2) returning a reference type. 197. Clang vs G++ lvalue to rvalue conversion. Thus, this syntax is now legal: T&& r = T(); rvalue references primarily provide for the following: Move semantics. , [expr. It's actually a cast. Yes it's possible, just add a const to your second overload: template<typename T> void overloaded (const T& x); template<typename T> void overloaded (const T&& x); // ^^^^^. Consider this similar question: "Is an integer an lvalue or an rvalue". const tells you if a variable can be modified or not. Note that there is one exception: there can be lvalue const reference binding to an rvalue. X& r = X(99); // ERRORI use forward declaration here to pass object of class B as parameter in class A. 3. 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). In C++, the cast result belongs to one of the following value categories:. 3. Note that the lvalue-to-rvalue conversion is not the only conversion that converts an lvalue to a prvalue: There's also the array-to-pointer conversion and the function-to-pointer conversion. 1 Answer. That means std::move could take both lvalue and rvalue, and convert them to rvalue unconditionally. assign values to the reference return type directly in c++. 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. This is because, in C programming, characters are internally stored as integer values known as ASCII Values. It's also echoed in 5. 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). The lvalue-to-rvalue conversion is covered in N3485 in section 4. Under the conditions specified in [dcl. You don't need universal reference here const T& source is enough and simpler. conv] 1 A simple-type-specifier or typename-specifier followed by a parenthesized optional expression-list or by a braced-init-list (the initializer) constructs a value of the specified type given the initializer. Intuitively, a typecast says "give me the value that this expression would have if it had some other type," so typecasting a variable to its own type still produces an rvalue and not an lvalue. 4. Both of these options are user-defined conversion functions, so neither is better in terms of overload resolution, thus an ambiguity. 9. A pointer is a type. You would need const_cast<char*&> (a) in order to have an lvalue to assign to, and that brings up the next problem. e. Update: The code is ill-formed in C++11. Here’s a much more concise rundown (assuming you know basic C++ already): Every C++ expression is either an lvalue or rvalue. 25, or 4 (leaving off the units for brevity). If an lvalue-to-rvalue conversion were performed on s, it would also call the copy constructor; [conv. lvalue:-. This is a changeable storage location. This is why you will see the C++ library provide what appears to be a single template, that works in both lvalue and rvalue contexts. 8. The list of languages that are currently supported includes C++, C#, Go, Java, Kotlin, PHP, Python, Ruby, Rust, TypeScript, and more. This function takes an lvalue reference and converts it to an rvalue reference. If T is an incomplete type, a program that necessitates this conversion is ill-formed. Both of g and h are legal and the reference binds directly. You are comparing two different things that are not really related. If U is t’s underlying non-reference type (namely std::remove_reference_t<decltype(t)>), then T will. According to the C++ specifications, it takes two rvalues as arguments and returns an rvalue. (This is somewhat of a simplification, in C++11 we have lvalues, xvalues and prvalues. An object is a region of storage that can be examined and stored into. An entity (such as an. @BЈовић: I did mean that (although I've since renamed the function baz). thanks a lot! I've just another question for you. If you can't, it's usually an rvalue. This function takes an lvalue reference and converts it to an rvalue reference. From C++11 4. (prvalue) The output of this example is: produces an answer of type int because both are integers. 2 Infinite. [3] Finally, this temporary variable is used as the value of the initializer. in Example 1 Lvalue-to-rvalue conversion is applied to the two operands ( x and 0) No. Note: The ISO C standard does not require this, but it is required for POSIX conformance. . A r-value is an expression, that can’t have a value assigned to it, which means r-value can appear on right but not on left hand side of an assignment operator (=). However it is prohibited to accept rvalues when forwarding as an lvalue, for the same reasons that a reference to non-const won't bind to an rvalue. Therefore, in the third line, they undergo an implicit lvalue-to-rvalue conversion. The type of the variable k is an r-value reference, but that's fine. If you compile with /W4 then the compiler will warn you. M. Correct. G. At the same time, we cannot move away from const values. This example might clarify it: 16. This type of static_cast is used to implement move semantics in std::move. Yes, rvalues are moved, lvalues are copied. If you pass an argument to a reference type parameter (whether lvalue or rvalue reference), the object will not be copied. reinterpret_cast reinterpret_cast converts any pointer type to any other pointer type, even of unrelated classes. Similarly, rhs in Gadget. If an lvalue or xvalue is used in a situation in which the compiler expects a (prvalue) rvalue, the compiler converts the lvalue or xvalue to a (prvalue) rvalue. ; In all other cases, the cast result is a (prvalue) rvalue. If element at this position doesn't exist, function. 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. Open the project's Property Pages dialog box. 4. Safe downcast may be done with dynamic_cast. 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. They are declared using the ‘&’ before the name of the variable. 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. 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. The only thing that can be an rvalue or an lvalue is an expression. This means the following is illegal: int main() { const int x { 5 }; int& ref { x }; return 0; } This is disallowed because it would allow us to modify a. It's not needed, and suppressed. An obvious example of an lvalue expression is an identifier with suitable type and storage class. Rvalue references are a feature of C++ that was added with the C++11 standard. But it is still a reference, which is a lvalue. You do pass an rvalue to some_function - but at the same time you create an argument rvalue_ref which is now an lvalue (so you can actually call the. You are returning a copy of A from test so *c triggers the construction of a copy of c. 10): An lvalue (so called, historically, because lvalues could appear on the left-hand side of an assignment expression) designates a function or an object. Related reference: “Pointers” on page 114. It can convert between pointers. 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. 左值可以出现在赋值号的左边或右边。. 3. 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 . You can convert an lvalue to an rvalue by casting it to an xvalue; this is conveniently encapsulated into the type-deducing cast. Lvalue-to-rvalue conversion. Overload resolution is used to select the conversion function to be invoked. To get a lvalue expression to the value pointed to by a pointer, just use the unary * operator. Return lvalue reference from temporary object. That is the historical origin of the letters l. In that case, consider processing only one argument at a time, leaving the remaining ones as rvalue-references. 1, a standard conversion sequence cannot be formed if it requires binding an lvalue reference other than a reference to a non-volatile const type to an rvalue or binding an rvalue reference to an lvalue other than a function lvalue. Assume a variable name as a label attached to its location in memory. Compiled with "g++ -std=c++0x". When an lvalue-to-rvalue conversion is applied to an expression e, and either. std::get returns an lvalue reference if its tuple argument is an lvalue. 3. You can disable this behaviour with the /Za (disable language extensions) compiler switch under. OK. Improve this answer. All lvalues that aren't arrays, functions or of. 1, 4. I guess you are reading the Rvalue References: C++0x Features in VC10, Part 2. @banana36 With that function, calling foo(std::move(my_ptr_var)) wont actually pass ownership. L-value: “l-value” refers to memory location which identifies. init. ”. C++20 the conversion restriction regarding designated initializer lists was applied even if the parameter is a reference not restricted in this case P2468R2:Postfix operator++ requires the value-category of the operand to be an l-value, regardless of the type of the operand. When you create a std::reference_wrapper<int> and pass it in, rvalues of that type can convert to int&. Forwarding references are very greedy, and if you don't pass in the exact same type (including. To convert an lvalue to an rvalue, you can also use the std::move() function. Hence we know that in int t = e; , the result of the conversion sequence is a prvalue, because int is a non-reference type. Operationally, the difference among these kinds of expressions is this:std::function can be move-constructed from rvalue of a functor object. ASCII defines a set of characters for encoding text in computers. Radius: 2 2 4. That means std::move could take both lvalue and rvalue, and convert them to rvalue unconditionally. lvalueはアドレスを取得できるがrvalueはアドレスを取得できない。 