Can reference and the variable it is pointing to have different values?

Question

I could not understand why in below code snippet reference and its associated variable are giving different results.

const int x = 10;
const int &ptr = x; 
int& y = const_cast<int&>(ptr);
y = 19;
std::cout << "x = " << x << " ptr=" << ptr << " y=" << y << std::endl;

Output:

x=10 ptr=19 y=19

As per my understanding references are just alias for variable so why ptr is 19 and x is 10? Is this something to do with const or const_cast ?

It's undefined behavior. Your fridge may explode as likely as you're getting that result. — πάντα ῥεῖ, May 05 '16 at 13:00
When you modify a constant value, the behaviour is undefined. — Kerrek SB, May 05 '16 at 13:00

erip · Accepted Answer · 2016-05-05T13:11:57.360

8

From 5.2.11 Const cast [expr.const.cast]:

Note: Depending on the type of the object, a write operation through the pointer, lvalue or pointer to data member resulting from a const_cast that casts away a const-qualifier may produce undefined behavior (7.1.6.1). — end note ]

You're performing undefined behavior, so you're quite (un)lucky that your computer didn't implode. :)

edited May 05 '16 at 13:11

answered May 05 '16 at 13:04

erip

16,374
11
66
121

@Humam Because everyone wants to have their computer broken? – LogicStuff May 05 '16 at 13:09
@HumamHelfawi I guess it depends on if you're the developer or the user. :) If I were a user whose software didn't work because dev relied on UB, I might hope his computer imploded, too. – erip May 05 '16 at 13:10
@LogicStuff imploding is better than uncaughted UB – Humam Helfawi May 05 '16 at 13:10
@erip another perspective... – Humam Helfawi May 05 '16 at 13:11
1

@HumamHelfawi Added a little note in there for you. ;) – erip May 05 '16 at 13:12
@erip Thanks :) +1 – Humam Helfawi May 05 '16 at 13:12
@HumamHelfawi A (mis)fortune would be if the code produced results expected to the OP and they would then rely on that behavior in the future. – LogicStuff May 05 '16 at 13:15
@LogicStuff That is the unluckiest possible situation. – Humam Helfawi May 05 '16 at 13:16
I have seen computers exploding (well, sort of. Not really exploding, and not the whole computer - just some white smoke from hard drive) but I am yet to see one **imploded**. – SergeyA May 05 '16 at 13:18
@SergeyA It's more elusive, but scarier... I guess. – erip May 05 '16 at 13:19
1

It might help to mention that something like `int x; int const & cx = x; const_cast(cx) = 19;` does not result in undefined behaviour. – jotik May 05 '16 at 13:40

Yu Hao · Answer 2 · 2016-05-05T13:46:31.220

5

Is this something to do with const or const_cast?

Yes.

Even with const_cast, it's still undefined behavior to modify an object that was originally const. Or, in other words, it's illegal to use const_cast to write to a const object.

edited May 05 '16 at 13:46

answered May 05 '16 at 13:05

Yu Hao

119,891
44
235
294

I know. That's why I said it's illegal to *modify* / *write*. – Yu Hao May 05 '16 at 13:25
The way I see it is, if you did the cast but didn't do the write operation, that means the write access isn't given. So I don't see how that statement is false. It's quite possible that I was wrong though, as English is not my native language, so feel free to edit my answer any time. – Yu Hao May 05 '16 at 13:37

Can reference and the variable it is pointing to have different values?

2 Answers2