Why I dont see a msg from message queue? - c

Question

I would like to understand how message queues in Unix work. I wrote a simple code which sends a short message to queue and then I can read that message. But my code shows :
And I dont know why - and I cant see a message I send to queue. Heres my code:
#include <sys/types.h>
#include <sys/ipc.h>
#include <sys/msg.h>
#include <stdio.h>
#include <stdlib.h>
#include <unistd.h>
#include <string.h>
#include <fcntl.h>
struct mymsgbuf {
long mtype;
char mtext[1024];
}msg;
int send_message(int qid, struct mymsgbuf *buffer )
{
int result = -1, length = 0;
length = sizeof(struct mymsgbuf) - sizeof(long);
if((result = msgsnd(qid, buffer, length, 0)) == -1)
return -1;
return result;
}
int read_message(int qid, long type, struct mymsgbuf *buffer)
{
int result, length;
length = sizeof(struct mymsgbuf) - sizeof(long);
if((result = msgrcv(qid, buffer, length, type, 0)) == -1)
return -1;
printf("Type: %ld Text: %s\n", buffer->mtype, buffer->mtext);
return result;
}
int main(int argc, char **argv)
{
int buffsize = 1024;
int qid = msgget(ftok(".", 0), IPC_CREAT | O_EXCL);
if (qid == -1)
{
perror("msgget");
exit(1);
}
msg.mtype = 1;
strcpy(msg.mtext, "my simple msg");
if((send_message(qid, &msg)) == -1)
{
perror("msgsnd");
exit(1);
}
if((read_message(qid, 1, &msg) == -1))
{
perror("msgrcv");
exit(1);
}
return 0;
}
When I changed a line with msgget for this line:
int qid = msgget(ftok(".", 0), IPC_CREAT | O_EXCL | 0600);
it shows:

Answer 1

From the documentation for msgget:
The low-order 9 bits of msg_perm.mode shall be set equal to the low-order 9 bits of msgflg.
You need to add some permissions to your queue, at least read and write. Do something like:
int qid = msgget(ftok(".", 0), IPC_CREAT | O_EXCL | 0600);