これは一見見分ける強力な手段に思える。しかし考えて欲しい。コードを書くときにいちいちアドレス演算子を書いてコンパイルしてみるなんて悠長なことをするだろうか、いいやしない。2 Answers. Example: std::unique_ptr<int> get_int() { auto p = std::make_unique<int>(1); // `p` is an lvalue but treated as an rvalue in the return statement. The idea is that if you have a reference binding that could have been a direct binding if only the reference were of the appropriate kind (i. Each expression in C (an operator with its arguments, a function call, a constant, a variable name, etc) is characterized by two independent properties: a type and a value category . The first constructor is the default one. 3 and of temporaries in 12. Assuming C++11 or later:. Yes, rvalues are moved, lvalues are copied. Otherwise, the reference you get behaves more. If an lvalue-to-rvalue conversion were performed on s, it would also call the copy constructor; [conv. Ternary conditional operator will yield an lvalue, if the type of its second and third operands is an lvalue. g. Otherwise, the reference shall be an lvalue reference to a non-volatile const type (i. std::string hello = "hello"; std::string planet. This allows you to explicitly move from an lvalue, using move. That would also solve the <T> issue BTW. Lvalue and rvalue are expressions that identify certain categories of values. オブジェクトという言葉が聞き慣れないなら. So sizeof (0, arr) = sizeof (arr) and which would be equal to 100* sizeof (char) and not = sizeof (char*). And an identifier "is an lvalue if the entity is a function or variable" (5. — Otherwise, the reference shall be an lvalue reference to a non-volatile const type (i. "cannot bind non-const lvalue reference of type ‘M&’ to an rvalue of type. The confusion you're having is pretty common. 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. Then std::forward<SomeClass&> (element) will be invoked, and the instantiation of std::forward would be. 1 is rvalue, it doesn't point anywhere, and it's contained within lvalue x. But you can take the address of an array, as with &arr. 5. An lvalue does not necessarily permit modification of the object it designates. Using lvalue or rvalue qualifiers to construct a correct interface for lvalue or rvalue objects is just the same as using const, and it should be approached the same way- each function should be considered for restriction. return 17;} int m=func2(); // C++03-style copying. 1, 4. Using it only makes sense inside of a template, where you choose whether to move or not depending on a template argument. The implementation of the language level is based on IBM's interpretation of the standard. 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. accesses its value), casts that value to T1, constructs a temporary of type T1 (with value 1, since that is the value of b and is a valid value of type T1 ), and binds it to an rvalue. rvalue references are considered lvalue (this part I understand) They are not. In C++, non-const references can bind to lvalues and const references can bind to lvalues or rvalues, but there is nothing that can bind to a non-const rvalue. rvalue/lvalue tells you the value category. 44. But in the circumstances of the linked question, the template instantiation of std::function cannot be inferred from the lambda type. Given all three functions, this call is ambiguous. 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 C++11, you can go one step further and obtain a non-const pointer to an temporary: template<typename T> typename std::remove_reference<T>::type* example (T&& t) { return &t; } Note that the object the return value points to will only still exist if this function is called with an lvalue (since its argument will turn out to be. 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. lval]/3. begin(), dataBlock. c++ template type matching with references [duplicate] Ask Question Asked 5 days ago. C++98 it was unspecified whether a temporary is created for an lvalue-to-rvalue conversion on the conditional operator always creates a temporary if the operator returns a class rvalue CWG 462: C++98 if the second operand of a comma operator is a temporary, it was unspecified whether its lifetime will be extended whenIt is used to convert an lvalue into an rvalue. e. (since C++11) 4) If new_type is the type void (possibly cv-qualified), static_cast discards the value of expression after. A modifiable lvalue may be used as the first (left) argument of the built-in assignment operator. 2. 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. @MikeMB the standard rarely prevents compilers from inserting for (int i = 0; i < 1 billion; ++i) at arbitrary points. c++11 decltype returns reference type. Thus, if the thickness is 1 inch, and the K-value is 0. The answer lies in the second property of expressions: the value category. 