We're talking about the temporary object created by Contrived(), it doesn't make sense to say "this object is an rvalue". 10/7 reads, Whenever an lvalue appears in a context where an rvalue is expected, the lvalue is converted to an rvalue; see 4. 2k 9 128 212 asked Jan 14, 2016 at 8:26 Simon X. All lvalues that aren't arrays, functions or of. 44. However, rvalues can't be converted to lvalues. Consider the following code where an lvalue reference is bound to an rvalue (the lambda): int main () { auto& f = [] () -> void {}; return 0; } gcc (4. Share. 106) This requires a conversion function (12. The Parent class stores a pointer, but due to rvalue to lvalue conversion, the Parent ends up storing a reference to a pointer. An entity (such as an. It could even do so with std::move only. Conversion operators are treated inconsistentlyAn lvalue can be converted to a value of an expression through lvalue conversion. The reason is simple; named rvalue reference is treated as lvalue (and implicit conversion from lvalue to rvalue reference is forbidden by standard). This is its value category. 23. No temporary is created, no copy is made, no constructors or. 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. The answer lies in the second property of expressions: the value category. If the C-value is 0. int rVal () { return 0; }. 1) Two possibly multilevel pointers to the same type may be converted between each other, regardless of cv-qualifiers at each level. 「右辺値」「左辺値」というのは 誤訳だ (正確には時代遅れ)、もう一度言うが直ちに脳内から消去するべきである。. 3. As for why the compile fails when you omit the move: When Stream& operator<< (Stream& s, Dummy) is called without the move, Stream will be std::fstream. init. ; T is not reference-related to U. Now an lvalue reference is a reference that binds to an lvalue. 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. 3. 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. A prvalue (“pure” rvalue) is an rvalue that is not an xvalue. 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. Read 5. 1. Lvalue to rvalue conversion. Example: int a. I think it's reasonable to call print_stream like this:. 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. (An xvalue is an rvalue). We are allowed to do that because the object is an rvalue, when the constructor finishes its job, t will be destructed. 3 -- Lvalue references ), we discussed how an lvalue reference can only bind to a modifiable lvalue. 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. 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. The pass-by-value version allows an lvalue argument and makes a copy of it. std::move performs a static_cast to an rvalue reference type and returns the rvalue reference. The pre-C++ origin of the terms "lvalue" and "rvalue" might be related to "left" and "right" side of assignment, but that meaning is only applicable in a small subset of the C++ language. 3) If new_type is an rvalue reference type, static_cast converts the value of expression to xvalue. For example, this code will not compile. Through an lvalue to rvalue conversion. If the target type is an inaccessible or ambiguous base of the. rvalues can bind to rvalue references and const lvalue references, e. 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. – 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. 1/4 "Primary expressions"). Basically, VS will allocate the space somewhere and just let the reference point to it, as if it was a reference-to- const without the constness (or in C++11 an rvalue reference). If t returns a local variable, then you get a dangling reference, since that variable is gone after the call. Convert any type to void, evaluating and discarding the value. int a = 0, b = 1; a = b; both a and b are lvalues, as they both potentially - and actually - designate objects, but b undergoes lvalue conversion on the right-hand side of the assignment, and the value of the expression b after lvalue conversion is 1. e. This is already done in some places. warning C4238: nonstandard extension used: class rvalue used as lvalue But the very same program compiles fine in gcc 11 and clang 12 with the options -std=c++20 -Wall, without any warnings. The example is interesting because it seems that only lvalues are combined. But then i got following error:. Converts between types using a combination of explicit and implicit conversions. 5. (Lvalue-to-rvalue conversions on class types are rare, but do occur in some places in the language, e. There is no implicit conversion as suggested in the title, the reference binds directly to the. 6 — Pass by const lvalue reference. 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. 2. I played a bit around with composite-patterns and inheritance in c++. When you pass a string literal a temporary std::string will be constructed from the string literal. Forwarding referece works with both lvalues and rvalues, with the help of template argument deduction. Among. In the case of object constructing is true but in the case of object assigning is false. Forwarding references are a special kind of references that preserve the value category of a function argument, making it. You might consider A& f () & { to ensure the call is happening on an lvalue object if you need to do something like this. An lvalue is a value bound to a definitive region of memory whereas an rvalue is an expression value whose existence is temporary and who does not necessarily refer to a definitive region of memory. And the lvalue-to-rvalue conversion always returns a prvalue value, not a (temporary) object. It's actually a cast. Found workaround how to use rvalue as lvalue. 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. [3] Finally, this temporary variable is used as the value of the initializer. IBM® continues to develop and implement the features of the new standard. You would then need to add a destructor to AttrDec and delete the pointer in it and add a copy constructor. That's an exception to the general rule that it is impossible for lvalues to be bound to rvalue. 