What is the difference between exit() and abort()?

In C and C++, what is the difference between exit() and abort()? I am trying to end my program after an error (not an exception).

Solution 1:

abort() exits your program without calling functions registered using atexit() first, and without calling objects' destructors first. exit() does both before exiting your program. It does not call destructors for automatic objects though. So

A a;
void test() { 
    static A b;
    A c;
    exit(0);
}

Will destruct a and b properly, but will not call destructors of c. abort() wouldn't call destructors of neither objects. As this is unfortunate, the C++ Standard describes an alternative mechanism which ensures properly termination:

Objects with automatic storage duration are all destroyed in a program whose function main() contains no automatic objects and executes the call to exit(). Control can be transferred directly to such a main() by throwing an exception that is caught in main().

struct exit_exception { 
   int c; 
   exit_exception(int c):c(c) { } 
};

int main() {
    try {
        // put all code in here
    } catch(exit_exception& e) {
        exit(e.c);
    }
}

Instead of calling exit(), arrange that code throw exit_exception(exit_code); instead.

Solution 2:

abort sends a SIGABRT signal, exit just closes the application performing normal cleanup.

You can handle an abort signal however you want, but the default behavior is to close the application as well with an error code.

abort will not perform object destruction of your static and global members, but exit will.

Of course though when the application is completely closed the operating system will free up any unfreed memory and other resources.

In both abort and exit program termination (assuming you didn't override the default behavior), the return code will be returned to the parent process that started your application.

See the following example:

SomeClassType someobject;

void myProgramIsTerminating1(void)
{
  cout<<"exit function 1"<<endl;
}

void myProgramIsTerminating2(void)
{
  cout<<"exit function 2"<<endl;
}

int main(int argc, char**argv)
{
  atexit (myProgramIsTerminating1);
  atexit (myProgramIsTerminating2);
  //abort();
  return 0;
}

Comments:

• If abort is uncommented: nothing is printed and the destructor of someobject will not be called.

• If abort is commented like above: someobject destructor will be called you will get the following output:

exit function 2
exit function 1