What is the difference between
int a = 42; and
int&& rvr = 42; ?
a is obviously an lvalue, but rvr is also an lvalue since it is a named variable, so can these expressions be considered exactly equivalent, and if not, in which scenarios the int&& declaration is preferred?
They're not exactly equivalent. You're right that both variables are lvalues, but for most types, initialisation of the form
T a = b; creates a temporary T object, and then constructs a from that temporary object. That temporary object may be elided, but it still requires the appropriate constructors to be available.
T &&a = b; on the other hand, binds a directly to b, but requires b to be an rvalue. So:
int a = 3; // int&&b = a; // error: && cannot bind to lvalues int b = a; // okay struct S { S(const S &) = delete; }; S f(); // S s = f(); // error: no copy or move constructor available S&&s = f(); // okay And more simply, decltype will also give different results for references.
[](auto&&x){ f(std::forward<decltype(x)>(x)); } is a practical example of the use of && and decltype.
auto, but the answers there might well clarify your questions.