The addition operator + (and all other binary operators) requires both operands to be rvalue, and the result is rvalue. Return lvalue reference from temporary object. There are operators that yield lvalues: for example, if E is an expression of pointer type, then *E is an lvalue expression referring to the object to which E points. Since the type of a is not an int, it cannot match the type that b. Introduction. ) is characterized by two independent properties: a type and a value category. b is just an alternative name to the memory assigned to the variable a. "lvalues are named variables and rvalues are temporaries" is a good enough heuristic for a beginner, and no more an "oversimplification" than "I before E except after C" is for English. Thus, both a rvalue and another value can be assigned to values. However, a (prvalue) rvalue cannot be converted implicitly to an lvalue or xvalue, except by user-defined conversions. But instead removing either reference overload results in ambiguity with f( int ). 3=i; is illegal. A so called 'rvalue-reference' can bind to a temporary , but anything with a name is an lvalue, so you need to forward<> () it if you need it's rvalueness back. i is named object, so it is lvalue. For example, when user tries to read a given position in the collection. 6. And most implementations do that. In the case of object constructing is true but in the case of object assigning is false. Each C++ expression (an operator with its operands, a literal, a variable name, etc. According to the rule of forwarding reference, when an lvalue is passed to add, the template type argument Element will be deduced as SomeClass&. 3. It's also echoed in 5. rvalue references are considered lvalue (this part I understand) They are not. foobar () is an rvalue because foobar () returns int. 8. To answer the titular question, "rvalue reference" is a kind of type, while "xvalue" is a kind of expression. In the second case that I've reported, in whch aString is A constructor is an LValue reference, the std::move operator will still convert it to an RValue reference and I should still. 左值可以出现在赋值号的左边或右边。. So in terms of the type analogy this means that cv T& and cv T&& are transformed to cv T if T is a class type and to T if T is a non-function non-array. I am trying to figure out the meaning of the following snippet: int main() { int&& a = 2; int& b = a; // (*) } I know a is an lvalue expression of type "rvalue reference to int", and b is a general variable with type "lvalue reference to int". cpp -std=c++11 -fno-elide-constructors. ASCII defines a set of characters for encoding text in computers. e. int a = 1; // a is an lvalue int b = 2; // b is an lvalue int c = a + b; // + needs rvalues, so a and b are converted to rvalues // and an rvalue is returned. You are returning a copy of A from test so *c triggers the construction of a copy of c. std::function's type is defined only by its target's signature(eg: void(int)) and std::function itself is defined by the. So you can write a couple of convert functions . The answer is: yes, we do. The object identified by an xvalue expression may be a nameless temporary, it may be a named object in scope, or any other kind of object, but if used as a function argument, xvalue will always bind to the rvalue reference overload if available. The implementation of the language level is based on IBM's interpretation of the standard. r-value references are designed to be the subject of a move-constructor or move-assignment. and some other people did a test on their C++ compiler ( please explain ) : says (time_t){time(NULL)} this will still be a rvalue which is opposite to the C. int f( int ); int f( int && ); int f( int const & ); int q = f( 3 ); Removing f( int ) causes both Clang and GCC to prefer the rvalue reference over the lvalue reference. This function takes an lvalue reference and converts it to an rvalue reference. When you use "Hello, World" in a context in which it is implicitly converted to a const char* pointing to its initial element, the resulting pointer is an rvalue (because it is a temporary object resulting from an implicit. This is a helper function to allow perfect forwarding of arguments taken as rvalue references to deduced types, preserving any potential move semantics involved. When an lvalue-to-rvalue conversion occurs within the operand of sizeof, the value contained in the referenced object is not accessed, since that operator does not evaluate its operand. That's right according also to the C++ Standard (talking about the lvalue-to-rvalue conversion): 4. An lvalue may get converted to an rvalue: that's something perfectly legit and it happens quite often. Also, xvalues do not become lvalues. Under the conditions specified in [dcl. If U is t’s underlying non-reference type (namely std::remove_reference_t<decltype(t)>), then T will. Both of g and h are legal and the reference binds directly. e. According to the C++ specifications, it takes two rvalues as arguments and returns an rvalue. In (static_cast<int&&> (3))++, the expression static. 