In my programming book, it shows me exit used without any parameters(exit();).
Unfortunately it does not work.
Some people have said use exit(0); while some say exit(1); exit(2); and exit(3);
What is the difference between them and is there even an exit(4); ?
The funny thing is that my compiler does not need stdlib.h to execute exit(0); and the rest.
Prior to the 1999 version of the ISO C standard, it was legal to call a function with no visible declaration. The compiler would assume that the function exists, creating an implicit declaration. (It would also assume that it returns a result of type int, which exit() does not.) If this implicit declaration doesn't match the actual definition of the function, the behavior is undefined.
As of the 1999 standard, the "implicit int" rule was dropped, and a call without a visible declaration (as provided, in this case, by #include <stdlib.h>) became invalid. Even though it's invalid, a compiler may still issue a non-fatal warning and handle it under the older rules; gcc does this by default.
Under any version of the language, exit requires a single argument of type int. Passing 0 or EXIT_SUCCESS (a macro defined in <stdlib.h> causes the program to terminate and pass a status to the environment indicating success. Passing EXIT_FAILURE causes the program to terminate with a status indicating failure.
The meanings of other argument values are not specified by the C language. You'll commonly see exit(1) to denote failure, but that's not entirely portable.
(exit may be some kind of built-in function in gcc, but that doesn't affect the rules of the language; it's still invalid to call exit with no visible declaration, or to call it without an int argument. If it's built-in, that might affect the level of detail in the diagnostic message.)