[c++] How do I make a C++ console program exit?

Is there a line of code that will terminate the program?

Something like python's sys.exit()?

Allowing the execution flow to leave main by returning a value or allowing execution to reach the end of the function is the way a program should terminate except under unrecoverable circumstances. Returning a value is optional in C++, but I typically prefer to return EXIT_SUCCESS found in cstdlib (a platform-specific value that indicates the program executed successfully).

#include <cstdlib>

int main(int argc, char *argv[]) {
  ...
  return EXIT_SUCCESS;
}

If, however, your program reaches an unrecoverable state, it should throw an exception. It's important to realise the implications of doing so, however. There are no widely-accepted best practices for deciding what should or should not be an exception, but there are some general rules you need to be aware of.

For example, throwing an exception from a destructor is nearly always a terrible idea because the object being destroyed might have been destroyed because an exception had already been thrown. If a second exception is thrown, terminate is called and your program will halt without any further clean-up having been performed. You can use uncaught_exception to determine if it's safe, but it's generally better practice to never allow exceptions to leave a destructor.

While it's generally always possible for functions you call but didn't write to throw exceptions (for example, new will throw std::bad_alloc if it can't allocate enough memory), it's often difficult for beginner programmers to keep track of or even know about all of the special rules surrounding exceptions in C++. For this reason, I recommend only using them in situations where there's no sensible way for your program to continue execution.

#include <stdexcept>
#include <cstdlib>
#include <iostream>

int foo(int i) {
  if (i != 5) {
    throw std::runtime_error("foo: i is not 5!");
  }
  return i * 2;
}

int main(int argc, char *argv[]) {
  try {
    foo(3);
  }
  catch (const std::exception &e) {
    std::cout << e.what() << std::endl;
    return EXIT_FAILURE;
  }
  return EXIT_SUCCESS;
}

exit is a hold-over from C and may result in objects with automatic storage to not be cleaned up properly. abort and terminate effectively causes the program to commit suicide and definitely won't clean up resources.

Whatever you do, don't use exceptions, exit, or abort/terminate as a crutch to get around writing a properly structured program. Save them for exceptional situations.

throw back to main which should return EXIT_FAILURE,

or std::terminate() if corrupted.

(from Martin York's comment)

simple enough..

exit ( 0 ); }//end of function

Make sure there is a space on both sides of the 0. Without spaces, the program will not stop.

exit(0); // at the end of main function before closing curly braces

Yes! exit(). It's in <cstdlib>.

else if(Decision >= 3)
    {
exit(0);
    }

While you can call exit() (and may need to do so if your application encounters some fatal error), the cleanest way to exit a program is to return from main():

int main()
{
    // do whatever your program does

} // function returns and exits program

When you call exit(), objects with automatic storage duration (local variables) are not destroyed before the program terminates, so you don't get proper cleanup. Those objects might need to clean up any resources they own, persist any pending state changes, terminate any running threads, or perform other actions in order for the program to terminate cleanly.

There are several ways to cause your program to terminate. Which one is appropriate depends on why you want your program to terminate. The vast majority of the time it should be by executing a return statement in your main function. As in the following.

int main()
{
     f();
     return 0;
}

As others have identified this allows all your stack variables to be properly destructed so as to clean up properly. This is very important.

If you have detected an error somewhere deep in your code and you need to exit out you should throw an exception to return to the main function. As in the following.

struct stop_now_t { };
void f()
{
      // ...
      if (some_condition())
           throw stop_now_t();
      // ...
}

int main()
{
     try {
          f();
     } catch (stop_now_t& stop) {
          return 1;
     }
     return 0;
 }

This causes the stack to be unwound an all your stack variables to be destructed. Still very important. Note that it is appropriate to indicate failure with a non-zero return value.

If in the unlikely case that your program detects a condition that indicates it is no longer safe to execute any more statements then you should use std::abort(). This will bring your program to a sudden stop with no further processing. std::exit() is similar but may call atexit handlers which could be bad if your program is sufficiently borked.

if you are in the main you can do:

return 0;  

or

exit(exit_code);

The exit code depends of the semantic of your code. 1 is error 0 e a normal exit.

In some other function of your program:

exit(exit_code)  

will exit the program.

This SO post provides an answer as well as explanation why not to use exit(). Worth a read.

In short, you should return 0 in main(), as it will run all of the destructors and do object cleanup. Throwing would also work if you are exiting from an error.

#include <cstdlib>
...
/*wherever you want it to end, e.g. in an if-statement:*/
if (T == 0)
{
exit(0);
}

In main(), there is also:

return 0;