2) If target-type is an rvalue reference type, static_cast converts the value of glvalue, class prvalue, or array prvalue (until C++17) any lvalue (since C++17) expression to xvalue referring to the same object as the expression, or to its base sub-object (depending on target-type). Thus you need only two overloads plus recursive calls, but the exact form depends on what you. In that sense, rvalue references are a new language feature that adds a generic rvalue-to-lvalue. This distinction is very important and seems to be overlooked by most when introduced to the topic. Then I use rvalue reference of class B's this pointer to pass it to A's constructor. Note that by binding a temporary to a rvalue-reference (or a const. Let's think of the addition +. It matches arguments of any value category, making t an lvalue reference if the supplied argument was an lvalue or an rvalue reference if the supplied argument was an rvalue. An lvalue is an expression that designates (refers to) an object. An lvalue reference (commonly just called a reference since prior to C++11 there was only one type of reference) acts as an alias for an existing lvalue (such as a variable). lvalue cannot be a function, expression (like a+b) or a constant (like 3 , 4 , etc. This is a helper function to allow perfect forwarding of arguments taken as rvalue references to deduced types, preserving any potential move semantics involved. If t returns by lvalue reference, the code does not compile because a rvalue reference cannot bind to it. Problems remaining in C++20 3. If you had. void f1(int& namedValue){. first is in your example's instantiation is a rvalue (specifically xvalue) regardless of the const. ; // not legal, so no lvalue. You could not pass it to a function accepting a const char*&& (i. 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. This is not an rvalue reference. So, the conversion from a A rvalue to something that P&& would accept in (1) calls the user defined conversion function operator P() &&. 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. 9/1: The result of the expression static_cast<T> (v) is the result of converting the expression v to type T. The value category of an expression (or subexpression) indicates whether an expression. Variables of type rvalue reference have to be initialized in their definition like variables of type lvalue reference. On the other hand lvalue references to const forbids any change to the object they reference and thus you may bind them to a rvalue. Whenever an lvalue is used in a position in which an rvalue is expected, the compiler performs an lvalue-to-rvalue conversion and then. Being an lvalue or an rvalue is a property of an expression; that is, every expression is either an lvalue or an rvalue. I would respect the first compiler more, it is at least. Done. 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. Nothing is changed except the value category. 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); } 1 Answer. Lvalue-to-rvalue conversion C++. But i=3; is legal if i is an integer. But I do not see how it is related to the warning, please explain. “If T1 is reference-related to T2 and the reference is an rvalue reference, the initializer expression shall not be an lvalue. Since the type of a is not an int, it cannot match the type that b. 3. As we've seen earlier, a and b are both lvalues. 左值（lvalue）：指向内存位置的表达式被称为左值（lvalue）表达式。. 19, 9th bullet, three sub-bullets). It's long-lived and not short-lived, and it points to a memory location where 1 is. The reason why you need to const is to make x not a forwarding reference. 2, and 4. template <typename T> StreamWriter& StreamWriter::operator<< (const T& source) { Write (&source); return *this; } Share. But for the third case i. B. cond]/7. foo now is null. 1. rvalue references are marked with two ampersands (&&). ConclusionFrom expr. 1 is rvalue, it doesn't point anywhere, and it's contained within lvalue x. 9. The goal of rvalue references is sparing copies and using move semantics. C++11 also invented the forwarding reference: that when there’s a deduced type T directly modified by &&, T can sometimes be deduced as an lvalue reference type. If t returns by rvalue reference, you obtain a reference to whatever was returned. This way you explicitly say T&& should not match an lvalue-reference. 2. In C++, an rvalue is a temporary object that does not have a stable location in memory. 9. x is not assignable, because it's an rvalue in 03, a prvalue in 11 and an xvalue in 14, but using a member function always allows you to convert rvalues to lvalues (because *this is always an lvalue). In C++, each expression, such as an operator with its operands, literals, and variables, has type and value. h and move. However, it's type will be const std::string or std::string depending on the choice of const in the MyPair type. 2. e. Types shall not be defined in a reinterpret_cast. (since C++11)20. 