5 Reference binding (3) and 12. It can convert lvalues to lvalue references and rvalues to rvalue references. But Args itself is either an lvalue reference or not a reference. end()) is a temporary object and cannot be bound to lvalue reference. — Otherwise, the reference shall be an lvalue reference to a non-volatile const type (i. But I do not see how it is related to the warning, please explain. The reason why you need to const is to make x not a forwarding reference. So a and b are converted to rvalues before getting summed. e. Select the Configuration Properties > C/C++ > Language property page. This example might clarify it:So we have a reference being initialized by an xvalue of type const foo. This is what std::move is for. Firstly, pre C++17, the result of A<double>(a2) is an rvalue. 1) does not accept such code (makes perfect sense). 5. call]/12, [expr. static_cast can do other things, as listed in 5. @banana36 With that function, calling foo(std::move(my_ptr_var)) wont actually pass ownership. c++ c++11 overload-resolution rvalue Share Follow edited Jan 14, 2016 at 8:52 ildjarn 62. Introduction. 1. 7. Note that there is one exception: there can be lvalue const reference binding to an rvalue. return 17; //an lvalue reference to an rvalue} 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): r-value references are designed to be the subject of a move-constructor or move-assignment. Set the Enforce type conversion rules property to /Zc:rvalueCast or /Zc:rvalueCast. It is used to convert an lvalue into an rvalue. This differs from ISO C, in. Lvalue to rvalue conversion. and includes the following bullet which the examle belongs to: the evaluation of e results in the evaluation of a member ex of the set of potential results of e, and ex names a variable x that is not odr-used by ex (3. 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. 1 Answer. If you compile with /W4 then the compiler will warn you. When such a binding occurs to a prvalue, a temporary object is materialized. オブジェクトという言葉が聞き慣れないなら. A function parameter such as T&& t is known as a forwarding reference. 1, 4. (for user-defined types): rvalue or lvalue?. Whenever an lvalue is used in a position in which an rvalue is expected, the compiler performs an lvalue-to-rvalue conversion and then. The following diagram illustrates the relationships between the. But you can take the address of an array, as with &arr. With argument deduction, parameter of make_tuple is deduced to be: int&, and in this case i can be bound. Nothing is being turned into a lvalue. Their very nature implies that the object is transient. C++ pass parameter by rvalue reference if possible, otherwise copy the lvalue reference. returning either a rvalue or an lvalue. h and move. The second one constructs the object with an lvalue reference which reads the argument, t. 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. e. It would capitalize std::strings, and display each parameter after they are capitalized. All lvalues that aren't arrays, functions or of incomplete types can be converted to rvalues. 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&). move simply returns an rvalue reference to its argument, equivalent to. 2) yield xvalues, such as a call to a function whose return type is an rvalue reference or a cast to an rvalue reference type. move simply returns an rvalue reference to its argument, equivalent to. As with all cast expressions, the result is: an lvalue if target-type is an lvalue reference type or an rvalue reference to function type(since C++11) ; an xvalue if target. C++0x, by contrast, introduces the following reference collapsing rules: The second rule is a special template argument deduction rule for function templates that take an argument by rvalue reference to a template argument: When foo is called on an lvalue of type A, then T resolves to A& and hence, by the reference collapsing rules above, the. specifically, the argument expression is an rvalue that is bound to the rvalue reference parameter. thus, this is legal: string&& s = foo (); // extends lifetime as before s += "bar"; baz (std::move (s)); // move the temporary into the baz function. C Server Side Programming Programming. The value of x is 1. Put simply, an lvalue is an object reference and an rvalue is a value. Since the type of a is not an int, it cannot match the type that b. We're talking about the temporary object created by Contrived(), it doesn't make sense to say "this object is an rvalue". This isn't strictly true in all cases; in unevaluated. – 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. If U is t’s underlying non-reference type (namely std::remove_reference_t<decltype(t)>), then T will. If we have a lvalue we can return it from a function, so we get a rvalue. 4. If an lvalue-to-rvalue conversion were performed on s, it would also call the copy constructor; [conv. Informally this conversion is "evaluating" or "taking the value of" the object that the lvalue refers to. double && does not work for lvalues. baz(1) by itself is not UB, but it would be UB to dereference the resulting pointer after the end of the full-expression containing baz(1). If I change func (unsigned int&) to func (Color&), compiler accept it. lvalue and rvalue as function parameters. e. However, there is no reason why converting from one reference type to another as a cast should do anything at run time. 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. 