freezer: unexport refrigerator() and update try_to_freeze() slightly

There is no reason to export two functions for entering the
refrigerator.  Calling refrigerator() instead of try_to_freeze()
doesn't save anything noticeable or removes any race condition.

* Rename refrigerator() to __refrigerator() and make it return bool
  indicating whether it scheduled out for freezing.

* Update try_to_freeze() to return bool and relay the return value of
  __refrigerator() if freezing().

* Convert all refrigerator() users to try_to_freeze().

* Update documentation accordingly.

* While at it, add might_sleep() to try_to_freeze().

Signed-off-by: Tejun Heo <tj@kernel.org>
Cc: Samuel Ortiz <samuel@sortiz.org>
Cc: Chris Mason <chris.mason@oracle.com>
Cc: "Theodore Ts'o" <tytso@mit.edu>
Cc: Steven Whitehouse <swhiteho@redhat.com>
Cc: Andrew Morton <akpm@linux-foundation.org>
Cc: Jan Kara <jack@suse.cz>
Cc: KONISHI Ryusuke <konishi.ryusuke@lab.ntt.co.jp>
Cc: Christoph Hellwig <hch@infradead.org>

fs/jfs/jfs_logmgr.c

@@ -2349,7 +2349,7 @@ int jfsIOWait(void *arg)
 
 		if (freezing(current)) {
 			spin_unlock_irq(&log_redrive_lock);
-			refrigerator();
+			try_to_freeze();
 		} else {
 			set_current_state(TASK_INTERRUPTIBLE);
 			spin_unlock_irq(&log_redrive_lock);

fs/jfs/jfs_txnmgr.c

@@ -2800,7 +2800,7 @@ int jfs_lazycommit(void *arg)
 
 		if (freezing(current)) {
 			LAZY_UNLOCK(flags);
-			refrigerator();
+			try_to_freeze();
 		} else {
 			DECLARE_WAITQUEUE(wq, current);
 
@@ -2994,7 +2994,7 @@ int jfs_sync(void *arg)
 
 		if (freezing(current)) {
 			TXN_UNLOCK();
-			refrigerator();
+			try_to_freeze();
 		} else {
 			set_current_state(TASK_INTERRUPTIBLE);
 			TXN_UNLOCK();