3. It cannot convert from an rvalue to an lvalue reference, even a const one. It shouldn't be something special so i coded that a component has a parent as composite, the composite should derrived from component and use the constructor from it's base class (Component). Taking it by rvalue reference would cause a headache to a user who has an existing lvalue or const reference to a function; they would need to std::move it (in. For the second overload, it would call operator const P&() const&. The terms are somewhat language-specific; they were first introduced in CPL. But in this particular case, the rules. A simpler case: template <typename T> void foo(T&& ) { } foo(1); // T is int int x; foo(x); // T is int& When you specify float for x, you are specifying that that particular argument will have type float&&, and you cannot implicitly convert an lvalue float to an rvalue. Found workaround how to use rvalue as lvalue. 3) If new_type is an rvalue reference type, static_cast converts the value of expression to xvalue. e. There are operators that yield lvalues: for example, if E is an expression of pointer type, then *E is an lvalue expression referring to the object to which E points. One could also say that an rvalue is any expression that is not an lvalue . 1 (page 85 for version 3485). However, note that when binding this to an rvalue reference, the value of this will be copied into a temporary object and the reference will instead be bound to that. 10. Once an entity has a name, it is clearly an lvalue! If you have a name for an rvalue reference, the entity with the name is not an rvalue but an lvalue. The reason why you need to const is to make x not a forwarding reference. You have three choices: (1) assign to rvalue reference, (2) assign to const lvalue reference, (3) return by value but implement move semantics in your class. It is really about rvalues vs. As well as the potentially dangling lvalue references you've identified, this led in C++03 to the situation where operator<< on a temporary ostream could be called with a char (member function operator) but not with a string (free operator); C++11 fixes this with free operator overloads for rvalue references and rvalue *this overload for member. lvalue simply means an object that has an identifiable location in memory (i. The lvalue is. rvalue rvalue lvalue. Therefore it makes sense that they are mutable. OK. The following table lists exceptions to this rule. The C++17 standard defines expression value categories as follows: A glvalue is an expression whose evaluation determines the identity of an object, bit-field, or function. 4. The address of operator (&) requires an lvalue because you can only take the address of something in memory. an lvalue reference). Would you ever mark a C++ RValue reference parameter as const. The pre-C++ origin of the terms "lvalue" and "rvalue" might be related to "left" and "right" side of assignment, but that meaning is only applicable in a small subset of the C++ language. Converts between types using a combination of explicit and implicit conversions. An lvalue may get converted to an rvalue: that's something perfectly legit and it happens quite often. X& r = X(99); // ERRORI use forward declaration here to pass object of class B as parameter in class A. So in terms of the type analogy this means that cv T& and cv T&& are transformed to cv T if T is a class type and to T if T is a non-function non-array. a non-const reference). Here, the developer is probably thinking - “I’ll pass in an int because it’ll get implicitly converted to an integer, and it’ll get incremented”. There are no references of references in C++. This is. As long as no const is involved, the expression T() is a modifiable rvalue, to be more precise. Lvalue references and rvalue references are syntactically and semantically similar, but. 12. 