On 7/22/24 20:28, Shuah Khan wrote:
On 7/15/24 05:49, Dev Jain wrote:
On 6/30/24 20:48, Oleg Nesterov wrote:
I see nothing wrong, but perhaps this test can be simplified?
Feel free to ignore.
Say,
On 06/27, Dev Jain wrote:
+void handler_usr(int signo, siginfo_t *info, void *uc)
+{
+ int ret;
+
+ /*
+ * Break out of infinite recursion caused by raise(SIGUSR1)
invoked
+ * from inside the handler
+ */
+ ++cnt;
+ if (cnt > 1)
+ return;
+
+ ksft_print_msg("In handler_usr\n");
This message isn't very useful. Why do you need this message?
There isn't any specific use; I am just showing the progress
of the test. If you think this is just waste output....
+
+ /* SEGV blocked during handler execution, delivered on return */
+ if (raise(SIGSEGV))
+ ksft_exit_fail_perror("raise");>>> +
+ ksft_print_msg("SEGV bypassed successfully\n");
You could simply do sigprocmask(SIG_SETMASK, NULL, &oldset) and
check if
SIGSEGV is blocked in oldset. SIG_SETMASK has no effect if newset ==
NULL.
IMHO, isn't raising the signal, and the process not terminating, a
stricter test? I have already included your described approach in
the last testcase; so, the test includes both ways: raising the
signal -> process not terminating, and checking blockage with
sigprocmask().
thanks,
-- Shuah
