Question

#include <unistd.h>
#include <sys/types.h>
#include <sys/wait.h>
#include "display.h"
#include <sys/ipc.h>
#include <sys/sem.h>

int main()
{
   int i;
   struct sembuf up = {0,1,0};
   struct sembuf down = {0,-1,0};
   int sem0 = semget(IPC_PRIVATE,1,0600);
   int sem1 = semget(IPC_PRIVATE,1,0600);
   if (fork())
   {
       for (i=0;i<10;i++)
           display("ab");
           semop(sem1,&up,1);
           semop(sem0,&down,1);
           wait(NULL);
   }
   else
   {
        for (i=0;i<10;i++)
           semop(sem1,&down,1);
           display("cd\n");
           semop(sem0,&up,1);
           semop(sem1,&down,1);
   }
   semctl(sem0,0,IPC_RMID);
   semctl(sem1,0,IPC_RMID);
   return 0;
}

我希望输出为
     ABCD
     ABCD
     ABCD
     ...
而不是没有信号量的混乱输出。上面的代码是我到目前为止所做的，但它似乎没有用，因为我没有正确地设置起伏。我是整个进程同步和互斥主题的新手，这是练习的一部分，因此任何非信号量相关的代码都无法更改。如果你能详细说明你的答案会很棒。

编辑：如果需要，可以从头文件中获取显示功能。

void display(char *str)
{
  char *p;
  for (p=str; *p; p++)
  {
    write(1, p, 1);
    usleep(100);
  }
}

Answer 1

您的代码中存在一些问题，请参阅代码中的注释。

int main()
{
   int i;
   struct sembuf up = {0,1,0};
   struct sembuf down = {0,-1,0};
   int sem0 = semget(IPC_PRIVATE,1,0600);
   int sem1 = semget(IPC_PRIVATE,1,0600);
   if (fork())
   {
       for (i=0;i<10;i++)
       {
           display("ab");
           semop(sem1,&up,1);
           semop(sem0,&down,1);
           //wait(NULL); //MAYUR: wait() suspends the execution until one of its children terminates. But this was unnecessary here. I removed it. 
       } //MAYUR: There was no brackets for "for loop". Added it in both the blocks.
   }
   else
   {
        for (i=0;i<10;i++)
        {
           semop(sem1,&down,1);
           display("cd\n");
           semop(sem0,&up,1);
           //semop(sem1,&down,1); //MAYUR: You are doing wait for sem1 in two places. This is was leading to deadlock. Removed it.
        }
   }
   semctl(sem0,0,IPC_RMID);
   semctl(sem1,0,IPC_RMID);
   return 0;
}

输出：

abcd
abcd
abcd
abcd
abcd
abcd
abcd
abcd
abcd
abcd

使用sys信号量输出同步

1 个答案: