e. Assignment to an rvalue doesn't really make sense, so it should be forbidden. An lvalue (so called, historically, because lvalues could appear on the left-hand side of an assignment. Hence, values bound to an rvalue reference can be moved from (not necessarily always going to be moved from, but it is allowed), and lvalues can be bound to lvalue references and can't be moved from. And it's on the value level that talking about rvalue/lvalue-ness makes sense (after all, those are called value categories). At the same time, we cannot move away from const values. But you might just let regular deduction occurs. return 17;} int m=func2(); // C++03-style copying. An lvalue is (simplifying a bit) something that refers to someplace in memory that can/does hold a value. 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. An lvalue can be converted to an rvalue. M. 2, and 4. e. Convert enum class values into integers or floating-point values. Type conversions on references. In fact, in terms of overload resolution, an rvalue prefers to be bound to an rvalue reference than to an lvalue const reference. 2. (Lvalue-to-rvalue conversions on class types are rare, but do occur in some places in the language, e. That means std::move could take both lvalue and rvalue, and convert them to rvalue unconditionally. std::function's type is defined only by its target's signature(eg: void(int)) and std::function itself is defined by the. The actual problem is instantiating Parent with a reference type to begin with; in C++11 this is generally avoided via application of std::decay. In (static_cast<int&&> (3))++, the expression static. 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. 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). 9. The parameter list for a move constructor, however, consists of an rvalue reference, like B&& x. 8. h and move. That works well with normal variables but uint8Vect_t(dataBlock. 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. It seems like std::array can be converted to an std::span when it's an rvalue only on clang. According to the rule of forwarding reference, when an lvalue is passed to add, the template type argument Element will be deduced as SomeClass&. The "my target must be copy-constructable" requirement of std::function is due to its own requirement of being copy-constructable. It is still not allowed per [dcl. If you had. I would respect the first compiler more, it is at least. 2 days ago · C++ Operator Overloading [ ] for lvalue and rvalue. 3 -- Lvalue references ), we discussed how an lvalue reference can only bind to a modifiable lvalue. Rvalue to lvalue conversion? 2. universal reference. If t returns by lvalue reference, the code does not compile because a rvalue reference cannot bind to it. 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 address-of operator can only be used on lvalues. The goal of rvalue references is sparing copies and using move semantics. the original code was int&& rref = n; which was ill-formed, as n is an lvalue and therefore cannot bind to an rvalue reference. I guess you are reading the Rvalue References: C++0x Features in VC10, Part 2. But for the third case i. That stops the move if it is an lvalue reference. I can't speak for the motivation behind having it work this way despite the tuple explicitly holding an. foobar () is an rvalue because foobar () returns int. const A& x = 1; //compile x = 2; //error! A&& xxx = 1; //compile A& xx = 1; //does not compile. An rvalue is constant, it cannot be changed. 9/1: The result of the expression static_cast<T> (v) is the result of converting the expression v to type T. No temporary is created, no copy is made, no constructors or. You could not pass it to a function accepting a const char*&& (i. @eerorika In your example y is an int, so it qualifies for rvalue conversion on return. The right constructors for the first two cases are called. That is special syntax for a so-called forwarding reference. This way you explicitly say T&& should not match an lvalue-reference. e. If t returns by rvalue reference, you obtain a reference to whatever was returned. The implicitly defined copy constructor takes an lvalue reference (i. std::move performs a static_cast to an rvalue reference type and returns the rvalue reference. Radius: 2 2 4. Lvalue to rvalue conversion changes the value category of an expression, without changing its type. Using it only makes sense inside of a template, where you choose whether to move or not depending on a template argument. Creating a temporary object is usually not the desired behavior. That would also solve the <T> issue BTW. Converts between types using a combination of explicit and implicit conversions. 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. Both of these options are user-defined conversion functions, so neither is better in terms of overload resolution, thus an ambiguity. 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. an rvalue reference). R-value to U-value Conversion Calculator; U-value, lower the number the better (U-0. Lvalues and xvalues can be of incomplete types, but (prvalue) rvalues must be of complete types or void types. std::move is there to allow for the casting. 