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. 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. Let's look at the following snippet: So we have a reference being initialized by an xvalue of type const foo. 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). With string as template argument you get string&& in the function parameter, which is a rvalue reference that doesn't accept lvalues. There is no implicit conversion as suggested in the title, the reference binds directly to the. 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. specifically, the argument expression is an rvalue that is bound to the rvalue reference parameter. When programming in C++03, we can't pass an unnamed temporary T () to a function void foo (T&);. 1Primary categories lvalue prvalue xvalue 2Mixed categories glvalue rvalue 3Special categories Pending member function call Void expressions Bit-fields Move. 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. I would respect the first compiler more, it is at least. You have to pass pointer to smart pointer, and pointer can have any type - lvalue/rvalue. [dcl. 4/1: The result is an lvalue if T is an lvalue reference type or an rvalue reference to function type and an xvalue if T is an rvalue reference to object type; otherwise the result is a prvalue. The first are categories for the type of a variable/member. Let’s turn it around a bit. Otherwise, the type of the rvalue (until C++11) prvalue (since C++11) is T. C++11 introduced the Rvalue reference for the first time, which is a tool that allows us to get permanent access to temporary objects in memory. Arrays are lvalues. Note that by binding a temporary to a rvalue-reference (or a const reference) you extend its lifetime. Share. Set the Enforce type conversion rules property to /Zc:rvalueCast or. 1) Is actually not so arbitrary. If inside foo no move operation happened like my example, then my_ptr_var will not actually be moved from. 1: A glvalue of a non-function, non-array type T can be. 1 Answer. void func (unsigned int& num) this function need quote type. 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. x is not assignable, because it's an rvalue in 03, a prvalue in 11 and an xvalue in 14, but using a member function always allows you to convert rvalues to lvalues (because *this is always an lvalue). 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. ConclusionFrom expr. 区分左值和右值是很重要的，这是使用C++11 move语义的基础。. 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. In (static_cast<int&&> (3))++, the expression static. Recall that there is a difference between the concept of an Lvalue and an Rvalue. 1. The type of the variable k is an r-value reference, but that's fine. It can convert lvalues to lvalue references and rvalues to rvalue references. 2. If you pass an prvalue, it isn't converted, the temporary is materialised into the parameter object. const foo&& can only bind to an rvalue, but const foo& can bind to both lvalues and rvalues. 3. So, the conversion from a A rvalue to something that P&& would accept in (1) calls the user defined conversion function operator P() &&. Rvalue references allow one to make classes that can be both moved and copied. B. During reference initialization, where the reference to cv1 T is bound to the lvalue or rvalue result of a conversion from the initializer expression from the class type cv2 S,. lvalues and rvalues lvalue –writable memory location; has an address int a; rvalue –data at readable memory location or readonly value that doesn’t have an address Transient (temporary) variable (register, means that we cannot change it’s value in C/C++, only to fetch) constant (including addresses) 5 = a; //An rvalue is a type of expression, not a type of object. 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. 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. uint8Vect_t encodeData(uint8Vect_t &dataBuff); Here you are taking a reference to a uint8Vect_t. Getting into all the details of the various value categories isn't going to be at all helpful to a beginner and will just serve to confuse and discourage. This is because, in C programming, characters are internally stored as integer values known as ASCII Values. Consider this similar question: "Is an integer an lvalue or an rvalue". The reference declared in the above code is lvalue. The expression that created the object is an rvalue expression, but that's different. Either have a single function taking by value and moving from it, or have two functions, one taking lvalue ref and copying and one taking rvalue ref and moving. The reason why you need to const is to make x not a forwarding reference. 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. It is still not allowed per [dcl. 3. cv]/4. According to the C++ specifications, it takes two rvalues as arguments and returns an rvalue. ; // not legal, so no lvalue. The discussion of reference initialization in 8. But for the third case i. I'm a bit confused about lvalue and rvalue bindings, I have the following code:. , Circle c3 (Circle (4)), I'd expect the third constructor, (copy constructor with rvalue referecne) to be called but it's not the case. The list of languages that are currently supported includes C++, C#, Go, Java, Kotlin, PHP, Python, Ruby, Rust, TypeScript, and more. e. 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。. is an rvalue reference to an object type, is an xvalue. 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. 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. 