pid;
  }


  sleep(3);
  //Registering the signal generator
  for(int i=0; i<1; i++) {
    if((pid = fork()) == 0) {

      if(signal(SIGUSR1, SIG_IGN) == SIG_ERR) {
        perror("1");
        return 1;
      }

      if(signal(SIGUSR2, SIG_IGN) == SIG_ERR) {
        perror("2");
        return 2;
      }
      srand(time(0));
      while(true) {
        volatile int signal_id = rand()%2 + 1;
        cout << "Generating the signal " << signal_id << endl;
        if(signal_id == 1) {
          killpg(getpgid(getpid()), SIGUSR1);
        } else {
              killpg(getpgid(getpid()), SIGUSR2);
        }     
        int r = rand()%10 + 1;
        double s = (double)r/100;  
        sleep(s);
      }

      exit(0); 
    } else {
      cout << "Registered the sender " << pid << endl;
      senders[i] = pid;

    } 

  }


  while(w = wait(&status)) {
    cout << "Wait on PID " << w << endl;
  }

}



void signal_catcher_int(int the_sig) {
  //cout << "Handling the Ctrl C signal " << endl;
  for(int i=0; i<1; i++) {
    kill(senders[i],SIGKILL);
  }

  for(int i=0; i<4; i++) {
    kill(handlers[i],SIGKILL);
  }

  kill(reporter,SIGKILL);

  exit(3);
}

有什么建议么？

这是输出的示例

在一开始的时候

Registerd the handler 9544
Registerd the handler 9545
1
Registerd the handler 9546
Registerd the handler 9547
2
3
0
Registered the sender 9550
Generating the signal 1
The caught signal is 10
The monitored signal is 10
The caught signal is 10
Generating the signal 1
The caught signal is 10
The monitored signal is 10
The caught signal is 10
Generating the signal 1
The caught signal is 10
The monitored signal is 10
The caught signal is 10
Generating the signal 1
The caught signal is 10
The monitored signal is 10
The caught signal is 10
Generating the signal 2
The caught signal is 12
The caught signal is 12
The monitored signal is 12
Generating the signal 2
Generating the signal 2
The caught signal is 12
The caught signal is 12
Generating the signal 1
The caught signal is 12
The monitored signal is 10
The monitored signal is 12
Generating the signal 1
Generating the signal 2
The caught signal is 12
Generating the signal 1
Generating the signal 2
10
The monitored signal is 10
The caught signal is 12
Generating the signal 1
The caught signal is 12
The monitored signal is GenThe caught signal is TheThe caught signal is 10
Generating the signal 2

稍后的

The monitored signal is GenThe monitored signal is 10
Generating the signal 1
Generating the signal 2
The caught signal is 10
The caught signal is 10
The caught signal is 10
The caught signal is 12
Generating the signal 1
Generating the signal 2
Generating the signal 1
Generating the signal 1
Generating the signal 2
Generating the signal 2
Generating the signal 2
Generating the signal 2
Generating the signal 2
Generating the signal 1
The caught signal is 12
The caught signal is 10
The caught signal is 10
Generating the signal 2
Generating the signal 1
Generating the signal 1
Generating the signal 2
Generating the signal 1
Generating the signal 2
Generating the signal 2
Generating the signal 2
Generating the signal 1
Generating the signal 2
Generating the signal 1
Generating the signal 2
Generating the signal 2
The caught signal is 10
Generating the signal 2
Generating the signal 1
Generating the signal 1

正如您最初看到的那样，信号是由我的信号处理程序和监视进程生成和处理的。但后来产生了很多信号，但它的处理量并不像以前那样大。此外，我可以看到监控过程的信号处理非常少

任何人都可以提供一些见解。这是怎么回事？

4

1 回答 1

1

如果有多个相同类型的信号处于挂起状态，Linux 默认只提供一个这样的信号。这与sigwait 文档一致：

如果在调用 sigwait() 之前存在单个信号编号的多个未决实例，则在成功返回后是否还有该信号编号的任何剩余未决信号取决于实现。

所以你的程序的输出取决于调度器，如果 kill 被多次调用并且调度器没有同时唤醒监控进程，相同类型的信号将被合并为一个。

Linux 允许更改默认行为。

于 2012-09-27T17:24:26.007 回答