1Primary categories lvalue prvalue xvalue 2Mixed categories glvalue rvalue 3Special categories Pending member function call Void expressions Bit-fields Move. lvalue references are marked with one ampersand (&). Compiled with "g++ -std=c++0x". Value categories. (This is as per my understanding, please correct it otherwise). Unless encountered in unevaluated context (in an operand of sizeof, typeid, noexcept, or decltype), this conversion effectively copy-constructs a temporary object of type T using the original glvalue as the. Our motivation for this is generally to use it as the source of a move operation, and that’s why the way to convert an lvalue to an rvalue is to use std::move. ; T is not reference-related to U. But then i got following error:. Temporary lifetime extension does not pass through functions so there is no way to get a lvalue from the rvalue you pass to the function. I think it's reasonable to call print_stream like this:. The result is that of *reinterpret_cast<T2*>(p), where p is a pointer of type “pointer to T1 ” to the object designated by expression. When you look at a parameter thing&& x its type is an rvalue reference, however, the variable named x also has a value category: it's an lvalue. Regarding the second question. (since C++11)20. If you pass an argument to a reference type parameter (whether lvalue or rvalue reference), the object will not be copied. Operationally, the difference among these kinds of expressions is this:std::function can be move-constructed from rvalue of a functor object. 12. Say we want to steal from an lvalue: int main() { Holder h1(1000); // h1 is an lvalue Holder h2(h1); // copy-constructor invoked (because of lvalue in input) } This will not work: since h2 receives an lvalue in input, the copy constructor is being triggered. 11 for the exact listing what the cast can do; what that section doesn't list, it can't do. An lvalue or xvalue is an expression that refers to such an object. Lvalue to rvalue conversion. An lvalue or xvalue is an expression that refers to such an object. 2, and 4. However, you don't have double && in your code, you have U && for a deduced U. std::get returns an lvalue reference if its tuple argument is an lvalue. G. What makes rvalue references a bit difficult to grasp is that when you first look at them, it is not clear what their purpose is or what problems they solve. What I found by using this "real world" example is that if want to use the same code for lvalue ref and rvalue ref is because probably you can convert one to the other! std::ostringstream& operator<<(std::ostringstream&& oss, A const& a){ return operator<<(oss, a); }4. 16. e. And the lvalue-to-rvalue conversion always returns a prvalue value, not a (temporary) object. 0. This is already done in some places. 3 Viable functions (4). They are declared using the ‘&’ before the name of the variable. Jun 27 at 7:34. Example: int a = 10; // Declaring lvalue reference int& lref = a; // Declaring rvalue reference int&& rref = 20; Explanation: The following code will print True as both the variable are pointing to the same memory location. An lvalue-to-rvalue conversion (converting the name of the object x to its value 2. In the previous question, I asked how this code should work: void f (const std::string &); //less efficient void f (std::string &&); //more efficient void g (const char * arg) { f (arg); } It seems that the move overload should probably be called because of the. The expression 0 is. –std::forward is usually the way to 'convert' value category. 4. I discovered that N3290 (identical to C++11 standard) contains non-normative example of binding double&& to rvalue generated from int lvalue, and the updated wording in §8. Lvalue to rvalue conversion. An lvalue may be used to initialize an lvalue reference; this associates a new name with the object identified by the expression. The type of b is an rvalue reference to int , but the expression b is an lvalue; it is a variable, you can take its address. Whenever an lvalue a glvalue appears in a context where an rvalue a prvalue is expected, the lvalue glvalue is converted to an rvalue a prvalue; see 4. (until C++11) When an lvalue-to-rvalue conversion is applied to an expression E, the value contained in the referenced object is not accessed if: In general, lvalue is: Is usually on the left hand of an expression, and that’s where the name comes from - “left-value”. In the previous question, I asked how this code should work: void f (const std::string &); //less efficient void f (std::string &&); //more efficient void g (const char * arg) { f (arg); } It seems that the move overload should probably be called because of the. Naming expressions are always lvlaues. 