Quick question: exit();

I apologize for asking a super-simple question, but my
book says nothing about it...(Well, at least I don't think
it does. I checked the index, but nothing.)

Quick question. What does the exit() function do? What
would it do in this snippet of code:

Code:

---

int main()
{
  //Question on exit()

  FILE *filestuff;
 
  filestuff=fopen("file","w");
  if(filestuff==NULL)
  {
    pritnf("Sorry. \n");
    exit(1);
  }
  return 0;
}

---

I imagine it exits from something. But what? And what
parameters does it accept? What do they mean?

Much obliged.:confused:

The exit() terminates the program much like the return statement in the main function. It return control to the operating system. It closes open files, flushes output buffers, and calls destructors before control is returned to the operating system. The parameter passed to the function are exit status values which communicate to the operating system the program terminated prematurely or everything happened as expected. For example exit(0) (same as return 0) means the program terminated normally. exit(1) (same as return 1) means the program terminated because of some error ( it could not open a file, allocate memory, etc. ).

So why would you use exit instead of return?

In that case you could just say

return EXIT_FAILURE;

in other cases where your inside a function like

void f(void)
{
exit(EXIT_FAILURE);
}

putting a return there would be a syntax error.

Quote:

---

So why would you use exit instead of return?

---

The return statement would work fine if the error occured in main(). exit() can cleanly terminate the program from within any function. Like in Nick's example.

Interesting though...i've seen a lot of old school and sometimes nonstandard C code that rely completely on the exit() function. no return statement at all...

Don't forget to include the relevent library

#include <stdlib.h>

for the exit function to work!!
-ali

Use exit in C and in C++ you can use exception handling instead.