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().