4 — Lvalue references to const. Conversion of a function pointer to void * shall not alter the representation. } or in . Arrays can only be lvalues, and whenever they are used in an lvalue they decay to a pointer to the first element. Forwarding references are very greedy, and if you don't pass in the. a non-const reference). So when. 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. It is a forwarding reference. Rvalues of type int cannot bind to int& (aka an lvalue reference to int) so the compiler rejects your code. The lvalue is. 4. h, the output is same as Clang output it's reasonable. – Corristo. If an lvalue-to-rvalue conversion from an incomplete type is required by a program, that program is ill-formed. For the second overload, it would call operator const P&() const&. In any assignment statement “lvalue” must have the capability to store the data. end()) is a temporary object and cannot be bound to lvalue reference. It can appear only on the right-hand side of the assignment operator. I have defined two type conversion operators, one for lvalue and one for rvalue. const A& ), and lvalue-to-rvalue conversion is suppressed when binding lvalue-reference. The entire point is that you know that this entity references an rvalue and you can legitimately move its content. Non-const rvalue references always refer to a type. 3. An lvalue may get converted to an rvalue: that's something perfectly legit and it happens quite often. 12. thus, this is legal: string&& s = foo (); // extends lifetime as before s += "bar"; baz (std::move (s)); // move the temporary into the baz function. So we declare a variable x: int x = 42; An expression x in this scope is now an lvalue (so also a glvalue). 9/1: The result of the expression static_cast<T> (v) is the result of converting the expression v to type T. A modifiable lvalue may be used as the first (left) argument of the built-in assignment operator. 3. Let's think of the addition + operator for example. 0. So in your example, the expression to the right of the = is an expression that happens to be an lvalue. 1. It's been part of the language since the beginning. std::string hello = "hello"; std::string planet. Allowing both rvalues and lvalues to be bound to an lvalue reference makes that impossible. You could also pass it to a function accepting a const char*& (i. Without lvalue-to-rvalue conversion, it cannot read it's value. 2 indicates the behavior of lvalues and rvalues in other significant contexts. 1, 4. 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. 2) non-modifiable lvalues, which are const. “If T1 is reference-related to T2 and the reference is an rvalue reference, the initializer expression shall not be an lvalue. In C, (time_t) { time (NULL) } is a compound literal C99, initialized by the return value of time. Improve this answer. 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). So when you bind the references the lvalue will have to be const. A function parameter such as T&& t is known as a forwarding reference. – T. In both cases, if the wrapper has been successfully constructed, we mark the status as value to indicate that we have a value. Both of these options are user-defined conversion functions, so neither is better in terms of overload resolution, thus an ambiguity. The locator value is called lvalue, while the value resulting from evaluating that location is called rvalue. 左值可以出现在赋值号的左边或右边。. Therefore the usage of const rvalue references communicates thatAn lvalue is an expression representing an object that can have its address taken. An rvalue reference is a new type. 0. From C++11 4. 4. L-Values are locations, R-Values are storable values (i. 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. However, Microsoft compiler does accept it meaning that. 1 (page 85 for version 3485). To get a lvalue expression to the value pointed to by a pointer, just use the unary * operator. In C++ results of conversions are always rvalues (unless you convert to reference type). 5, then the R-value is 2. std::forward is a conditional std::move. 3 Pointer Types): All function pointer types shall have the same representation as the type pointer to void. 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. It is illegal in C++ to attach non-const references to rvalues. Open the project's Property Pages dialog box. 6. Lvaluesand Rvalues Lvalues and rvalues aren’t really language features. 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. The expressions f (), f (). Regarding the second question. 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. Naming expressions are always lvlaues. 1: (5. 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. Read 5. 1 Answer. 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). For example second type of the pair should be std::string , not const std::string * and all your problems would go away. 9. Otherwise, the reference shall be an lvalue reference to a non-volatile const type (i. It doesn't need to get the value of. ; The value of i is implicitly converted to integer by constructor. When the template gets resolved, baz is going to be either an lvalue or an rvalue reference, depending on the call situation. e. 2. 12. I think I'm missing something basic regarding the lvalue-to-rvalue standard conversion. , Circle c3 (Circle (4)), I'd expect the third constructor, (copy constructor with rvalue referecne) to be called but it's not the case. C++0x rvalue reference template argument deduction. 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. Lvalue to rvalue conversion A glvalue of any non-function, non-array type T can be implicitly converted to a prvalue of the same type . 3. call]/12, [expr. 