Ubuntu的信号量：分段故障（核心转储） [英] Ubuntu Semaphore: Segmentation fault (core dumped)

问题描述

我使用Ubuntu现在学习信号灯的温度。教授只是把我们这个code和要求我们对其进行研究和观察。当我整理我得到的ctime警告（安培; sem_buf.sem_ctime）返回 INT ，而不是的char * ，但没有重大。当我运行它的输出就是：信号灯标识符：0分割故障（核心转储）。我作为什么地方出了错很迷茫，我不知道在这个code回事。一些帮助将非常AP preciated。

下面是code：

 的#include＆LT; SYS / types.h中＆GT;
＃包括LT＆; SYS / ipc.h＆GT;
＃包括LT＆; SYS /＆sem.h中GT;
＃包括LT＆;＆stdio.h中GT;
＃包括LT＆;＆stdlib.h中GT;
＃包括LT＆;＆fcntl.h GT;
＃包括LT＆;＆unistd.h中GT;
＃包括LT＆;＆semaphore.h GT;
＃定义NS 3
工会semun {
   INT VAL;
   结构为semid_ds * BUF;
   USHORT *阵列; //无符号短整型。
};INT主要（无效）
{
   INT sem_id，sem_value我;
   的key_t ipc_key;
   结构为semid_ds sem_buf;
   静态USHORT sem_array [NS] = {3,1，4};
   工会semun ARG;
   ipc_key = ftok（，S。）; //创建密钥。
   / *创建信号量* /
   如果（（sem_id = semget子（ipc_key，NS，IPC_CREAT | 0666））== -1）{
      PERROR（了semget：IPC | 0666）;
      出口（1）;
   }
   的printf（旗语识别％d个\\ N，sem_id）;
   / *设置ARG（工会）的存储位置为*地址/
   / *返回的semid_ds值* /
   arg.buf =安培; sem_buf;
   如果（了semctl（sem_id，0，IPC_STAT，ARG）== -1）{
      PERROR（了semctl：IPC_STAT）;
      出口（2）;
   }
   的printf（创建％S的ctime（安培; sem_buf.sem_ctime））;
   / *设置ARG（工会）来初始化向量*的地址/
   arg.array = sem_array;
   如果（了semctl（sem_id，0，SETALL，ARG）== -1）{
      PERROR（了semctl：SETALL）;
      出口（3）;
   }
   对于（i = 0; I＆LT; NS; ++ I）{
      如果（（sem_value =了semctl（sem_id，I，GETVAL，0））==  -  1）{
         PERROR（了semctl：GETVAL）;
         出口（4）;
      }
      的printf（旗语％d个有％d \\ n的值，我，sem_value）;
   }
   / *删除信号量* /
   如果（和semctl（sem_id，0，IPC_RMID，0）==  -  1）{
      PERROR（了semctl：IPC_RMID）;
      出口（5）;
   }
}
 

解决方案

您需要包括 time.h中编译器识别的ctime 功能。该警告是因为编译器不知道的ctime 是一个函数，返回一个的char * 。默认情况下GCC假设未知函数返回一个 INT 。

I am learning semaphores in C using Ubuntu right now. The professor just throw us this code and ask us to study it and observe. When I compiled I get a warning that ctime(&sem_buf.sem_ctime) returns an int, not a char * but nothing major. When I run it the output is just: Semaphore identifier: 0 Segmentation fault (core dumped). I am very confused as of what went wrong and I have no idea what is going on in this code. Some help would be very much appreciated.

Here is the code:

#include <sys/types.h>
#include <sys/ipc.h>
#include <sys/sem.h>
#include <stdio.h>
#include <stdlib.h>
#include <fcntl.h>
#include <unistd.h>
#include <semaphore.h> 
# define NS 3
union semun { 
   int val; 
   struct semid_ds *buf; 
   ushort *array; // Unsigned short integer.
}; 

int main(void) 
{ 
   int sem_id, sem_value, i; 
   key_t ipc_key; 
   struct semid_ds sem_buf; 
   static ushort sem_array[NS] = {3, 1, 4}; 
   union semun arg; 
   ipc_key = ftok(".", 'S'); // Creating the key.
   /* Create semaphore */ 
   if ((sem_id = semget(ipc_key, NS, IPC_CREAT | 0666)) == -1) { 
      perror ("semget: IPC | 0666"); 
      exit(1); 
   } 
   printf ("Semaphore identifier %d\n", sem_id); 
   /* Set arg (the union) to the address of the storage location for */ 
   /* returned semid_ds value */ 
   arg.buf = &sem_buf; 
   if (semctl(sem_id, 0, IPC_STAT, arg) == -1) { 
      perror ("semctl: IPC_STAT"); 
      exit(2); 
   } 
   printf ("Create %s", ctime(&sem_buf.sem_ctime)); 
   /* Set arg (the union) to the address of the initializing vector */ 
   arg.array = sem_array; 
   if (semctl(sem_id, 0, SETALL, arg) == -1) { 
      perror("semctl: SETALL"); 
      exit(3); 
   } 
   for (i=0; i<NS; ++i) { 
      if ((sem_value = semctl(sem_id, i, GETVAL, 0)) == -1) { 
         perror("semctl : GETVAL"); 
         exit(4); 
      } 
      printf ("Semaphore %d has value of %d\n",i, sem_value); 
   } 
   /*remove semaphore */ 
   if (semctl(sem_id, 0, IPC_RMID, 0) == -1) { 
      perror ("semctl: IPC_RMID"); 
      exit(5); 
   } 
} 

解决方案

You need to include time.h to the compiler recognize ctime function. The warning is because the compiler don't know ctime is a function and that returns an char*. By default GCC assumes the unknown function returns an int.

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