lvalue-to-rvalue

I asked this question: static_assert of const Variable And apparently it comes down to the question does a floating point lvalue get converted to an rvalue for the purposes of comparison? So in this code does an lvalue-to-rvalue conversion occur? const float foo = 13.0F; static_assert(foo > 0.0F, "foo must be greater than 0."); Yes, it is performed. Basically, it's all because 3.0 > 1.2 is a well-formed expression, that contains nothing but prvalues for operands. First, [expr]/9 states (emphasis mine) that Whenever a glvalue expression appears as an operand of an operator that expects a

I suppose when a universal reference parameter is matched with an rvalue reference argument, an rvalue reference argument is returned. However, my testing shows that the rvalue reference is turned into a lvalue reference by the universal reference function template. Why is it so? #include <iostream> #include <type_traits> using namespace std; template <typename T> T f1(T&&t) { //<-----this is a universal reference cout << "is_lvalue reference:" << is_lvalue_reference<T>::value << endl; cout << "is_rvalue reference:" << is_rvalue_reference<T>::value << endl; cout << "is_reference:" << is

In the following code, I'm not able to pass a temporary object as argument to the printAge function: struct Person { int age; Person(int _age): age(_age) {} }; void printAge(Person &person) { cout << "Age: " << person.age << endl; } int main () { Person p(50); printAge(Person(50)); // fails! printAge(p); return 0; } The error I get is: error: invalid initialization of non-const reference of type ‘Person&’ from an rvalue of type ‘Person’ I realize that this is something to do with passing an lValue to a function expecting a rValue... Is there a way to convert my lValue to rValue by using std:

This question already has an answer here: Rvalue Reference is Treated as an Lvalue? 4 answers Lvalue reference constructor is called instead of rvalue reference constructor 1 answer Say I have the following function void doWork(Widget && param) // param is an LVALUE of RRef type { Widget store = std::move(param); } Why do I need to cast param back to an rvalue with std::move() ? Shouldn't it be obvious that the type of param is rvalue since it was declared in the function signature as an rvalue reference? Shouldn't the move constructor be automatically invoked here on this principle alone? Why

C++03 §4.2 N°1: An lvalue or rvalue of type “array of N T” or “array of unknown bound of T” can be converted to an rvalue of type “pointer to T.” The result is a pointer to the first element of the array. What has been confusing in this statement for a long time for me was that I didn't quite understand what an rvalue of array type would mean. That is, I couldn't come up with an expression whose type were an array and the result were an rvalue. I read this thread, which basically asks the same question and the accepted answer is "no, there is no rvalue of array type". I think I just might have

I've been reading quite many on the Internet and it seems that many people mentioned the following rules (but i couldn't find it in the standard), The addition operator + (and all other binary operators) requires both operands to be rvalue, and the result is rvalue. And so on.. I checked the C++ standard, and it clearly states that (clause 3.10/2), Whenever a glvalue appears in a context where a prvalue is expected, the glvalue is converted to a prvalue (clause 5/9), Whenever a glvalue expression appears as an operand of an operator that expects a prvalue for that operand, the lvalue-to-rvalue