Question

This question asks if all temporaries are rvalue.

The answer is no, because if we consider this expression:

const int &ri = 2 + 3;

then, the very same temporary (2 + 3), which is an rvalue here, can be used
as an lvalue in a subsequent expression:

const int *pi = &ri;

so this temporary is not (only) an rvalue.

The logic statement temporary ==> rvalue is then false.

However, we cannot write

const int &ri = &(2 + 3); // illegal, 2 + 3 -> temporary -> rvalue

or

int *i = &4; // illegal, 4 is an rvalue (literal)

or

int foo();
int *i = &foo(); // illegal, foo() -> temporary -> rvalue

Thus my question is, can we generate an rvalue in a certain expression without
having a temporary or a literal? Is rvalue ==> (temporary or literal) true?

Answer 1

Expressions that yield temporary objects are r-values. There’s a special rule which allows const-references and r-value references to bind to r-values, and this extends the lifetime of the temporary object to that of the reference (see 12.2(5)), but that does not make the temporary-object expression any less of an r-value.

However, once bound to a reference, the reference variable itself has a name, and thus the reference expression is an l-lvalue.

Don’t confuse expressions, variables and objects.