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. Note that there is one exception: there can be lvalue const reference binding to an rvalue. An lvalue (so called, historically, because lvalues could appear on the left-hand side of an assignment. Creating a temporary object is usually not the desired behavior. This allows you to explicitly move from an lvalue, using move to. And it's on the value level that talking about rvalue/lvalue-ness makes sense (after all, those are called value categories). According to the C++ specifications, it takes two rvalues as arguments and returns an rvalue. e. (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”. 53 If T is an incomplete type, a program that necessitates this conversion is ill-formed. 10) of a non-function, non-array type T can be converted to a prvalue. int & a = b * 5 is invalid. For reference: The relevant standard sections are 12. 4. But then i got following error:. But is not an lvalue that the reference can be bound to because of the wrong type. 197. r-value references are designed to be the subject of a move-constructor or move-assignment. Yes, if you pass an lvalue const char* to a function accepting a const char*, that'll work. This distinction is very important and seems to be overlooked by most when introduced to the topic. You could also pass it to a function accepting a const char*& (i. An lvalue or xvalue is an expression that refers to such an object. 1: (5. –std::forward is usually the way to 'convert' value category. const foo&& can only bind to an rvalue, but const foo& can bind to both lvalues and rvalues. So when. (This is somewhat of a simplification, in C++11 we have lvalues, xvalues and prvalues. 4. 左值可以出现在赋值号的左边或右边。. 3. has an address). Read 5. 1) modifiable lvalues. 3. That's an exception to the general rule that it is impossible for lvalues to be bound to rvalue. An identifier that refers to an object is an lvalue, but an. Both of these options are user-defined conversion functions, so neither is better in terms of overload resolution, thus an ambiguity. An rvalue is a prvalue or an xvalue. So sizeof (0, arr) = sizeof (arr) and which would be equal to 100* sizeof (char) and not = sizeof (char*). In fact, in terms of overload resolution, an rvalue prefers to be bound to an rvalue reference than to an lvalue const reference. It could even do so with std::move only. Move semantics relies on a new feature of C++11, called rvalue references, which you'll want to understand to really appreciate what's going on. This implies that the compilers that accept the above code without a diagnostic are non-conforming (i. Forwarding references are very greedy, and if you don't pass in the. It seems like std::array can be converted to an std::span when it's an rvalue only on clang. The r-value reference is a reference to the original object, so converting it to a l-value reference will just make a reference to the original object. "Hello, World" is not of type const char*. For the second overload, it would call operator const P&() const&. According to the rule of forwarding reference, when an lvalue is passed to add, the template type argument Element will be deduced as SomeClass&. , [expr. 4. Cast to reference type. c++11标准基本上是通过举例来说明一个表达式是否是一个lvalue还是rvalue的。. An lvalue-to-rvalue conversion is a conversion from a non-function, non-array lvalue or xvalue of type cv T to a prvalue of either type cv T if T is a class type or T if T is not a class type. Hot Network QuestionsSorted by: 19. C++ does not allow you to get an r-value reference to a variable without an explicit conversion. A pointer is not the kind of thing that can be an rvalue or an lvalue. )In the third line, they undergo an implicit lvalue-to-rvalue conversion. The name “lvalue” comes from the assignment expression E1 = E2 in which the. 1 Answer. , Circle c3 (Circle (4)), I'd expect the third constructor, (copy constructor with rvalue referecne) to be called but it's not the case. 14159, are rvalues. In C++ class and array prvalues can have cv-qualified types. 区分左值和右值是很重要的，这是使用C++11 move语义的基础。. 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. template <typename T> StreamWriter& StreamWriter::operator<< (const T& source) { Write (&source); return *this; } Share. Improve this answer. 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. 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. foobar () is an rvalue because foobar () returns int. 1) Is actually not so arbitrary. Like this: template <typename T> void foo (T &&value) { f (std::forward<T> (value)); } Here, T &&value is called a forwarding reference (as long T is deduced by the compiler. A modifiable lvalue may be used as the first (left) argument of the built-in assignment operator. What you're referring to is the fact that if an expression. After C++11, the compiler did some work for us, where the lvalue temp is subjected to this implicit rvalue conversion, equivalent to static_cast<std::vector<int> &&>(temp), where v here moves the value returned by foo locally. U is a class type. Category 4 used to be a bit different in C++11, but I believe this wording is correct for C++14. Template argument deduction deduces T to be X, so the parameter has type X&&. 1:. The && syntax is either referring to a rvalue-reference or a universal-reference. Visual Studio warning disappears if one removes std::move. Note that when we say lvalue or rvalue, it refers to. 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. An lvalue or xvalue is an expression that refers to such an object. As long as no const is involved, the expression T() is a modifiable rvalue, to be more precise. So: since foo () returns a reference ( int& ), that makes it an lvalue itself. If you write arg+1 inside the function, the lvalue expression arg of type int would undergo this conversion to produce a prvalue expression of type int, since that's what built-in + requires. 