2) non-modifiable lvalues, which are const. e. 3. 1/2: The value contained in the object indicated by the lvalue is the rvalue result. type. C++03, section §3. cv]/4. arg the variable has type int&& and no value category. D'uh. 99 * @return The parameter cast to an rvalue-reference to allow moving it. 4 — Lvalue references to const. I recently filed a bug against MSVC which relates to this, where the non-standard behavior caused standard-compliant code to fail to compile and/or compile with a deviant behavior. You are returning a copy of A from test so *c triggers the construction of a copy of c. If you really want to or need to specify the parameters, you can use std::move to convert an lvalue to an rvalue at the calling site. But it is still a reference, which is a lvalue. Let's look at the following snippet: So we have a reference being initialized by an xvalue of type const foo. There is no implicit conversion as suggested in the title, the reference binds directly to the. rvalue references are marked with two ampersands (&&). I believe this code is both well-formed and well-defined. Without lvalue-to-rvalue conversion, it cannot read it's value. 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). In fact, that's the origin of the names: an lvalue was (originally) anything that could appear on the Left side of an assignment, and. 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&). If you pass an prvalue, it isn't converted, the temporary is materialised into the parameter object. 4. int&& x = 3; x is now an lvalue. ref], a reference can be bound directly to the result of applying a conversion function to an initializer expression. For example, if you’ve declared a variable int x;, then x is an lvalue, but 253 and x + 6 are rvalues. So, the conversion from a A rvalue to something that P&& would accept in (1) calls the user defined conversion function operator P() &&. 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. A simpler case: template <typename T> void foo(T&& ) { } foo(1); // T is int int x; foo(x); // T is int& When you specify float for x, you are specifying that that particular argument will have type float&&, and you cannot implicitly convert an lvalue float to an rvalue. The difference between lvalues and rvalues plays a role in the writing and understanding of expressions. The expressions f (), f (). Temporary lifetime extension does not pass through functions so there is no way to get a lvalue from the rvalue you pass to the function. lvalue = rvalue; 对于以上的语句，lvalue是我. The Microsoft documentation is wrong. Correct, the epxression T() is always an rvalue for scalar and user-defined types T. So, when you type const int& ref = 40. Let’s turn it around a bit. , Circle c3 (Circle (4)), I'd expect the third constructor, (copy constructor with rvalue referecne) to be called but it's not the case. having an address). For the second overload, it would call operator const P&() const&. If T is a non-class type, the type of the prvalue is the cv-unqualified version of T. uint8Vect_t encodeData(uint8Vect_t &dataBuff); Here you are taking a reference to a uint8Vect_t. A nice feature of this heuristic is that it helps you remember that the type of an expression is independent of. But for the third case i. 1. Compiled with "g++ -std=c++0x". 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 . Prior VC++ version example VC10 had two versions, one to accept an lvalue and another an rvalue reference; Rvalue reference cannot be used to initialize a non const reference i. using g++. 3) If new_type is an rvalue reference type, static_cast converts the value of expression to xvalue. 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. It shouldn't. 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. I could have used std::move to convert the lvalue to rvalue reference and the call would be successful. An object is a region of storage that can be examined and stored into. This type of static_cast is used to implement move semantics in std::move. Types shall not be defined in a reinterpret_cast. 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). An lvalue or xvalue is an expression that refers to such an object. When an lvalue-to-rvalue conversion is applied to an expression e, and either. lvalueとrvalueとは いずれもオブジェクトだ 。. c++ base constructor lvalue to parameter. I played a bit around with composite-patterns and inheritance in c++. 1 Answer. If I change func (unsigned int&) to func (Color&), compiler accept it. FWIW, the POSIX 2008 standard says (System Interfaces, §2. To declare an lvalue reference type, we use an ampersand (&) in the type declaration: int // a normal int type int& // an lvalue reference to an int object double& //. Whenever an lvalue a glvalue appears in a context where an rvalue a prvalue is expected, the lvalue glvalue is converted to an rvalue a prvalue; see 4. by unconditionally casting its argument—which might be an lvalue—to an rvalue reference, it enables the compiler to subsequently move, rather than copy, the value passed in Arg if its type is. With argument deduction, parameter of make_tuple is deduced to be: int&, and in this case i can be bound. 