Const reference to a casted int from unsigned int

Question

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?

Live example

Yes, I think it should be that, but I want to know why this particular snippet behaves this way. — KKOrange, Jul 31 '16 at 14:12
...is it undefined? I cannot find if that is true, the other answers make it seem like it is defined behaviour too. — KKOrange, Jul 31 '16 at 14:22
No, it's not UB. The answers explain pretty well what's actually happening. — πάντα ῥεῖ, Jul 31 '16 at 14:23

score 33 · Accepted Answer · answered Jul 31 '16 at 14:11

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

songyuanyao · Answer 2 · 2016-07-31T14:24:50.053

The second print statement also gives 2, but I think it should give 100?

Because a temporary int is created here.

For const int &r = i;, i (unsigned int) needs to be converted to int at first, means a temporary int will be created and then be bound to r (temporary could be bound to lvalue reference to const), it has nothing to do with the original variable i any more.

If I make variable i into type int instead of unsigned int, it works as I expect.

Because no conversion and temporary is needed, i could be bound to r directly.

Const reference to a casted int from unsigned int

2 Answers2