Are literal numbers mutable or not?

Question

Naturally, this won't compile:

int &z = 3; // error: invalid initialization of non-const reference ....

and this will compile:

const int &z = 3; // OK

Now, consider:

const int y = 3;
int && yrr = y; // const error (as you would expect)
int && yrr = move(y); // const error (as you would expect)

But these next lines do compile for me. I think it shouldn't.

int && w = 3;
int && yrr = move(3);
void bar(int && x) {x = 10;}
bar(3);

Wouldn't those last two lines allow the literal 3 to be modified? What is the difference between 3 and a const int? And finally, Is there any danger with 'modifying' literals?

(g++-4.6 (GCC) 4.6.2 with -std=gnu++0x -Wall -Wextra)

Answer 1

The rvalue reference to the literal 3:

int && w = 3;

is actually bound to a temporary that is the result of evaluating the expression 3. It's not bound to some Platonic literal 3.

(all the following standards references are from the March 2011 draft, n3242)

3.10/1 "Lvalues and rvalues"

The value of a literal such as 12, 7.3e5, or true is also a prvalue

Then 8.5.3 "References" gives the rules for how a reference is bound falls through to the last case, which says:

Otherwise, a temporary of type “cv1 T1” is created and initialized from the initializer expression using the rules for a non-reference copy-initialization (8.5). The reference is then bound to the temporary.

and gives as one example something very close to what's in your question:

double&& rrd = 2; // rrd refers to temporary with value 2.0