2. If something happens to the temporary being referenced by a , b still holds a valid reference to a in the current scope. You are comparing two different things that are not really related. Nothing is changed except the value category. You could disallow rvalues, but not sure if that would be acceptable. 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. in . Arrays are lvalues. However, Microsoft compiler does accept it meaning that. However, a (prvalue). C++0x rvalue reference template argument deduction. An lvalue (until C++11) A glvalue (since C++11) of any non-function, non-array type T can be implicitly converted to an rvalue. –6. 1 Answer. Rvalue references allow one to make classes that can be both moved and copied. If we have a lvalue we can return it from a function, so we get a rvalue. (I found that via this StackOverflow question: Rvalues in C++03 ) Here's a demo of this working at run-time. 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. In the op's example y is actually a reference to the sub-object of some unnamed object the structured binding declared. 0. 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. " What this is saying in layman's terms is that you can't (and shouldn't) store an address reference to an rvalue. Here is a silly code that doesn't compile: int x; 1 = x; // error: expression must be a modifyable lvalue. call]/12, [expr. b is just an alternative name to the memory assigned to the variable a. 4. Except for an implicit object parameter, for which see 13. In k++, the expression k is an l-value (roughly speaking, it has a name), which is its value-category. In that case, consider processing only one argument at a time, leaving the remaining ones as rvalue-references. Each C++ expression (an operator with its operands, a literal, a variable name, etc. The answer is: yes, we do. 2. If you pass an argument to a reference type parameter (whether lvalue or rvalue reference), the object will not be copied. IBM® continues to develop and implement the features of the new standard. lvalue cannot be a function, expression (like a+b) or a constant (like 3 , 4 , etc. If encodeData() does not change dataBuff then the simplest. An object is a region of storage that can be examined and stored into. This is a follow-on question to C++0x rvalue references and temporaries. 1. Loosely speaking, think of lvalue as some sort of container, and rvalue as the value contained in the container. 1/2 (your. The Parent class stores a pointer, but due to rvalue to lvalue conversion, the Parent ends up storing a reference to a pointer. 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. C++ does not allow you to get an r-value reference to a variable without an explicit conversion. assign values to the reference return type directly in c++. References in C++ are nothing but the alternative to the already existing variable. It would capitalize std::strings, and display each parameter after they are capitalized. But instead removing either reference overload results in ambiguity with f( int ). If inside foo no move operation happened like my example, then my_ptr_var will not actually be moved from. 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. To convert an lvalue to an rvalue, you can also use the std::move() function. 9. 3. Thus, if the thickness is 1 inch, and the K-value is 0. However, if the value is const than the compiler can convert the rvalue to an lvalue duringThe title of the question you linked is a little misleading. But in this particular case, the rules. 2) returning a reference type. std::forward is a conditional std::move. The value of x is 1. That's an exception to the general rule that it is impossible for lvalues to be bound to rvalue. 3. But in the circumstances of the linked question, the template instantiation of std::function cannot be inferred from the lambda type. Let's look at (T1&&)t2 first. 4. When I discovered this, it seemed odd to me, so I tried. 5. An rvalue can also be bound to a const lvalue reference, i. As we've seen earlier, a and b are both lvalues. Informally, "lvalue-to-rvalue conversion" means "reading the value". That is because arr is indeed an lvalue, as it is not a function designator, the result of [], or the. Does template argument resolution convert L-values to R-values or like how does this work? c++; c++11; templates;. void f(A *&&p) {} function which accept rvalue ref to pointer which points to A; but p is still lvalue which has type r-value reference to a pointer, so u have to "cast"(std::move - does nothing just cast l-value to r-value) std::shared_ptr(std::move(p));C++ Function taking lvalue and rvalue parameters transparently. Lvalue and rvalue are expressions that identify certain categories of values. e. The C++11 standard for lvalue and rvalue conversion can be found at chapter 4. Roughly, it’s an lvalue if you can “take its address”, and an rvalue otherwise. Category 4 used to be a bit different in C++11, but I believe this wording is correct for C++14. I think I'm missing something basic regarding the lvalue-to-rvalue standard conversion. Indeed it does. The name “lvalue” comes from the assignment expression E1 = E2 in which the. Return lvalue reference from temporary object. move simply returns an rvalue reference to its argument, equivalent to. If type is an lvalue reference type or an rvalue reference to a function type, the cast result is an lvalue. An rvalue reference is a new type. ) is characterized by two independent properties: a . (since C++11) 4) If new_type is the type void (possibly cv-qualified), static_cast discards the value of expression after. std::apply perfect-forwards the tuple to std::get to access elements of the tuple. Therefore, I will not jump right in and explain what rvalue references are. That is expected. 