hidden g++ reference error - c++

Question

Compiling with GCC 4.8.1 (MinGW) with command: g++ -o test.exe test.cpp
test.cpp:
#include <iostream>
int a = 1, b = 2;
void F()
{
int &r = a;
std::cout << r;
r = b;
std::cout << r;
}
int main ( int, char** )
{
F();
F();
return 0;
}
The expected output is: 1212, but real is: 1222
It works as expected when r is defined as pointer.
void F()
{
int *r = &a;
std::cout << *r;
r = &b;
std::cout << *r;
}
Does someone have any idea where is "the key from the tent"?

Answer 1

In the example with a pointer:
r = &b;
This re-seats the pointer r to point to b. In the example with the reference:
r = b;
You probably think that this is analogous to the example with the pointer, but it is not. References are not re-seatable the way pointers are. Once they are initialized, they become aliases for the thing which they reference. So the line above is exactly equivalent to this:
a = b;
That is, it copies the value stored in b over to a.
The reference example is could be written like this:
void F()
{
std::cout << a;
a = b;
std::cout << a;
}
While the pointer example could be written like this:
void F()
{
std::cout << a;
std::cout << b;
}

Answer 2

Your "problem" is the following line:
int &r = a;
This will create a so called reference pointing to a. While references don't require additional logic when accessing them (i.e. dereferencing; compared to pointers), they still work the same way behind the scenes.
So once you hit the line
r = b;
You'll actually use the reference and update the value of a. In this case r never has it's own value. It will always point at a and essentially be an alias.
In your second example, this is different. The line
int *r = &a;
creates a pointer pointing at a and
r = &b;
will reassign that address, making the pointer now point at b.
In short, when you assign some other value (address) to a pointer, you change the pointer and it will point somewhere else. However, when assigning a new value to a reference, you're actually updating the inital value where it's referencing to, not the reference itself.

Answer 3

//global variables! remember that this will be like a state
//and will be freed on (data?BSS?stack?) only after end of program.
int a = 1, b = 2;
void F()
{
//You are referencing here,
//meaning, once you change the referee ( r )
//the reference will changed permanently.
//Permanently because you are referring to a global variable.
int &r = a;
std::cout << r;
//You have changed the value of ( r )
//forever!
r = b; //I think you are assuming the you have changed the reference address here?
std::cout << r;
}
void F()
{
//You have declared and initialized a pointer
//that points to global variable.
//w/c means you can also change its value forever!
int *r = &a;
std::cout << *r;
//You have changed the ( r ) address pointing to, not its value!
r = &b;
std::cout << *r;
}

Answer 4

Your code is absolutely fine and it's working according to your instruction.
Suppose a variable has 1000 address b variable 1004 address, now your first line is
int &r = a;
that means r has 1000 address and you did cout<< r //print 1
In second statement r = b that means 1000 address hold value of b means 2;
Now , your scenario has been changed 1000---> 2 and 1004--->2
and again you are calling F();
that's why compiler print twice 2 value. . There is a concept referencing in c++. Hope it
would be useful for u