e. If type is an lvalue reference type or an rvalue reference to a function type, the cast result is an lvalue. for the same reason as that example. 1) If the reference is an lvalue reference. (This is somewhat of a simplification, in C++11 we have lvalues, xvalues and prvalues. An rvalue reference is a new type. Here's why. e. Sorted by: 7. To set this compiler option in the Visual Studio development environment. 5. 1:. 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. An entity (such as an. lvalue references are marked with one ampersand (&). 12. Being an lvalue or an rvalue is a property of an expression; that is, every expression is either an lvalue or an rvalue. And an identifier "is an lvalue if the entity is a function or variable" (5. The entire point is that you know that this entity references an rvalue and you can legitimately move its content. Found workaround how to use rvalue as lvalue. The confusion you're having is pretty common. From the linked documentation. You might consider A& f () & { to ensure the call is happening on an lvalue object if you need to do something like this. 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. If something happens to the temporary being referenced by a , b still holds a valid reference to a in the current scope. The third constructor is called move constructor. The biggest difference between a C++03 reference (now called an lvalue reference in C++11) is that it can bind to an rvalue like a temporary without having to be const. It is really about rvalues vs. An rvalue can also be bound to a const lvalue reference, i. — 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. Abbreviations of constructors, operators and destructors: Dc — Default constructorA{} is always an rvalue per [expr. An rvalue reference is a new type. return 17;} int m=func2(); // C++03-style copying. c++ base constructor lvalue to parameter. Until IBM's implementation of all the features of the C++11 standard is. 2), then: the value contained in the referenced. Conversion of a function pointer to void * shall not alter the representation. 3. 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. if you were to use an local variable instead). I have tried to simulate the assignment of the object (pair. The Microsoft documentation is wrong. lvalue simply means an object that has an identifiable location in memory (i. ; In all other cases, the cast result is a (prvalue) rvalue. A function parameter such as T&& t is known as a forwarding reference. You could not pass it to a function accepting a const char*&& (i. An lvalue is, according to §3. Whenever an lvalue appears in a context where an rvalue is expected, the lvalue is converted to an rvalue; see 4. it is a reference only to rvalues. It could be an rvalue of course, but it doesn't have to be. The "l" and "r" in "lvalue reference" and "rvalue reference" refers to the categories of values to which the reference can bind, not to the category of the id-expression naming a variable of this reference type. Roughly, it’s an lvalue if you can “take its address”, and an rvalue otherwise. 99 * @return The parameter cast to an rvalue-reference to allow moving it. 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”. Example: int a. lvalue and rvalue as function parameters. 4. An lvalue is an expression that designates (refers to) an object. Or the compiler could convert said references to pointers, push a pointer on the stack, pop the identical pointer off, and call it std::move. arg the variable has type int&& and no value category. lvalue cannot be a function, expression (like a+b) or a constant (like 3 , 4 , etc. So I know why the compiler is complaining (because of trying to bind rvalue to lvalue reference -- at least this is what I think is happening -- please correct me if I am wrong). It cannot convert from an rvalue to an lvalue reference, even a const one. You don't need universal reference here const T& source is enough and simpler. 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. Convert to rvalue references. In C, (time_t) { time (NULL) } is a compound literal C99, initialized by the return value of time. 10. To convert an lvalue to an rvalue, you can also use the std::move() function. According to the rule of forwarding reference, when an lvalue is passed to add, the template type argument Element will be deduced as SomeClass&. It makes sure a thing& x is passed as a value category lvalue, and thing&& x passed as an rvalue. int rVal () { return 0; }. thus, this is legal: string&& s = foo (); // extends lifetime as before s += "bar"; baz (std::move (s)); // move the temporary into the baz function. The choice of copy or move constructor only occurs when passing an object by value. 1) Two possibly multilevel pointers to the same type may be converted between each other, regardless of cv-qualifiers at each level. @YueZhou Function lvalues may be bound to rvalue references. The second are value categories for expressions. having an address). cv]/4. The relevant part is only that prvalues become xvalues by temporary materialization conversion and that both xvalues and lvalues (collectively glvalues) share a lot of behavior, in particular that they refer to objects or functions (which prvalues don't). 