3. 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. This is a helper function to allow perfect forwarding of arguments taken as rvalue references to deduced types, preserving any potential move semantics involved. An lvalue is, according to §3. 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. C Server Side Programming Programming. 12. Is it normal that I can't bind a lvalue to a rvalue reference ? EDIT: same thing for a function : void f(int && v) { } int v; f(v); //won't compile I'm a little bit confused because I think std::forward is usefull to detect if a method is called with a lvalue or a rvalue reference. 5, then the R-value is 2. template <class T, class Other = T> T exchange(T& val, Other&& new_val). 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. But one important rule is that: one can. You can use str as a variable, which also implies that it is an lvalue, not a temporary rvalue. For fundamental types, the copy approach is reasonable. Share. 1. In this case, the conversion function is chosen by overload resolution. e. Secondly, the compiler will look for a move assignment operator or copy assignment operator implementation then, failing that, will fall back to the copy constructor which has been implemented. h, the output is same as Clang output it's reasonable. The confusion you're having is pretty common. Unless encountered in unevaluated context (in an operand of sizeof, typeid, noexcept, or decltype), this conversion effectively copy-constructs a temporary object of type T using the original glvalue as the. Template argument deduction deduces T to be X, so the parameter has type X&&. 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. 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. You need to pass in an rvalue, and for that you need to use std::move: Insert(std::move(key), Value()); // No compiler error any more I can see why this is. const tells you if a variable can be modified or not. 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. The lvalue or xvalue refers to an object not of the type of the (prvalue) rvalue, nor of a type derived from the type of the (prvalue) rvalue. – super. 98 * @param __t A thing of arbitrary type. 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. ref]/5. This article also mentioned that issue. It cannot convert from an rvalue to an lvalue reference, even a const one. Follow. However, you don't have double && in your code, you have U && for a deduced U. 2 1). @user2308211: I think what I might have meant to say (back when I didn't know any C++!) was that vec4(). The result is an lvalue if T is an lvalue reference type or an rvalue reference to function type (8. 20 and lower) & R-value, higher the number the better (R-5 and higher). The only thing that can be an rvalue or an lvalue is an expression. In the introduction to "Effective Modern C++" it says: A useful heuristic to determine whether an expression is an lvalue is to ask if you can take its address. 25, or 4 (leaving off the units for brevity). C++0x: rvalue reference versus non-const lvalue. ; T is not reference-related to U. 2 Answers. Ternary conditional operator will yield an lvalue, if the type of its second and third operands is an lvalue. thus, this is legal: string&& s = foo (); // extends lifetime as before s += "bar"; baz (std::move (s)); // move the temporary into the baz function. Non-const rvalue references always refer to a type. 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. Understanding Lvalues and Rvalues. , values that can be assigned: namespaces, for instance, are not assignable; thanks to @Maggyero for the edit suggestion). However once the const keyword was added to the C++, lvalues were split into —. Since int() isn't an lvalue, you can't assign to int(). e. Forwarding references are a special kind of references that preserve the value category of a function argument, making it. an lvalue reference instead of an rvalue reference) and had the appropriate cv-qualification, then it's probably the programmer's mistake. 6. 106) This requires a conversion function (12. "cannot bind non-const lvalue reference of type ‘M&’ to an rvalue of type. @MikeMB the standard rarely prevents compilers from inserting for (int i = 0; i < 1 billion; ++i) at arbitrary points. 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. However what matters here is the expression and: Each C++ expression (an operator with its operands, a literal, a variable name, etc. If we have a rvalue we can assign it to a variable, or take a reference, hence becoming a lvalue. 