Where in the C++ Standard is it stated that a const rvalue reference doesn't bind to an lvalue?

Question

Where in the C++ Standard is it stated that a const rvalue reference doesn't bind to an lvalue?

For example the code below doesn't compile:

#include <iostream>

int i = 10;

int f(const int&& j) { return j; }
int main()
{
    std::cout << f(i) << '\n';
}

Answer 1

In the last bullet of [dcl.init.ref]/5 (quoting n3485):

If T1 [the type of the initialized reference] is reference-related to T2 [the type of the initializer expression] and the reference is an rvalue reference, the initializer expression shall not be an lvalue.

The cv-qualification is irrelevant in this case.

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The first (main) bullet of /5 doesn't apply since the reference is not an lvalue reference:

If the reference is an lvalue reference and the initializer expression [...]

The second (main) bullet point applies:

Otherwise, the reference shall be an lvalue reference to a non-volatile const type (i.e., cv1 shall be const), or the reference shall be an rvalue reference

[emphasis mine]

The first sub bullet point of this doesn't apply since the initializer is is not an xvalue or function prvalue and doesn't have class type.

The second sub-bullet-point is an unconditional "otherwise", so that last sub bullet point applies.

---

If the initializer is not reference-related to the referred type (T1), the example compiles:

#include <iostream>

double i = 10;

int f(const int&& j) { return j; }
int main()
{
    std::cout << f(i) << '\n';
}

Live example

Answer 2

[C++11: 8.5.3/3] lists how initialisers for references work; what you're looking for is there. It's too exhaustive to quote here verbatim, though.