8. must error, because you're trying to pass an rvalue argument, std::move(n), to a lvalue reference parameter, T&. 23. L-value: “l-value” refers to memory location which identifies. The purpose of r-value reference parameters is to detect specifically when an object is an r-value. 3. All lvalues that aren't arrays, functions or of. 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. Thus, if the thickness is 1 inch, and the K-value is 0. "3" is an integer, and an rvalue. Visual Studio warning disappears if one removes std::move. Rvalue references are a feature of C++ that was added with the C++11 standard. 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. 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. When I discovered this, it seemed odd to me, so I tried. 3. However, as far as class objects are concerned. 1. 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. 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. An lvalue may get converted to an rvalue: that's something perfectly legit and it happens quite often. 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. The difference is that &i is OK but &5 is not. How to cast/convert pointer to reference in C++. In the function, the argument has a name and thus is an lvalue. The compiler will synthesize a move constructor only for such class that doesn't define any of its own copy-control members (copy-constructor, copy-assignment, or destructor), and if all the non- static members can be moved. An rvalue is a prvalue or an xvalue. 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. 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. @BЈовић: I did mean that (although I've since renamed the function baz). (An xvalue is an rvalue). 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 (=). If element on this position doesn't exist, it should throw exception. g. Your terminology needs improvement. "Hello, World" is not of type const char*. So sizeof (0, arr) = sizeof (arr) and which would be equal to 100* sizeof (char) and not = sizeof (char*). Indeed it does. This is. If you compile with /W4 then the compiler will warn you. The copy constructor uses the lvalue references which are marked with one ampersand (&) while the move constructor uses the rvalue references are marked with two ampersands (&&). In both cases, if the wrapper has been successfully constructed, we mark the status as value to indicate that we have a value. , with extensions: pointer or reference to a is additionally allowed to be cast to pointer or reference to unambiguous base class (and vice versa) even if the base class is (that is, this cast ignores the private inheritance specifier). A reference (“lvalue reference” since C++11) is a type of C++ variable that can act as an alias to another value. 1. This is a changeable storage location. 1 is rvalue, it doesn't point anywhere, and it's contained within lvalue x. 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. C++03, section §3. int&& x = 3; x is now an lvalue. The following table lists exceptions to this rule. . HI Enlico, Thank's for the awesome answer, now I have a clearer idea of how to use RValue and LValue references. undefined behavior: The lvalue or xvalue is a nonclass type, qualified by either const or volatile. 1. It would capitalize std::strings, and display each parameter after they are capitalized. In C++, each expression, such as an operator with its operands, literals, and variables, has type and value. h, the output is same as Clang output it's reasonable. If an lvalue-to-rvalue conversion were performed on s, it would also call the copy constructor; [conv. 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. 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. 1: (5. e. 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. Now an lvalue reference is a reference that binds to an lvalue. (prvalue) The output of this example is: produces an answer of type int because both are integers. 10) of a non-function, non-array type T can be converted to a prvalue. The first constructor is the default one. The. Officially, C++ performs an lvalue-to-rvalueconversion. I would like to move an object into a std::vector using std::vector::push_back(). 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. If this was allowed, then it would look something like: The expression i in increment(i) is casted to an rvalue via lvalue-to-rvalue conversion. IBM® continues to develop and implement the features of the new standard. However, you don't have double && in your code, you have U && for a deduced U. Otherwise, the reference you get behaves more. Hence, the end result is the attempted binding of the rvalue. This is already done in some places. This article Understanding lvalues and rvalues in C and C++ probably is one of the better detailed explanations. But in this particular case, the rules. (since C++11) 4) If new_type is the type void (possibly cv-qualified), static_cast discards the value of expression after. オブジェクトという言葉が聞き慣れないなら. If an l-value could bind to an r-value reference, that would mean the detection I was talking about. auto (* p) [42] = & a; is valid if a is an lvalue of type int [42]. 