return of a local variable by ref works - variables

Question

Take a look at this C++ code:
#include <iostream>
using namespace std;
class B{
public:
int& f() {
int local_n = 447;
return local_n ;
} // local_n gets out of scope here
};
int main()
{
B b;
int n = b.f(); // and now n = 447
}
I don't understand why n = 447 at the end of main, because I tried to return a reference to a local_n, when it should be NULL;

Answer 1

Returning a reference to a local variable invokes undefined behavior - meaning you might get lucky and it might work... sometimes... or it might format your hard drive or summon nasal demons. In this case, the compiler generated code that managed to copy the old value off the stack before it got overwritten with something else. Oh, and references do not have a corresponding NULL value...
Edit - here's an example where returning a reference is a bad thing. In your example above, since you copy the value out of the reference immediately before calling anything else, it's quite possible (but far from guaranteed) that it might work most of the time. However, if you bind another reference to the returned reference, things won't look so good:
extern void call_some_other_functions();
extern void lucky();
extern void oops();
int& foo()
{ int bar = 0;
return bar;
}
main()
{ int& x = foo();
x = 5;
call_some_other_functions();
if (x == 5)
lucky();
else
oops();
}