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. The entire point is that you know that this entity references an rvalue and you can legitimately move its content. 2. The terms "lvalue/rvalue reference" and "lvalue/rvalue" are related but not interchangeable or one a shortened form of the other. Using lvalue references where rvalue references are required is an error: int& func2(){//compilation error: cannot bind. Using it only makes sense inside of a template, where you choose whether to move or not depending on a template argument. If you pass an prvalue, it isn't converted, the temporary is materialised into the parameter object. 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. Type conversions on references. Let's think of the addition +. But it is still a reference, which is a lvalue. To set this compiler option in the Visual Studio development environment. . 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. 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). You will often find explanations that deal with the left and right side of an assignment. lvalue-- an expression that identifies a non-temporary object. The reference declared in the above code is lvalue. std::forward is a conditional std::move. rvalues are defined by exclusion. 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. 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. 9/1: The result of the expression static_cast<T> (v) is the result of converting the expression v to type T. The Lvalue refers to a modifiable object in c++ that can be either left or right side of the assignment operator. b is just an alternative name to the memory assigned to the variable a. The reference declared in the above code is lvalue. @eerorika In your example y is an int, so it qualifies for rvalue conversion on return. 25, or 4 (leaving off the units for brevity). When being passed an lvalue, the template parameter would be deduced as lvalue-reference, after reference. c++ base constructor lvalue to parameter. 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. The expression *this is an lvalue; A {} is an rvalue (prvalue) even though they designate the same temporary object. Note that this must wait until construction is complete for two reasons. The difference between lvalues and rvalues plays a role in the writing and understanding of expressions. In C++, each expression, such as an operator with its operands, literals, and variables, has type and value. Even though the object in question is a temporary object, its lifetime has been extended. When I discovered this, it seemed odd to me, so I tried. There are two common ways to get an xvalue expression: Use std::move to move an object. 9. The C++ language says that a local const reference prolongs the lifetime of temporary values until the end of the containing scope, but saving you the cost of a copy-construction (i. An lvalue may get converted to an rvalue: that's something perfectly legit and it happens quite often. In C++ results of conversions are always rvalues (unless you convert to reference type). The locator value is called lvalue, while the value resulting from evaluating that location is called rvalue. 2 1). References. The implementation of the language level is based on IBM's interpretation of the standard. g. But i=3; is legal if i is an integer. @YueZhou Function lvalues may be bound to rvalue references. Rvalue references enable you to distinguish an lvalue from an rvalue. However what matters here is the expression and: Each C++ expression (an operator with its operands, a literal, a variable name, etc. static_cast<Type> (expression) belongs to one of the following value categories: or an rvalue reference to a function type is an lvalue. This allows you to explicitly move from an lvalue, using move. 3. Allowing both rvalues and lvalues to be bound to an lvalue reference makes that impossible. Rvalue references are a feature of C++ that was added with the C++11 standard. static_cast<typename remove_reference<T>::type&&> (t) The result of the function call is an rvalue (specifically, an xvalue ), so it can be bound to an rvalue reference where the function argument couldn't. [2] Then, the resulting value is placed in a temporary variable of type T. static_cast can do other things, as listed in 5. The right constructors for the first two cases are called. cast (this is applicable from C++11 and later). lvalue and rvalue in C. Through an lvalue to rvalue conversion. Naming expressions are always lvlaues. When you create a std::reference_wrapper<int> and pass it in, rvalues of that type can convert to int&. It is of type const char [13] and it is an lvalue, not an rvalue. It boils down to an lvalue assignment - references as function arguments refer to objects that may exist for longer than a function call, and as such are lvalues even when the argument type is an rvalue. But you might just let regular deduction occurs. why std::forward converts both as rvalue reference. 12. const T& still binds happily to both lvalues and rvalues. We are allowed to do that because the object is an rvalue, when the constructor finishes its job, t will be destructed. 14′. Now enter C++11 with rvalue references and move semantics. An lvalue may be used to initialize an lvalue reference; this associates a new name with the object identified by the expression. N. It can appear only on the right-hand side of the assignment operator. Returning an explicit rvalue-reference. first is in your example's instantiation is a rvalue (specifically xvalue) regardless of the const. c++11 decltype returns reference type. There are no references of references in C++. An lvalue does not necessarily permit modification of the object it designates. 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. [ Note: If T is a non-class type that is cv. 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. Conversion of a function pointer to void * shall not alter the representation. You do not need workaround on how to use rvalue as lvalue, but rather fix your code that you do not need this workaround. Open the project's Property Pages dialog box. All you have to do here is make sure you get a pointer to an array, rather than a pointer to the first element of the array. An lvalue (pronounced “ell-value”, short for “left value” or “locator value”, and sometimes written as “l-value”) is an expression that evaluates to an identifiable object or function (or bit-field). 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. const A& x = 1; //compile x = 2; //error! A&& xxx = 1; //compile A& xx = 1; //does not compile. While the type of _x is 'r-value reference to std::vector<int>', it is still an l-value as it has a name. Such an expression is always an lvalue, even if x is an rvalue and even if y is an rvalue reference. e. If element on this position doesn't exist, it should throw exception. 99 * @return The parameter cast to an rvalue-reference to allow moving it. 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. , values that can be assigned: namespaces, for instance, are not assignable; thanks to @Maggyero for the edit suggestion). I would like to move an object into a std::vector using std::vector::push_back(). 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. Yes, rvalues are moved, lvalues are copied. Lvalue reference and rvalue reference are both types; because the names are so similar, it is easy to confuse the two. By tracing slt_pair. Indeed it does. If you pass an argument to a reference type parameter (whether lvalue or rvalue reference), the object will not be copied. So when you bind the references the lvalue will have to be const. 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. We create two types of access: one const and one not const. For example, this code will not compile. Convert to rvalue references. A void * value resulting from such a conversion can be converted back to the original function pointer type, using an explicit cast, without loss of information. If you would fix the copy constructor to: DriverPoint(const DriverPoint& driverPoint) both adding lvalue and adding rvalue to the vector by calling push_back would work, but both would go through the copy ctor and not through move, as you didn't implement move and the default move is implicitly deleted if you declare any single one. Both of g and h are legal and the reference binds directly. I expect that when using a temporary instance of a Wraper object, the conversion operator defined for rvalue will always be used. Under the conditions specified in [dcl. Why?The C++ standard specifies that such expressions do not undergo lvalue to rvalue conversion, and that the type of the dereferenced object may be incomplete. rvalue references are considered lvalue (this part I understand) They are not. Being an lvalue or an rvalue is a property of an expression. In the function, the argument has a name and thus is an lvalue. c++ template type matching with references [duplicate] Ask Question Asked 5 days ago. 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. According to the rule of forwarding reference, when an lvalue is passed to add, the template type argument Element will be deduced as SomeClass&. Otherwise, the reference you get behaves more. The result is an lvalue if T is an lvalue reference type or an rvalue reference to function type (8. Then std::forward<SomeClass&> (element) will be invoked, and the instantiation of std::forward would be. An example of an rvalue would be a literal constant – something like ’8′, or ’3. 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. The following table lists exceptions to this rule. @MikeMB the standard rarely prevents compilers from inserting for (int i = 0; i < 1 billion; ++i) at arbitrary points. 1 Answer. e. This article also mentioned that issue. 3. Consider this similar question: "Is an integer an lvalue or an rvalue". 1) Two possibly multilevel pointers to the same type may be converted between each other, regardless of cv-qualifiers at each level. 3. An rvalue is any expression that has a value, but cannot have a value assigned to it. e. Except for an implicit object parameter, for which see 13. For example, when user tries to read a given position in the collection. An lvalue-to-rvalue conversion (converting the name of the object x to its value 2. 6. An lvalue may get converted to an rvalue: that's something perfectly legit and it happens quite often. One that returns an int used when a rvalue is needed. 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. Only the following conversions can be done with const_cast. In C++, the cast result belongs to one of the following value categories:. Read 5. 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. Every lvalue is, in turn, either modifiable or non-modifiable. , Circle c3 (Circle (4)), I'd expect the third constructor, (copy constructor with rvalue referecne) to be called but it's not the case. So in your example, the expression to the right of the = is an expression that happens to be an lvalue. 