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. If the C-value is 0. Temporary materialization thus occurs in both of the OP's examples: The first temporary (with value 10) will be. Nothing is being turned into a lvalue. b is just an alternative name to the memory assigned to the variable a. That's to protect people from changing the values of temporaries that are destroyed before their new value can be used . 10/2), Whenever a glvalue appears in a context where a prvalue is expected, the glvalue is converted to a prvalue. I couldn't find an example of l2r applicable to class types myself; in all the seemingly applicable examples there's usually a function involved that takes lvalue-ref (like copy-ctor), for which l2r seems to be suppressed (see. Since the type of a is not an int, it cannot match the type that b. However, rvalues can't be converted to lvalues. Once a move constructor is called upon the reference, the original object should be reset to the origin state, and so does any reference to it. We could categorize each expression by type or value. In the case of object constructing is true but in the case of object assigning is false. 14′. The parameter list for a move constructor, however, consists of an rvalue reference, like B&& x. You would need const_cast<char*&> (a) in order to have an lvalue to assign to, and that brings up the next problem. rvalue references are sausage-making devices added later after nobody could find a. To mark the place(s) where you want to take advantage of the licence to ruthlessly plunder it, you have to convert it to an rvalue-reference on passing it on, for example with std::move or std::forward, the latter mostly for templates. 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. N. The lvalue to rvalue conversion isn't being done either, of course, but that's rather intuitive and normal. 5. As we've seen earlier, a and b are both lvalues. You will often find explanations that deal with the left and right side of an assignment. A move constructor and move assignment operator can now. This allows you to explicitly move from an lvalue, using move. Sorted by: 1. 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. 2, and 4. And so on. test prep. lvalue-- an expression that identifies a non-temporary object. 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. The address of operator (&) requires an lvalue because you can only take the address of something in memory. lvalue = rvalue; 对于以上的语句，lvalue是我. 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. And an rvalue reference is a reference that binds to an rvalue. 1, 4. Every expression belongs to one of three value categories: lvalue, non-lvalue object (rvalue), and function designator. When being passed an lvalue, the template parameter would be deduced as lvalue-reference, after reference. Abbreviations in this article. But Args itself is either an lvalue reference or not a reference. As an example, the operand of unary & must be a function designator, the result of [], the result of unary *, or an lvalue (C 2018 6. Lvalues and Rvalues. m, static_cast<A&&> (a), and a + a are xvalues. An identifier that refers to an object is an lvalue, but an. 1, a standard conversion sequence cannot be formed if it requires binding an lvalue reference to non-const to an rvalue or binding an rvalue reference. 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. Values return by functions/methods and expression are temporary values, so you do have to use std::move to move them (C++ standard to convert to rvalue) when you pass them to functions/methods. 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. Most operators require lvalue-to-rvalue conversion because they use the value of the object to calculate a result. 5. , Circle c3 (Circle (4)), I'd expect the third constructor, (copy constructor with rvalue referecne) to be called but it's not the case. Applying the lvalue-to-rvalue conversion to x reads the value of the mutable global variable globx, which makes it not a constant expression as the value of globx is subject to change (and, even if it were const, there would be the issue of its value not being known at compile time). Similarly, rhs in Gadget. Rvalue references enable you to distinguish an lvalue from an rvalue. An object is a region of storage that can be examined and stored into. Informally, "lvalue-to-rvalue conversion" means "reading the value". The result is an lvalue if T is an lvalue reference type or an rvalue reference to function type (8. In C++03, Boost's Foreach, using this interesting technique, can detect at run-time whether an expression is an lvalue or an rvalue. But one important rule is that: one can. Something that points to a specific memory location. The address-of operator can only be used on lvalues. References. 