why is the semaphore not working?

Question

#include <stdio.h>
#include <sys/types.h>
#include <iostream>
#include <unistd.h>
#include <fstream>
#include <string>
#include <semaphore.h>

using namespace std;

int main(int argc, char *argv[]){
  int pshared = 1;
  unsigned int value = 0;
  sem_t sem_name;
  sem_init(&sem_name, pshared, value);

  int parentpid = getpid();
  pid_t  pid = fork();

  if (parentpid == getpid()){
    cout << "parent id= " << getpid() << endl;
    sem_wait(&sem_name);
    cout << "child is done." << endl;
  }

  if (parentpid != getpid()){
    cout << "child id= " << getpid() << endl;
    for (int i = 0; i < 10; i++)
      cout << i << endl;

    sem_post(&sem_name);
} 
  sleep(4);
  return 0; 
}

the result should be:

parent id 123456.
child id 123457.
0
1
2
3
4
5
6
7
8
9
child is done.

Program exits, but instead it never signals the semaphore.

Answer 1

XY Problem?

If what you want to do is to wait for your child process to be done, you do not need semaphores, but wait or waitpid. The following C code has your expected output.

#include <stdio.h>
#include <sys/types.h>
#include <unistd.h>
#include <sys/wait.h>

int main(void){
    pid_t  pid;
    pid = fork();
    if (pid < 0) {
        fprintf(stderr, "fork failed!\n");
        return 1;
    }
    if (pid == 0) {
        int i;
        printf("child id= %d\n", getpid());
        for (i = 0; i < 10; i++) {
            printf("%d\n",i);
        }
    }
    else {
        int status;
        printf("parent id= %d\n", getpid());
        waitpid(-1, &status, 0);
        printf("child is done\n");
    }
    return 0; 
}

NOTE: I did in C because the only C++ you were using was initial declaration in a for loop and prints with cout << "blah" << endl;

Answer 2

Instead of using sem_init(), use sem_open(). This is because the semaphore needs to be in shared address space rather than on the process stack where it gets duplicated through the fork().

#include <fcntl.h>
...

sem_t *sem_ptr;
sem_ptr = sem_open("my_semaphore", O_CREAT, 0644, value);
...

sem_wait(sem_ptr);
...

sem_post(sem_ptr);
...

Taken from http://blog.superpat.com/2010/07/14/semaphores-on-linux-sem_init-vs-sem_open/

Answer 3

From sem_init's manpage:

If pshared is nonzero, then the semaphore is shared between processes, and should be located in a region of shared memory (see shm_open(3), mmap(2), and shmget(2)). (Since a child created by fork(2) inherits its parent's memory mappings, it can also access the semaphore.) Any process that can access the shared memory region can operate on the semaphore using sem_post(3), sem_wait(3), etc.

POSIX semaphores are on-the-stack structs. They aren't reference-counted references to a kernel-maintained struct like filedescriptors are. If you want to share a POSIX semaphore with two processes, you need to take care of the sharing part yourself.

This should work:

#include <fstream>
#include <iostream>
#include <semaphore.h>
#include <stdio.h>
#include <string>
#include <sysexits.h>
#include <sys/mman.h>
#include <sys/types.h>
#include <unistd.h>


int main(int argc, char *argv[]){
  using namespace std;
  sem_t* semp = (sem_t*)mmap(0, sizeof(sem_t), PROT_READ|PROT_WRITE, MAP_ANONYMOUS|MAP_SHARED, 0, 0 );
  if ((void*)semp == MAP_FAILED) { perror("mmap");  exit(EX_OSERR); } 

  sem_init(semp, 1 /*shared*/, 0 /*value*/);

  pid_t  pid = fork();
  if(pid < 0) { perror("fork");  exit(EX_OSERR); } 

  if (pid==0){ //parent
    cout << "parent id= " << getpid() << endl;
    sem_wait(semp);
    cout << "child is done." << endl;
  }else { //child
    cout << "child id= " << getpid() << endl;
    for (int i = 0; i < 10; i++)
      cout << i << endl;
    sem_post(semp);
  } 
  return 0; 
}

Note: If you want just this behavior, then waitpid is obviously the way to go. I'm assuming what you want is to test out POSIX semaphores.