Fix incorrect open count check in release function

In the original code, the producer stops producing characters
when atomic_dec_and_test(&open_cnt) returns zero, but it will
return zero when the resulting value of open_cnt is nonzero.
Therefore, in the original code, if the open_cnt changes from
2 to 1, it will stop producing characters, but it is supposed
to consistently produce characters until the open_cnt becomes
zero.

Therefore, we modify the code so that it stops producing characters
when atomic_dec_and_test(&open_cnt) returns one.

Author: rota1001

simrupt.c

@@ -298,7 +298,7 @@ static int simrupt_open(struct inode *inode, struct file *filp)
 static int simrupt_release(struct inode *inode, struct file *filp)
 {
     pr_debug("simrupt: %s\n", __func__);
-    if (atomic_dec_and_test(&open_cnt) == 0) {
+    if (atomic_dec_and_test(&open_cnt)) {
         del_timer_sync(&timer);
         flush_workqueue(simrupt_workqueue);
         fast_buf_clear();