1. One that returns an int& used when a lvalue is expected, for storing a value at a given position. Otherwise your compiler will throw an error: obj & a1 = bar (); invalid initialization of non-const reference of type ‘obj&’ from an rvalue of type ‘obj’. The reason why you need to const is to make x not a forwarding reference. Given all three functions, this call is ambiguous. An lvalue may get converted to an rvalue: that's something perfectly legit and it happens quite often. 2, and 4. Read 5. You need to pass in an rvalue, and for that you need to use std::move: I can see why this is counter-intuitive! x is lvalue (as we know it). Forwarding references are a special kind of references that preserve the value category of a function argument, making it. Note that this must wait until construction is complete for two reasons. 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. Creating a temporary object is usually not the desired behavior. 10/7 reads, Whenever an lvalue appears in a context where an rvalue is expected, the lvalue is converted to an rvalue; see 4. The compiler will synthesize a move constructor only for such class that doesn't define any of its own copy-control members (copy-constructor, copy-assignment, or destructor), and if all the non- static members. static_cast<Type> (expression) belongs to one of the following value categories: or an rvalue reference to a function type is an lvalue. I. A compiler can optimize the call to copy constructor and directly call the matching constructor. You would need to provide const string& as template argument for T to make T&& also const string&. 右值（rvalue）：. Convert enum class values into integers or floating-point values. Nothing is changed except the value category. Consider this similar question: "Is an integer an lvalue or an rvalue". Read it along with, §4. There is a very important distinction to be made between expressions which are rvalues and expressions whose type is an rvalue reference. If t returns by rvalue reference, you obtain a reference to whatever was returned. It cannot convert from an rvalue to an lvalue reference, even a const one. If inside foo no move operation happened like my example, then my_ptr_var will not actually be moved from. Unlike a reference to non-const (which can only bind to modifiable lvalues), a reference to const can bind to modifiable lvalues, non-modifiable lvalues, and rvalues. C++ 中有两种类型的表达式：. e. 4. Lvalues and rvalues are fundamental to C++ expressions. 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. 4. When C++11 invented rvalue references, none of this behavior changed at all. The && syntax is either referring to a rvalue-reference or a universal-reference. Lvalue to rvalue conversion changes the value category of an expression, without changing its type. is an rvalue reference to an object type, is an xvalue. Let's look at (T1&&)t2 first. 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). The discussion of reference initialization in 8. void f1(int& namedValue){. Although the syntax of a compound literal is similar to a cast, the important distinction is that a cast is a non-lvalue. “If T1 is reference-related to T2 and the reference is an rvalue reference, the initializer expression shall not be an lvalue. 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. array), and function-to-pointer (conv. Let's think of the addition + operator for example. Set the Enforce type conversion rules property to /Zc:rvalueCast or /Zc:rvalueCast. Lvalue-to-rvalue conversion. If you pass an argument to a reference type parameter (whether lvalue or rvalue reference), the object will not be copied. Is there a way to write a function in C++ that accepts both lvalue and rvalue arguments, without making it a template? For example, suppose I write a function print_stream that reads from an istream and prints the data that was read to the screen, or something. An lvalue does not necessarily permit modification of the object it designates. Among. In C++, the cast result belongs to one of the following value categories:. assign values to the reference return type directly in c++. 1. In the next example, we first use the addition operator + (→//3) to add two Lvalues and then the assignment operator = to assign the result to another Lvalue. call]/12, [expr. 4. (for user-defined types): rvalue or lvalue?. Ternary conditional operator will yield an lvalue, if the type of its second and third operands is an lvalue. From C++11 4. ref], a reference can be bound directly to the result of applying a conversion function to an initializer expression. A glvalue of a non-function, non-array type T can be converted to a prvalue. ref]/5. Except for an implicit object parameter, for which see 13. Rvalue references are types, types are not expressions and so cannot be "considered lvalue". The value of x is 1. I would respect the first compiler more, it is at least. The terms "lvalue/rvalue reference" and "lvalue/rvalue" are related but not interchangeable or one a shortened form of the other. 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. 1, 4. Yes. The goal was providing a function that both accepts lvalue and rvalue references, I did not want to write two functions or to really care about lvalue/rvalue on the caller's side. Note: The ISO C standard does not require this, but it is required for POSIX conformance. Note that by binding a temporary to a rvalue-reference (or a const. 