I am having some trouble understanding the behaviour in this snippet:
unsigned int i = 2;
const int &r = i;
std::cout << r << "\n";
i = 100;
std::cout << r << "\n";
The first print statement gives 2 as I expect, but when I change the value of the referenced variable, it is not reflected in the reference. The second print statement also gives 2, but I think it should give 100?
If I make variable i
into type int
instead of unsigned int
, it works as I expect. What is going on here?
You can only have a reference to an object of the same type.
You cannot have an int
reference to an unsigned int
.
What is happening here is, essentially:
const int &r = (int)i;
A new int
temporary gets constructed, a new temporary object, and a const
reference is bound to it.
Using your debugger, you should be able to observe the fact that the reference is referring to a completely different object:
(gdb) n
6 const int &r = i;
(gdb)
7 std::cout << r << "\n";
(gdb) p i
$1 = 2
(gdb) p &i
$2 = (unsigned int *) 0x7fffffffea0c
(gdb) p &r
$3 = (const int *) 0x7fffffffe9fc
(gdb) q