Why signal handler for SIGSEGV doesn't catch my C++ throw exception?

Question

I was trying to see if SIGSEGV signal hander could help to deal with unhandled exception of C++, I experiment it:

#include <stdio.h>
#include <execinfo.h>
#include <signal.h>
#include <stdlib.h>
#include <unistd.h>
void handler(int sig) {
  void *array[10];
  size_t size = backtrace(array, 10);
  fprintf(stderr, "Error: signal %d:\n", sig);
  backtrace_symbols_fd(array, size, STDERR_FILENO);
  exit(1);
}
int main()
{
  signal(SIGSEGV, handler);
  throw 1;
}

$ g++ -g h.cpp -rdynamic && ./a.out

terminate called after throwing an instance of 'int' Aborted (Core dump)

Well, the program doesn't print crash call stack back trace as I expected. My question is:

As long as it terminates, does it through out any signals like SIGSEGV? Is there a system call or posix api that could catch even C++ exceptions and prints out the call stack?

Thanks!

Possible duplicate of [c++ stack trace from unhandled exception?](http://stackoverflow.com/questions/3355683/c-stack-trace-from-unhandled-exception) + an exception is not a segmentation violation (segmentation fault). — user1810087, Sep 30 '16 at 11:11
@Hind Forsum - you should install the signal handler for SIGABRT instead. — Rudolfs Bundulis, Sep 30 '16 at 11:28
A signal is not an exception. You catch exceptions with `throw ... catch()`. — Pete Becker, Sep 30 '16 at 13:25

Rudolfs Bundulis · Accepted Answer · 2016-09-30T11:27:04.903

You can install a termination handler with std::set_terminate - the installed handler will then be invoked for unhandled exception. There is no strong guarantee, but there is a great chance that the invocation of abort that causes invocation of the termination handler is occurring on the top of the stack that threw the exception and thus you can still collect a backtrace.

Also, you can install a handler for SIGABRT, that will work as well:

#include <execinfo.h>
#include <signal.h>
#include <stdlib.h>
#include <unistd.h>
#include <stdio.h>

void handler(int sig, siginfo_t*, void*) {
  void *array[10];
  size_t size = backtrace(array, 10);
  fprintf(stderr, "Error: signal %d:\n", sig);
  backtrace_symbols_fd(array, size, STDERR_FILENO);
  exit(1);
}

void terminate_handler()
{
}

int main()
{
  struct sigaction new_sigaction, old_sigaction;
  new_sigaction.sa_flags = SA_SIGINFO;
  new_sigaction.sa_sigaction = handler;
  sigaction(SIGABRT, &new_sigaction, &old_sigaction);
  throw 1;
}

I'd personally prefer the termination handler, since in that case you explicitly know the cause. If the SIGABRT handler is invoked you must understand if this is because an unhandled exception invoked abort() or the signal was sent by other means.

score -2 · Answer 2 · answered Sep 30 '16 at 11:25

-2

I think you need to do

raise(SIGSEGV)

if you want to end up in your handler:

#include <stdio.h>
#include <execinfo.h>
#include <signal.h>
#include <stdlib.h>
#include <unistd.h>
void handler(int sig) {
  void *array[10];
  size_t size = backtrace(array, 10);
  fprintf(stderr, "Error: signal %d:\n", sig);
  backtrace_symbols_fd(array, size, STDERR_FILENO);
  exit(1);
}
int main()
{
  signal(SIGSEGV, handler);
  raise(SIGSEGV);
}

answered Sep 30 '16 at 11:25

Rik

1
2

OP wants to catch an unhandled C++ exception. – Rudolfs Bundulis Sep 30 '16 at 11:27
Why do you want to handle c++ exception with C functions ? – Simon PA Sep 30 '16 at 11:39
I don t know... I just noticed he installed a handler for SIGSEGV and assumed he wanted to end up in that handler. Hence raise SIGSEGV – Rik Oct 01 '16 at 19:20

Why signal handler for SIGSEGV doesn't catch my C++ throw exception?

2 Answers2