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. The rvalue-reference version can't be called with an lvalue argument. If the function argument is an rvalue, the compiler deduces the argument to be an rvalue reference. Their very nature implies that the object is transient. An rvalue (so-called, historically, because rvalues could appear on the right-hand side of an assignment expression) is an xvalue, a temporary object or subobject thereof, or a value that is not associated with an object. This is indeed a temporary materialization; what happens is that the compiler performs lvalue-to-rvalue conversion on t2 (i. An lvalue may get converted to an rvalue: that's something perfectly legit and it happens quite often. The addition operator + (and all other binary operators) requires both operands to be rvalue, and the result is rvalue. 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. 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. 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. Hence, the end result is the attempted binding of the rvalue. int a = 1, b; a + 1 = b; int *p, *q; cppreference wrote:; An xvalue is an expression that identifies an "eXpiring" object, that is, the object that may be moved from. 1. 197. const A& x = 1; //compile x = 2; //error! A&& xxx = 1; //compile A& xx = 1; //does not compile. An rvalue reference is a new type. e. 1 is rvalue, it doesn't point anywhere, and it's contained within lvalue x. @YueZhou Function lvalues may be bound to rvalue references. 4. i by itself is an lvalue. 6. In C++, each expression, such as an operator with its operands, literals, and variables, has type and value. Therefore the usage of const rvalue references communicates thatAn lvalue is an expression representing an object that can have its address taken. 1. Most operators require lvalue-to-rvalue conversion because they use the value of the object to calculate a result. const T& is the O. It can appear only on the right-hand side of the assignment operator. 5. 3. So sizeof (0, arr) = sizeof (arr) and which would be equal to 100* sizeof (char) and not = sizeof (char*). Whenever a glvalue expression. The difference is that &i is OK but &5 is not. )In the third line, they undergo an implicit lvalue-to-rvalue conversion. But the third one steals the goalKeeper object of t. Improve this answer. If t returns by lvalue reference, the code does not compile because a rvalue reference cannot bind to it. –std::forward is usually the way to 'convert' value category. 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). Alex November 11, 2023. Without this, the compiler will complain that you "cannot bind non-const lvalue reference of type 'std::string&' to an rvalue. You can use an lvalue almost anywhere where an rvalue is required and an implicit lvalue to rvalue conversion will occur automatically. Set the Enforce type conversion rules property to /Zc:rvalueCast or /Zc:rvalueCast. e. Note that by binding a temporary to a rvalue-reference (or a const reference) you extend its lifetime. class XAttr : public AttrDec { public: XAttr (const std::wstring& name) :AttrDec (new Attr (name)) // create a pointer here {} }; And then get rid of the rvalue constructor in AttrDec. Radius: 2 2 4. That means you can't call non-const functions on the object, but if you want to pass rvalues such as temporaries, then calling non-const functions wouldn't necesarily make much sense anyway. I would like to move an object into a std::vector using std::vector::push_back(). goo<int> is an lvalue of function type, but expressions of function type are. 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. 6 — Pass by const lvalue reference. A minimal example:This is because of copy elision in C++. lval), array-to-pointer (conv. We are allowed to do that because the object is an rvalue, when the constructor finishes its job, t will be destructed. 3. 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 reference could be bound to the result of the implicit conversion if it wasn't non-const because the result of that implicit conversion is an rvalue i. A compiler can optimize the call to copy constructor and directly call the matching constructor. std::auto_ptr<Foo> foo(new Foo()); // auto_ptrs are deprecated btw bar(std::move(foo)); // changed ownership. 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. Lvalue to rvalue conversion changes the value category of an expression, without changing its type. Given all three functions, this call is ambiguous. It's not needed, and suppressed. For example second type of the pair should be std::string, not const std::string * and all your problems would go away. This is a follow-on question to C++0x rvalue references and temporaries. 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. 18. Lvalue reference and rvalue reference are both types; because the names are so similar, it is easy to confuse the two. undefined behavior: The lvalue or xvalue is a nonclass type, qualified by either const or volatile. 1, 4. test prep. It's long-lived and not short-lived, and it points to a memory location where 1 is. It is VC++'s evil extension.