How to Handle SIGABRT signal?

As others have said, you cannot have abort() return and allow execution to continue normally. What you can do however is protect a piece of code that might call abort by a structure akin to a try catch. Execution of the code will be aborted but the rest of the program can continue. Here is a demo:

#include <csetjmp>
#include <csignal>
#include <cstdlib>
#include <iostream>

jmp_buf env;

void on_sigabrt (int signum)
{
  signal (signum, SIG_DFL);
  longjmp (env, 1);
}

void try_and_catch_abort (void (*func)(void))
{
  if (setjmp (env) == 0) {
    signal(SIGABRT, &on_sigabrt);
    (*func)();
    signal (SIGABRT, SIG_DFL);
  }
  else {
    std::cout << "aborted\n";
  }
}    

void do_stuff_aborted ()
{
  std::cout << "step 1\n";
  abort();
  std::cout << "step 2\n";
}

void do_stuff ()
{
  std::cout << "step 1\n";
  std::cout << "step 2\n";
}    

int main()
{
  try_and_catch_abort (&do_stuff_aborted);
  try_and_catch_abort (&do_stuff);
}

Although you can replace handler for SIGABRT and abort() will pay attention to the handler, the abort is only inhibited if the signal handler does not return. The relevant quote in C99 is in 7.20.4.1 paragraph 2:

The abort function causes abnormal program termination to occur, unless the signal SIGABRT is being caught and the signal handler does not return. ...

Your signal handler does return and thus the program is aborted.

You get those symptoms i.e. the popup debug dialog, when you have a debug build (with windows and Visual Studio- I'm testing with 2012 version), since it sets a debug break, in the debug implementation of abort() ). If you pick "ignore" you get that message "Function triger"

If you do a release build, then you don't get the debug popup dialog, and you get the message, as expected