3. first) as same as the implementation of std_pair. 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. Such an expression is always an lvalue, even if x is an rvalue and even if y is an rvalue reference. const T& still binds happily to both lvalues and rvalues. auto (* p) [42] = & a; is valid if a is an lvalue of type int [42]. But an rvalue reference can't bind to an lvalue because, as we've said, an rvalue reference refers to a value whose contents it's assumed we don't need to preserve (say, the parameter for a move constructor). Even if the variable's type is rvalue reference, the expression consisting of its name is an lvalue expression; vector has two overloads of assignment operator, one for Lvalue reference. fstream file{"filename"}; print_stream(file);I would like to write a variadic template function that accepts rvalues and lvalue references. For example, assume you pass an rvalue reference to an object of type X to a function template that takes type T&& as its parameter. The initializer for a const T& need not be an lvalue or even of type T. returning either a rvalue or an lvalue. Here is a silly code that doesn't compile: int x; 1 = x; // error: expression must be a modifyable lvalue. But when there's no according move operation, rvalues are copied as well. Conversely, d = static_cast<float> (j)/v; produces an. The result of std::move is an xvalue [1], which is a type of glvalue; and converting a glvalue to an lvalue reference with reinterpret_cast appears to be allowed by the wording. 2 Answers. L-Values are locations, R-Values are storable values (i. Since int() isn't an lvalue, you can't assign to int(). Therefore, in the third line, they undergo an implicit lvalue-to-rvalue conversion. ). It could even do so with std::move only. 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. (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”. So instead of A a = A (10); what gets called is this A a (10); If you want to disable copy elision, compile the above program with. the name of a variable, a function, a template parameter object (since C++20), or a data member, regardless of type, such as std::cin or std::endl. Which basically triggers the non-const rvalue to non-const lvalue conversion and makes all the difference in the example above. I. There's no benefit in this case. static_cast can do other things, as listed in 5. Forwarding references are a special kind of references that preserve the value category of a function argument, making it. 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. 1 Answer. 53 If T is an incomplete type, a program that necessitates this conversion is ill-formed. I could have used std::move to convert the lvalue to rvalue reference and the call would be successful. e. You decided to add a move. The term “identity” is used by the C++ standard, but is not well-defined. about undefined behaviorIf T is a reference an lvalue-reference type, the result is an lvalue; otherwise, the result is an rvalue and the lvalue-to-rvalue (conv. 1. Otherwise, the reference you get behaves more. For the second overload, it would call operator const P&() const&. Write a function template to convert rvalues to lvalues: template<typename T> T &as_lvalue (T &&val) { return val; } Now, use it: deref (&as_lvalue (42)); Warning: this doesn't extend the lifetime of the temporary, so you mustn't use the returned reference after the end of the full-expression in which the temporary was. If you write arg+1 inside the function, the lvalue expression arg of type int would. Would you ever mark a C++ RValue reference parameter as const. Rvalue references work in principle similarly to Lvalue references: We declare them by writing the data type of the rvalue followed by && and an identifier. Temporary materialization thus occurs in both of the OP's examples: The first temporary (with value 10) will be. The locator value is called lvalue, while the value resulting from evaluating that location is called rvalue. By make_tuple<int> you make make_tuple signature look like: make_tuple(int&&). 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. During reference initialization, where the reference to cv1 T is bound to the lvalue or rvalue result of a conversion from the initializer expression from the class type cv2 S,. It satisfies the requirements in 4. You will often find explanations that deal with the left and right side of an assignment. There is no lvalue-to-rvalue conversion in this scenario. r-value references are designed to be the subject of a move-constructor or move-assignment. It was introduced specifically to allow temporary streams to be usable without resorting to tricks. c++ base constructor lvalue to parameter. 