Lines Matching refs:is

2 	If ps_tq is scheduled, ps_tq_active is 1.  ps_tq_int() can be called
3 	only when ps_tq_active is 1.
8 	anything except B, since it is under if (!ps_tq_active) under
9 	ps_spinlock.  C is always preceded by B, since we can't reach it
11 	IOW, the sequence is A?(BA|BC|B)*.  OTOH, number of B can not exceed
12 	the sum of numbers of A and C, since each call of ps_tq_int() is
14 	A and each B is preceded by either A or C.  Moments when we enter
18 	In other words, the sequence of events is (A or C set ps_tq_active to
19 	1 and schedule ps_tq, ps_tq is executed, ps_tq_int() is entered,
25 	  case when pd_req is NULL.
51 	2) When there is such a thread, pd_busy is set or pd_lock is held by
53 	3) When there is such a thread, ps_tq_active is 0 or ps_spinlock is
55 	4) When there is such a thread, all PIA belonging to pd.c have NULL
56 	->claim_cont or pi_spinlock is held by thread in question.
58 Proof:	consider the first moment when the above is not true.
60 (1) can become not true if some thread enters that area while another is there.
67 	c) pi_do_claimed() is called by pd.c only from the area.
68 	d) ps_tq_int() can enter the area only when the thread is holding
69 	   ps_spinlock and ps_tq_active is 1 (due to Lemma 1).  It means that
72 	   case that needs consideration is call from pi_wake_up() and there
77 	f) pi_wake_up() can enter the area only when the thread is holding
78 	   pi_spinlock and ->claimed_cont is non-NULL for PIA belonging to
81 (2) can become not true only when pd_lock is released by the thread in question.
82 	Indeed, pd_busy is reset only in the area and thread that resets
83 	it is holding pd_lock.	The only place within the area where we
84 	release pd_lock is in pd_next_buf() (called from within the area).
92 For similar reasons, (3) can become not true only when ps_spinlock is released
96 (4) is done the same way - all places where we release pi_spinlock within
99 also in the area.  The only place where ->claimed_cont is made non-NULL is
107 check its value is in ps_set_intr() and if it had been non-zero at that
113 that the only possible contention is between scheduling ps_tq followed by
118 can be killed.  Indeed, we are not holding pd_lock and thus pd_busy is already
122 ps_continuation, since the latter is changed only from the area.
126 reason - the value is guaranteed to be set by the last ps_set_intr() and