calling a function indirectly from main with an particular format output - c++

Question

Here is my program below.
void fun1(void);
int main(int argc, char *argv[])
{
cout<<"In main"<<endl;
atexit(fun1); //atexit calls the function at the end of main
cout<<"Exit main"<<endl;
return 0;
}
void fun1(void)
{
cout<<"fun1 executed"<<endl;
}
The output is
In main
Exit main
fun1 executed
But I intend the output like this:
In main
fun1 executed
Exit main
I don't want to call "fun1" function directly or use any other functions which would call my function "fun1".
Is there any way I can acheive this this output? Any help is most welcome.

Answer 1

No. †
Only atexit and static object destruction happen "on their own", and both these things occur after returning from main.
And this makes sense, when you think about it: something that should occur during main should be written in main. If you need a function to be invoked at a given time, write it into your program. This is what you write the program for. main is for your program.
† Probably, with a "trick" or "hack" that'll get you fired from my team.

Answer 2

Here is a hack using a scope:
#include <iostream>
class call_on_destructed {
private:
void (*m_callee)();
public:
call_on_destructed(void (*callee)()): m_callee(callee) {}
~call_on_destructed() {
m_callee();
}
};
void fun() {
std::cout << "fun\n";
}
int main() {
{
call_on_destructed c(&fun);
std::cout << "In main" << std::endl;
}
std::cout<<"Exit main"<<std::endl;
}
Outputs:
In main
fun
Exit main
The scope end causes the class destructor to be invoked, which in turn calls the fun function, registered within the class.

Answer 3

Do you want something like:
void fun1(void)
{
std::cout << "fun1 executed" << std::endl;
}
void call(void f())
{
f();
}
int main(int argc, char *argv[])
{
std::cout << "In main" << std::endl;
call(fun1); //call calls the given function directly
std::cout << "Exit main" << std::endl;
return 0;
}

Answer 4

Clearly, if you want to have your function executed, something is going to call it! That is, calling your function without calling your function won't work. atexit() is also just calling your function: you register what's being called eventually.
From the sounds of it, your assignment asks to pass off a function object off and have that facility call your function. For example, you could use a function caller:
template <typename F>
void caller(F f) {
f();
}
int main() {
std::cout << "entering main\n";
caller(fun1);
std::cout << "exiting main()\n";
}
Clearly, caller() does call fun1 although without mentioning that name.