2, and 4. An lvalue is (simplifying a bit) something that refers to someplace in memory that can/does hold a value. rvalue (until C++11) / prvalue (since C++11)Since you are giving your rvalue reference a name in the parameter list, it indeed becomes an lvalue. If the operator accepts paramters by value, whenever you use an lvalue expression, there needs to be lvalue-to-rvalue conversion, which is copy initialising the parameter object from the argument. reinterpret_cast reinterpret_cast converts any pointer type to any other pointer type, even of unrelated classes. D'uh. It can convert lvalues to lvalue references and rvalues to rvalue references. I don't really understand why an rvalue and a non-modifiable lvalue would be the same. L-value: “l-value” refers to memory location which identifies. If element on this position doesn't exist, it should throw exception. It would capitalize std::strings, and display each parameter after they are capitalized. As we've seen earlier, a and b are both lvalues. Assignment involving scalar types requires a modifiable lvalue on the left hand side of the assignment operator. 25, then the R-value is 1 divided by 0. 1. So, when you type const int& ref = 40. It seems like std::array can be converted to an std::span when it's an rvalue only on clang. In the op's example y is actually a reference to the sub-object of some unnamed object the structured binding declared. const T& still binds happily to both lvalues and rvalues. e. ; // not legal, so no lvalue. From reference - value categories. 1) modifiable lvalues. In example 4, a is an lvalue, becuase it has a name and I can take its address so it's ok to bind a lvalue reference b to an lvalue (int&& a) that happens to be a rvalue reference. An rvalue can be bound to an rvalue reference (T&&) to prolong its lifetime, and to lvalue references to const (const T&), but not to plain lvalue references (T&). The reference declared in the above code is lvalue. This implies that the compilers that accept the above code without a diagnostic are non-conforming (i. Note that there is one exception: there can be lvalue const reference binding to an rvalue. Variables of type rvalue reference have to be initialized in their definition like variables of type lvalue reference. Radius: 2 2 4. 3. You are comparing two different things that are not really related. Since int() isn't an lvalue, you can't assign to int(). Suppose r is an rvalue reference or non-volatile const lvalue reference to type T, and r is to be initialized by an expression e of type U. In short: every named object is Lvalue, and even if v is reference to Rvalue you need to use move to force move ctor to be called. e. But Args itself is either an lvalue reference or not a reference. For example in the following instructions. int&& x = 3; x is now an lvalue. In this case 2*b is an rvalue since it does not persist beyond the expression. Except for an implicit object parameter, for which see 13. I still can't figure out which one is correct though :(–In your specific case, since you are calling the function immediately you don't need to worry about taking ownership of it, so it would be better to take the function by const reference. Correct. lvalue and rvalue as function parameters. ConclusionFrom expr. Well, neither. end()) is a temporary object and cannot be bound to lvalue reference. Let's think of the addition +. Lvalue-to-rvalue conversion. Therefore it makes sense that they are mutable. Since you can call the function object std::bind gives you multiple times, it cannot “use up” the captured argument so it will be passed as an lvalue reference. References. 197. Similarly, rhs in Gadget. There is no implicit conversion as suggested in the title, the reference binds directly to the. 5. 2), an xvalue if T is an rvalue reference to object type, and a prvalue otherwise. The output is: Copy constructor with lvalue reference. Understanding Lvalues and Rvalues. The return of a new is a prvalue not an lvalue, because you cannot write: new T (arg) =. Only the following conversions can be done with const_cast. If you really want to pass i to g (), you have two options: provide a temporary object which is a copy of i (then considered as a rvalue) g (int {i}) force the conversion to rvalue reference with std::move (); then the original i must not. Even though the object in question is a temporary object, its lifetime has been extended. I played a bit around with composite-patterns and inheritance in c++. 5. 3. The effect of any implicit conversion is the same as performing the corresponding declaration and initialization and then using the temporary variable as the result of the conversion. And let’s define our storage to be either one of those cases: template<typename T> using Storage = std::variant<Value<T>, ConstReference<T>, NonConstReference<T>>; Now we need to give access to the underlying value of our variant, by providing a reference. It is VC++'s evil extension. Note that by binding a temporary to a rvalue-reference (or a const reference) you extend its lifetime. c++ base constructor lvalue to parameter. Function to pointer An lvalue that is a function can be converted to a C++11 (prvalue) C++11 rvalue that is a pointer to a function of the same type, except when the expression is used as the operand of the &(address) operator, the () (function call) operator, or the sizeof operator. } or in . So a and b are converted to rvalues before getting summed. 