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. 98 * @param __t A thing of arbitrary type. 5. C++98 assigning a value to a volatile variable might result in an unnecessary read due to the lvalue-to-rvalue conversion applied to the assignment result introduce discarded-value expressions and exclude this case from the list of cases that require the conversion CWG 1343: C++98 sequencing of destructor calls inExcept for an implicit object parameter, for which see 13. lval]/3. The discussion of reference initialization in 8. (This is as per my understanding, please correct it otherwise). A glvalue of a non-function, non-array type T can be converted to a prvalue. Then std::forward<SomeClass&> (element) will be invoked, and the instantiation of std::forward would be. 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. It is illegal in C++ to attach non-const references to rvalues. It is a forwarding reference. In this case 2*b is an rvalue since it does not persist beyond the expression. It makes sure a thing& x is passed as a value category lvalue, and thing&& x passed as an rvalue. And the lvalue-to-rvalue conversion always returns a prvalue value, not a (temporary) object. 右值 (rvalue, right value) ，右边的值，是指表达式结束后就不再存在的临时对象。. } it evaluates, no matter what, to an lvalue. type. Each C++ expression (an operator with its operands, a literal, a variable name, etc. When C++11 invented rvalue references, none of this behavior changed at all. A lvalue overload can accept both lvalues and rvalues, but an rvalue overload can only accept rvalues. Conversely, d = static_cast<float> (j)/v; produces an. A prvalue (“pure” rvalue) is an rvalue that is not an xvalue. e. The C++11 standard for lvalue and rvalue conversion can be found at chapter 4. It's actually a cast. I could have used std::move to convert the lvalue to rvalue reference and the call would be successful. My guess is that this restriction has historical roots in the C++98 standard where rvalues were limited to temporaries, that were fully managed by the compiler. The result is an lvalue if T is an lvalue reference type or an rvalue reference to function type (8. assign values to the reference return type directly in c++. 10. If t returns by lvalue reference, the code does not compile because a rvalue reference cannot bind to it. cond]/7. 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. I think I'm missing something basic regarding the lvalue-to-rvalue standard conversion. lvalue references are marked with one ampersand (&). ) In very broad and simple terms, an lvalue refers to. rvalue/lvalue tells you the value category. Zero or one conversion from the following set: lvalue-to-rvalue conversion, array-to-pointer conversion, and function-to-pointer conversion. 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. @whY because for an rvalue a const reference is not an exact match for template deduction. 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. std::auto_ptr<Foo> foo(new Foo()); // auto_ptrs are deprecated btw bar(std::move(foo)); // changed ownership. If type is an lvalue reference type or an rvalue reference to a function type, the cast result is an lvalue. For fundamental types, the copy approach is reasonable. B. If you can, it typically is. That's to protect people from changing the values of temporaries that are destroyed before their new value can be used . If something happens to the temporary being referenced by a , b still holds a valid reference to a in the current scope. 2, and 4. However, a (prvalue). –6. In C, (time_t) { time (NULL) } is a compound literal C99, initialized by the return value of time. Whenever an lvalue appears in a context where an rvalue is expected, the lvalue is converted to an rvalue; see 4. The difference is that &i is OK but &5 is not. fstream file{"filename"}; print_stream(file);I would like to write a variadic template function that accepts rvalues and lvalue references. 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); // ^^^^^. The fact that you pass bind itself an rvalue only means that there is. 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. rvalues can bind to rvalue references and const lvalue references, e. Both of these options are user-defined conversion functions, so neither is better in terms of overload resolution, thus an ambiguity. In C++, it is illegal to implicitly convert an rvalue to an lvalue reference. Every expression in C and C++ is either an lvalue or an rvalue. 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. you cannot change the integer 5, fact. If an lvalue-to-rvalue conversion were performed on s, it would also call the copy constructor; [conv. 1 Lvalue-to-rvalue conversion paragraph 1 and says (emphasis mine going forward): A glvalue (3. 6) An lvalue (until C++11) glvalue (since C++11) expression of type T1 can be converted to reference to another type T2. The most common lvalue is just a variable, so in something like x=10, x is an lvalue, and 10 is an rvalue. 2) non-modifiable lvalues, which are const. 0. Deciding whether a function must take an argument by value, lvalue reference or rvalue reference depends very much on what it does. You could disallow rvalues, but not sure if that would be acceptable. Yes, the type of the variable r is indeed int&&. In (static_cast<int&&> (3))++, the expression static. 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). Abbreviations in this article. 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.