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");
+
+       /* 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().

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