2 Answers. 5. (If you insist to know, the result of subscripting into an rvalue array used to be an lvalue in C++11, but is an xvalue in C++14 - issue 1213 . Therefore, in the third line, they undergo an implicit lvalue-to-rvalue conversion. The right constructors for the first two cases are called. That is the historical origin of the letters l. The rvalue reference is bound to the temporary materialized from the prvalue conversion of arr. The locator value is called lvalue, while the value resulting from evaluating that location is called rvalue. Correct, the epxression T() is always an rvalue for scalar and user-defined types T. Because a non-const reference is always a lvalue, so the code works and result in a lvalue (i. Convert any type to void, evaluating and discarding the value. it can be passed to a copy constructor or copy assignment operator as well (although overload resolution will prefer passing to a function which takes a rvalue reference). All lvalues that aren't arrays, functions or of. In C++03 copying the rvalue to an lvalue is the preferred choice (in some cases you can bind an lvalue reference to const to achieve a similar effect): int func2(){ // an rvalue expression. 7. undefined behavior: The lvalue or xvalue is a nonclass type, qualified by either const or volatile. You can: int&& x = 3; x is now an lvalue. You might want to use it more than once in your constructor, so it shouldn't be moved from on first use unless you explicitly want to. Then I use rvalue reference of class B's this pointer to pass it to A's constructor. 53 If T is an incomplete type, a program that necessitates this conversion is ill-formed. The choice of copy or move constructor only occurs when passing an object by value. Yes, the result of a cast to an object type is an rvalue, as specified by C++11 5. , cv1 shall be const), or the reference shall be an rvalue reference. A prvalue (“pure” rvalue) is an rvalue that is not an xvalue. C Server Side Programming Programming. 5 Reference binding (3) and 12. Among. 1 is rvalue, it doesn't point anywhere, and it's contained within lvalue x. If something happens to the temporary being referenced by a , b still holds a valid reference to a in the current scope. It is very easy to preserve the "lvalueness" of pre-increment: just increment the operand and return it as an lvalue. For example second type of the pair should be std::string, not const std::string * and all your problems would go away. Rvalue references enable you to distinguish an lvalue from an rvalue. const A& x = 1; //compile x = 2; //error! A&& xxx = 1; //compile A& xx = 1; //does not compile. 4. How to cast/convert pointer to reference in C++. This is not an rvalue reference. 1) does not accept such code (makes perfect sense). The expressions f (), f (). " What this is saying in layman's terms is that you can't (and shouldn't) store an address reference to an rvalue. warning C4238: nonstandard extension used: class rvalue used as lvalue But the very same program compiles fine in gcc 11 and clang 12 with the options -std=c++20 -Wall, without any warnings. Here's what happens when we call move with lvalue: Object a; // a is lvalue Object b = std::move (a); and corresponding move instantiation:3. Enums are different in C and C++, for example, if someColor is enum, 'someColor = 1' is legal C, but not C++. arg the expression (it is an expression at lines 3 and 4) has type int and value category "lvalue". The returned lvalue will contain exactly the result it is supposed to. Being an lvalue or an rvalue is a property of an expression. This is its value category. The value of x is 1. goo<int> is an lvalue of function type, but expressions of function type are. I'm not sure if this is the root of the issue but here's MSVC's implementation of std::array -related constructors of std::span . 