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. a non-const reference). g. 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. (C++14) Assigns a new value to an object and returns its old value. For example, if you’ve declared a variable int x;, then x is an lvalue, but 253 and x + 6 are rvalues. e. const A& ), and lvalue-to-rvalue conversion is suppressed when binding lvalue-reference. 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. – Corristo. You. write_Rvalue will only accept an rvalue. 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. 1 for an lvalue-to-rvalue conversion. Done. Select the Configuration Properties > C/C++ > Language property page. Address of an lvalue may be taken: &++i and &std::endl are valid expressions. An rvalue is any expression that isn't an lvalue. The C++ standard does not specify explicitly that it is lvalue to rvalue conversion that is responsible for causing an access. The example is interesting because it seems that only lvalues are combined. Yes, rvalues are moved, lvalues are copied. Type conversions on references. 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). and write_Lvalue will only accept an lvalue. When programming in C++03, we can't pass an unnamed temporary T () to a function void foo (T&);. – 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. Set the Enforce type conversion rules property to /Zc:rvalueCast or. In ASCII code, the character 'a' has integer value 97 , that's why the character 'a' is automatically converted to integer 97 . We can take the address of an lvalue, but not of an rvalue. 2. Types shall not be defined in a reinterpret_cast. 1. An rvalue is constant, it cannot be changed. 1 (page 85 for version 3485). The only thing that can be an rvalue or an lvalue is an expression. 1 Lvalue-to-rvalue conversion paragraph 1 and says (emphasis mine going forward): A glvalue (3. [ Note: If T is a non-class type that is cv. According to the C++ specifications, it takes two rvalues as arguments and returns an rvalue. 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. e. end()) is a temporary object and cannot be bound to lvalue reference. Assignment involving scalar types requires a modifiable lvalue on the left hand side of the assignment operator. Otherwise, the type of the prvalue is T. If t returns a local variable, then you get a dangling reference, since that variable is gone after the call. 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 lvalue reference, the code does not compile because a rvalue reference cannot bind to it. Correct, the epxression T() is always an rvalue for scalar and user-defined types T. 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. Answer below is for C++14. 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. It is still not allowed per [dcl. 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. 3. 1 Answer. Note that there is one exception: there can be lvalue const reference binding to an rvalue. cpp -std=c++11 -fno-elide-constructors. You will often find explanations that deal with the left and right side of an assignment. Oct 31, 2016 at 20:29. 2 days ago · C++ Operator Overloading [ ] for lvalue and rvalue. Then I use rvalue reference of class B's this pointer to pass it to A's constructor. In particular, only const_cast may be used to cast away (remove) constness or volatility. double && does not work for lvalues. " What this is saying in layman's terms is that you can't (and shouldn't) store an address reference to an rvalue. Perhaps the most significant new feature in C++11 is rvalue references; they’re the foundation on which move semantics and perfect forwarding are built. 3. Compiled with "g++ -std=c++0x". What you're referring to is the fact that if an expression. The question related to this one. It's also echoed in 5. They are declared using the ‘&’ before the name of the variable. (Lvalue-to-rvalue conversions on class types are rare, but do occur in some places in the language, e. init. The value of x is 1. 2), an xvalue if T is an rvalue reference to object type. Arrays are lvalues. This differs from ISO C, in. 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 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. You are comparing two different things that are not really related. Assume a variable name as a label attached to its location in memory. 2. Class rvalues prvalues]. 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). It shouldn't. an lvalue reference instead of an rvalue reference) and had the appropriate cv-qualification, then it's probably the programmer's mistake. 3. An lvalue-to-rvalue conversion (converting the name of the object x to its value 2. The standard defines (§3. Both of g and h are legal and the reference binds directly. If we have a rvalue we can assign it to a variable, or take a reference, hence becoming a lvalue. Operationally, the difference among these kinds of expressions is this:std::function can be move-constructed from rvalue of a functor object.