In a service routine, if copymsg(9F) fails (line 6), the message is put back on the queue (line 7) and a routine, tryagain, is scheduled to be run in one tenth of a second. Then the service routine returns.

When the timeout(9F) function runs, if there is no message on the front of the queue, it just returns. Otherwise, for each message block in the first message, check to see if an allocation would succeed. If the number of message blocks equals the number we can allocate, then enable the service procedure. Otherwise, reschedule tryagain to run again in another tenth of a second. Note that tryagain is merely an approximation. Its accounting may be faulty. Consider the case of a message comprised of two 1024-byte message blocks. If there is only one free 1024-byte message block and no free 2048-byte message blocks, then testb() will still succeed twice. If no message blocks are freed of these sizes before the service procedure runs again, then the copymsg(9F) will still fail. The reason testb() is used here is because it is significantly faster than calling copymsg. We must minimize the amount of time spent in a timeout() routine.

1 xxxsrv(q) 2 queue_t *q; 3 { 4 mblk_t *mp; 5 mblk_t *nmp; . . . 6 if ((nmp = copymsg(mp)) == NULL) { 7 putbq(q, mp); 8 timeout(tryagain, (intptr_t)q, drv_usectohz(100000)); 9 return; 10 } . . . 11 } 12 13 tryagain(q) 14 queue_t *q; 15 { 16 register int can_alloc = 0; 17 register int num_blks = 0; 18 register mblk_t *mp; 19 20 if (!q->q_first) 21 return; 22 for (mp = q->q_first; mp; mp = mp->b_cont) { 23 num_blks++; 24 can_alloc += testb((mp->b_datap->db_lim - 25 mp->b_datap->db_base), BPRI_MED); 26 } 27 if (num_blks == can_alloc) 28 qenable(q); 29 else 30 timeout(tryagain, (intptr_t)q, drv_usectohz(100000)); 31 }