1 Answer. init. In return w, the implicitly movable entity w is treated as an rvalue when the return type of the function is RRefTaker as in example three, but it is treated as an lvalue when the return type of the function is Widget && as in example four. 3. Basically, VS will allocate the space somewhere and just let the reference point to it, as if it was a reference-to- const without the constness (or in C++11 an rvalue reference). ref], a reference can be bound directly to the result of applying a conversion function to an initializer expression. This is the place where compiler complains, because i as lvalue cannot be bound to rvalue reference. C++ (as opposed to C) is a devoted lvalue-preserving language: it strives to painstakingly preserve the "lvalueness" of an expression whenever it is possible. – Corristo. If you write arg+1 inside the function, the lvalue expression arg of type int would. The Microsoft documentation is wrong. If I understand correctly what do you want, you can use std::reference (to wrap a l-value reference so that std::make_tuple() produce std::tuple with a reference in the corresponding position), and std::forward, to get the correct type of reference from a variadic list of arguments. In fact, in terms of overload resolution, an rvalue prefers to be bound to an rvalue reference than to an lvalue const reference. The value category of a compound literal is lvalue (its address can be taken). (prvalue) The output of this example is: produces an answer of type int because both are integers. I'm a bit confused about lvalue and rvalue bindings, I have the following code:. — Otherwise, the reference shall be an lvalue reference to a non-volatile const type (i. This example might clarify it: 16. 3. 2. User-defined conversion function and casting to reference. This example might clarify it:So we have a reference being initialized by an xvalue of type const foo. An entity (such as an. Yes. 3. Therefore, I will not jump right in and explain what rvalue references are. must error, because you're trying to pass an rvalue argument, std::move(n), to a lvalue reference parameter, T&. For the class type A, f (a); causes the copy constructor of A to be invoked. To convert an rvalue to an lvalue, you can use this lvalue helper function: template<class T> T& lvalue_ref (T&& x) { return x; } And then the call becomes: scan (lvalue_ref (std::ifstream ("myfile")), lvalue_ref (Handler ())); This is safe as the temporaries (the ifstream and Handler) aren't destructed until the end of. Say we want to steal from an lvalue: int main() { Holder h1(1000); // h1 is an lvalue Holder h2(h1); // copy-constructor invoked (because of lvalue in input) } This will not work: since h2 receives an lvalue in input, the copy constructor is being triggered. "Hello, World" is not of type const char*. If T is an lvalue reference type or an rvalue reference to function type, the result is an lvalue; if T is an rvalue reference to object type, the result is an xvalue; otherwise, the result is a prvalue. thus, this is legal: string&& s = foo (); // extends lifetime as before s += "bar"; baz (std::move (s)); // move the temporary into the baz function. Since your t variable is an lvalue, std::apply calls product with an int& instead of an int&&. 5. e. An rvalue can also be bound to a const lvalue reference, i. C++98 the rhs in built-in pointer-to-member access operators could be an lvalue can only be an rvalue CWG 1800: C++98 when applying & to a non-static data member of a member anonymous union, it was unclear whether the anonymous union take a part in the result type the anonymous union is not included in the result type CWG. Whenever a glvalue expression. Practically every example of lvalue-to-rvalue conversion I've seen on the web relates to fundamental types like int etc. @eerorika In your example y is an int, so it qualifies for rvalue conversion on return. Of course, this is not surprising: no one would expect. Even if the variable's type is rvalue reference, the expression consisting of its name is an lvalue expression; now your data member m_v is vector which contains. So in this case, this should be a prvalue B* and perfectly bindable to B*&&. using g++. To convert an lvalue to an rvalue, you can also use the std::move() function. As we've seen earlier, a and b are both lvalues. If t returns by lvalue reference, the code does not compile because a rvalue reference cannot bind to it. If inside foo no move operation happened like my example, then my_ptr_var will not actually be moved from. 11 for the exact listing what the cast can do; what that section doesn't list, it can't do. call]/12, [expr. 12. 2. universal reference. The question related to this one. I could have used std::move to convert the lvalue to rvalue reference and the call would be successful. I have tried to simulate the assignment of the object (pair. So MSVC++ is giving incorrect result (in case of C++ code). For non-class types you cannot assign to rvalues. A constant lvalue reference to a temporary doesn't lead to trouble, a non-constant reference to a temporary can: the receiver might be treating it as an out-parameter, and the caller might not notice the conversion that means a temporary is being passed. lval]/3. Safe downcast may be done with dynamic_cast. const tells you if a variable can be modified or not. Being an lvalue or an rvalue is a property of an expression; that is, every expression is either an lvalue or an rvalue. 