How to register a non-void function with atexit()?

Question

I'm trying to register a function that returns an int to be called at the end of a program using the atexit() function. (Specifically, the endwin() function from ncurses.)

But since atexit() needs a pointer to a void function, I ran into a problem. I tried the following:

static_cast<void (*)()>(endwin)

but static_casting from an int function to a void function doesn't seem to be allowed.

Is what I'm trying to accomplish possible at all, and if yes, how?

Note: I'm willing to just ignore the return value of the function.

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Edit: I also tried creating a lambda function, which seems to do what I want:

atexit([]{ endwin(); });

Is this a good solution compared to a wrapper/forwarding function? (Other than that it needs C++11 and avoids defining a new function whose sole purpose is just forwarding another function.)

Answer 1

Function pointers can't be converted. Just register a forwarding function:

#ifdef __cplusplus
extern "C"
#endif
void endwin_wrapper() { endwin(); }
...
atexit(endwin_wrapper);

Since you tagged your question C++: if you define the forwarding function in C++ you need to declare it to be extern "C" to have the proper language linkage.

Answer 2

With the current C++ standard, a lambda is probably the best solution to this:

atexit([]{ endwin(); });

Now there's no need to define a whole new named function with the sole purpose of just forwarding another function.

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And if you one day decide you need to have more functions called at program exit, you can either define them with a new call to atexit():

atexit(anotherCleanupFunction);

or, you can just add them to the lambda body:

atexit([]{
  anotherCleanupFunction();
  endwin();
});

However, when you have multiple functions to be registered with atexit(), then a separate wrapper function to hold all those functions may be a better solution. There's no one right answer though, just write good and clear code. (A call to atexit() with by a lambda full of function calls can look rather messy.)