Linux環境下のでセマフォ

Linux環境でセマフォを使いたいと思い，下記のサンプルコードを実行しようと思ったのですが「inivalの記憶域が不明です」と言われ実行できません。
サンプルコードなので間違いはないはずなのですが，何か環境に不備があるのでしょうか?

#コード

C
1#include <stdio.h>
2#include <stdlib.h>
3#include <sys/types.h>
4#include <sys/ipc.h>
5#include <sys/shm.h>
6#include <sys/sem.h>
7
8int main (int argc, char *argv [])
9{
10	int id, tmp , sid, i;
11	int *counter;
12	int num;
13	union semun inival;
14	struct sembuf semlock;
15	struct shmid_ds buf;
16
17	semlock.sem_num=0;
18	semlock.sem_flg=0;
19	
20	printf("test");
21
22	srand((unsigned) time (NULL));
23	
24	if (argc == 1)
25	{
26		if ((id = shmget (IPC_PRIVATE, sizeof (int), IPC_CREAT|0666)) == -1)
27		{
28			perror ("shmget");
29			exit(EXIT_FAILURE);
30		}
31		if ((sid = semget (id, 1,IPC_CREAT | 0700)) == -1)
32		{
33			perror ("semget");
34			exit (EXIT_FAILURE);
35		}
36		inival.val = 1;
37		semctl (sid, 0, SETVAL, inival);
38		printf("shared memory ID = %d\n", id);
39	}
40	else if (argc == 2)
41	{
42		id = atoi (argv[1]);
43		if((sid = semget (id, 1,0)) == -1)
44		{
45			perror ("semget");
46			exit (EXIT_FAILURE);
47		}
48	}
49	printf ("Input the number to add. >");
50	scanf ("%d", &num) ;
51	
52	counter = (int *) shmat (id,0,0);
53	if (argc == 1) *counter = 0;
54	
55	for (i = 0; i < 3; i++)
56	{
57		semlock.sem_num=0;
58		semlock.sem_op=-1;
59		semlock.sem_flg=0;
60		if (semop (sid, & semlock, 1) < 0) break;
61											/* セマフォを1減算(ロック) */
62		tmp = *counter;
63		fprintf (stderr, "counter=%d", tmp);
64		sleep(rand()%4);
65		tmp += num; *counter = tmp;
66		semlock.sem_num=0;
67		semlock.sem_op=1;
68		semlock.sem_flg=0;
69		if (semop(sid, &semlock, 1) < 0) break;
70										/* セマフォを1加算 (アンロック) */
71		fprintf (stderr," -> %d\n", *counter);
72		sleep(rand () % 4);
73	}
74	
75	shmctl (id, IPC_STAT, &buf);
76		
77	if (shmdt (counter) == -1) perror ("shmdt");
78	if (buf.shm_nattch == 1)
79	{
80		if (shmctl (id, IPC_RMID,0) == -1) perror ("chmctl");
81	}
82	
83	if (semctl (sid,IPC_RMID,0) == -1) perror ("semctl") ;
84}