3 -- Lvalue references ), we discussed how an lvalue reference can only bind to a modifiable lvalue. The discussion of reference initialization in 8. But then i got following error:. func) standard conversions are performed on the the expression v. One more step. It's been part of the language since the beginning. For example, this code will not compile. b is just an alternative name to the memory assigned to the variable a. Both rvalues and lvalues can be modified. You can use the function template is_lvalue (below) to find out if an operand is an lvalue and use it in the function template isTernaryAssignable to find out if it can be assigned to. 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". Of course, this is not surprising: no one would expect. The question related to this one. void func (unsigned int& num) this function need quote type. So: since foo () returns a reference ( int& ), that makes it an lvalue itself. Otherwise, the type of the rvalue (until C++11) prvalue (since C++11) is T. Yes, the result of a cast to an object type is an rvalue, as specified by C++11 5. int&& x = 3; x is now an lvalue. rvalue rvalue lvalue. All lvalues that aren't arrays, functions or of incomplete types can be converted to rvalues. const A& x = 1; //compile x = 2; //error! A&& xxx = 1; //compile A& xx = 1; //does not compile. Also, xvalues do not become lvalues. To convert an lvalue to an rvalue, you can also use the std::move() function. In the previous lesson ( 12. An rvalue is a prvalue or an xvalue. Don't mix the two patterns. Allowing non-const references to bind to r-values leads to extremely confusing code. here, X is copied into a temporary tuple, then the copy is passed to thread_exrcutor as a rvalue. For example in an expression. 3. cond]/7. This allows you to explicitly move from an lvalue, using move to. 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. You can't assign to an object that is const. arg the variable has type int&& and no value category. But in this particular case, the rules. So when you bind the references the lvalue will have to be const. Although the syntax of a compound literal is similar to a cast, the important distinction is that a cast is a non-lvalue. m, static_cast<A&&> (a), and a + a are xvalues. It can convert lvalues to lvalue references and rvalues to rvalue references. In such cases: [1] First, implicit type conversion to T is applied if necessary. cpp -std=c++11 -fno-elide-constructors. Visual Studio warning disappears if one removes std::move. An lvalue (so called, historically, because lvalues could appear on the left-hand side of an assignment. 1:. 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. @eerorika In your example y is an int, so it qualifies for rvalue conversion on return. 3/5 of the C++11 Standard: A reference to type “cv1 T1” is initialized by an expression of type “cv2 T2” as follows: — If the reference is an lvalue reference and the initializer expression — is an lvalue (but is not a bit-field), and “cv1 T1” is reference-compatible with “cv2 T2,” orAn expression has a possibly cv-qualified non-reference type, and has value category: lvalue, xvalue, or prvalue. Their very nature implies that the object is transient. I played a bit around with composite-patterns and inheritance in c++. std::move performs a static_cast to an rvalue reference type and returns the rvalue reference. 1. When I discovered this, it seemed odd to me, so I tried. You could disallow rvalues, but not sure if that would be acceptable. Because a non-const reference is always a lvalue, so the code works and result in a lvalue (i. 0. – T. 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”. init. — even if the implicit object parameter is not const-qualified, an rvalue can be bound to the parameter as long as in all other respects the argument can be converted to the type of the implicit object parameter. Explicitly call a single-argument constructor or a conversion operator. 3) If new_type is an rvalue reference type, static_cast converts the value of expression to xvalue. 12. call]/12, [expr. , [expr. The quote doesn't say anything about the result of &, which in fact is an rvalue. the original code was int&& rref = n; which was ill-formed, as n is an lvalue and therefore cannot bind to an rvalue reference. There are two common ways to get an xvalue expression: Use std::move to move an object. The value of x is 1. lvalue. Radius: 2 2 4. 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. Understanding Lvalues and Rvalues.