I have done some experimentation on rvalue references with the TDM-GCC 4.6.1 compiler and made some interesting observations that I cannot explain away with theories. I would like experts out there to help me explain them.

I have a very simple program that does not deal with objects but int primitives and that has defined 2 functions:
foo1 (returning a local variable by rvalue reference) and
foo2 (returning a local variable by value)

#include <iostream>

using namespace std;

int &&foo1();
int foo2();

int main()

int&& variable1 = foo1();
//cout << "My name is softwarelover." << endl;
cout << "variable1 is: " << variable1 << endl; // Prints 5.
cout << "variable1 is: " << variable1 << endl; // Prints 0.

int&& variable2 = foo2();
cout << "variable2 is: " << variable2 << endl; // Prints 5.
cout << "variable2 is still: " << variable2 << endl; // Still prints 5!

return 0;

int &&foo1() {

int a = 5;
return static_cast<int&&>(a);

int foo2() {

int a = 5;
return a;

It seems the value returned by foo1 and received by variable1 dies out after some time - perhaps, a brief period of some milliseconds. Notice that I have prevented cout from printing "My name is softwarelover" by commenting it out. If I allow that statement to run, the result is different. Instead of printing 5, 0 it prints 0, 0. Seems like it is because of the time-delay introduced by "cout << "My name is softwarelover." that 5 turns into 0.

Is the above how an rvalue reference is supposed to behave when referring to a primitive integer which a function returned by reference as opposed to return-by-value? By the way, why is it 0, why not garbage?

Notice also that variable2 never seems to die out, no matter how many times I print it with cout! variable2 refers to a primitive integer which a function returned by value, not return-by-reference.