4/1: The result is an lvalue if T is an lvalue reference type or an rvalue reference to function type and an xvalue if T is an rvalue reference to object type; otherwise the result is a prvalue. The new version creates a temporary of type double for the conversion int -> double and binds. 5 (I only have the current draft, your paragraph number may vary) we read : An lvalue for an object is necessary in order to modify the object except that an rvalue of class type can also be used to modify its referent under certain circumstances. class XAttr : public AttrDec { public: XAttr (const std::wstring& name) :AttrDec (new Attr (name)) // create a pointer here {} }; And then get rid of the rvalue constructor in AttrDec. Related reference: “Pointers” on page 114. A move constructor and move assignment operator can now. why std::forward converts both as rvalue reference. for the same reason as that example. For fundamental types, the copy approach is reasonable. 106) This requires a conversion function (12. There's a special rule in C++ template deduction rules which says that when the parameter type is T&& where T is a deduced type, and the argument is an lvalue of type. The rvalue variant can already bind to this because you're already passing a temporary and the lvalue variant can bind to. That would also solve the <T> issue BTW. I recently filed a bug against MSVC which relates to this, where the non-standard behavior caused standard-compliant code to fail to compile and/or compile with a deviant behavior. Another example of conversion: int c = 6; &c = 4; //ERROR: &c is an rvalue On the contrary you cannot convert an rvalue to an lvalue. Lvalue-to-rvalue conversion C++. This is apprently gcc's interpretation, and since arr is not an rvalue (it is an lvalue), this tiebreaker is skipped and no subsequent tiebreakers apply. 2 1). Temporary materialization thus occurs in both of the OP's examples: The first temporary (with value 10) will be. Explicitly call a single-argument constructor or a conversion operator. 9. But you can take the address of an array, as with &arr. 27 Non-Modifiable Lvalueslvalue_cast(const T& rvalue) {return const_cast<T&>(rvalue);} converts a rvalue to a lvalue, by changing const reference to a non-const reference (removing const qualification on the variable). But due to the the existence of std::vector::push_back(value_type const & val), which copies and would be the overriding call, I need to convert the lvalue object to an rvalue. ). lvalues and rvalues are expression categories, not flavours of object. It can appear only on the right-hand side of the assignment operator. move simply returns an rvalue reference to its argument, equivalent to. But in this particular case, the rules. This distinction is very important and seems to be overlooked by most when introduced to the topic. 1, 4. Numeric literals, such as 3 and 3. If type is an lvalue reference type or an rvalue reference to a function type, the cast result is an lvalue. 1. Therefore, I thought of providing some macro/function that wraps a parameter so it can be passed whether it's an l/rvalue - in this case get_addr. Conversion of a function pointer to void * shall not alter the representation. 3. We can take the address of an lvalue, but not of an rvalue. N. I think I'm missing something basic regarding the lvalue-to-rvalue standard conversion. Here is a silly code that doesn't compile: int x; 1 = x; // error: expression must be a modifyable lvalue. g++ t. To set this compiler option in the Visual Studio development environment. Lvalue-to-rvalue can be considered the reading of a value from an object in memory. That is any named parameter of a function cannot be implicitly casted or used to initialize another rvalue reference; it only copies to lvalue references; but static_cast can explicitly cast the valueness of the reference. You could also pass it to a function accepting a const char*& (i. an lvalue reference instead of an rvalue reference) and had the appropriate cv-qualification, then it's probably the programmer's mistake. For example, this means, that when rvalue reference is passed to a function, an lvalue reference overload will be chosen: T&& x=T(); f(x); Links: C++ lvalue rvalue xvalue glvalue prvalue Value categories in C++ 17 Value categories.