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. Expressions don't have return types, they have a type and - as it's known in the latest C++ standard - a value category. Note that there is one exception: there can be lvalue const reference binding to an rvalue. The usual arithmetic conversions required by many arithmetic operators do invoke an lvalue-to-rvalue conversion indirectly via the standard conversion used. C++ (as opposed to C) is a devoted lvalue-preserving language: it strives to painstakingly preserve the "lvalueness" of an expression whenever it is possible. You will often find explanations that deal with the left and right side of an assignment. It can be useful if I am writing a function which expects either an lvalue or rvalue in a parameter and wants to pass it to another function. 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. This allows you to explicitly move from an lvalue, using move to. Since your t variable is an lvalue, std::apply calls product with an int& instead of an int&&. References. This is what std::move is for. That works well with normal variables but uint8Vect_t(dataBlock. 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. 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. We can take the address of an lvalue, but not of an rvalue. Correct. 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. Would you ever mark a C++ RValue reference parameter as const. These get their names from the types of items that can go on the left-hand-side and right-hand-side of an assignment statement. 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. c++ base constructor lvalue to parameter. How to pass lvalue to function taking rvalue only without templates. 3. 1) does not accept such code (makes perfect sense). "lvalues are named variables and rvalues are temporaries" is a good enough heuristic for a beginner, and no more an "oversimplification" than "I before E except after C" is for English. Nothing is being turned into a lvalue. 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. template <class T, class Other = T> T exchange(T& val, Other&& new_val). 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. A nice feature of this heuristic is that it helps you remember that the type of an expression is independent of. This ensures that you never actually modify the original this value. It shouldn't. Let's look at (T1&&)t2 first. e. C++0x: rvalue reference versus non-const lvalue. 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. 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. 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. Each C++ expression (an operator with its operands, a literal, a variable name, etc. 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. void f1(int& namedValue){. a glvalue (“generalized” lvalue) is an expression whose. The rvalue-reference version can't be called with an lvalue argument. Stripping away the const using const_cast doesn't fix the issue. Conversely, d = static_cast<float> (j)/v; produces an. key here is Key&& key - this is an lvalue! It has a name, and you can take its address. 1:. If an lvalue-to-rvalue conversion were performed on s, it would also call the copy constructor; [conv. 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’. 1/2: The value contained in the object indicated by the lvalue is the rvalue result. Whenever an lvalue appears in a context where an rvalue is expected, the lvalue is converted to an rvalue; see 4. 6) An lvalue (until C++11) glvalue (since C++11) expression of type T1 can be converted to reference to another type T2. 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. For the class type A, f (a); causes the copy constructor of A to be invoked. Therefore, I will not jump right in and explain what rvalue references are. By make_tuple<int> you make make_tuple signature look like: make_tuple(int&&). The output is: Copy constructor with lvalue reference. But one important rule is that: one can. That being said, and assuming you don't want to overload doStuff (otherwise see Hinnant's answer), you can write a utility. This example might clarify it:So we have a reference being initialized by an xvalue of type const foo. @banana36 With that function, calling foo(std::move(my_ptr_var)) wont actually pass ownership. Jun 27 at 7:34. 4. (since C++11) 4) If new_type is the type void (possibly cv-qualified), static_cast discards the value of expression after. The reason why you need to const is to make x not a forwarding reference. 20 hours ago · String Templates (a preview feature introduced in Java 21) greatly improves how we create strings in Java by merging constant strings with variable values. std::function's type is defined only by its target's signature(eg: void(int)) and std::function itself is defined by the. Select the Configuration Properties > C/C++ > Language property page. 1 Answer. . An lvalue may get converted to an rvalue: that's something perfectly legit and it happens quite often. So you can write a couple of convert functions . Done. The C++ standard does not specify explicitly that it is lvalue to rvalue conversion that is responsible for causing an access. static_cast<Type> (expression) belongs to one of the following value categories: or an rvalue reference to a function type is an lvalue. Informally, "lvalue-to-rvalue conversion" means "reading the value". That is the whole point of references. undefined behavior: The lvalue or xvalue is a nonclass type, qualified by either const or volatile. const A& x = 1; //compile x = 2; //error! A&& xxx = 1; //compile A& xx = 1; //does not compile. 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). I expect that when using a temporary instance of a Wraper object, the conversion operator defined for rvalue will always be used. Assignment involving scalar types requires a modifiable lvalue on the left hand side of the assignment operator. 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. Otherwise, the reference you get behaves more. 2, and 4. The value of x is 1. You are comparing two different things that are not really related. const T& is the O. Types shall not be defined in a reinterpret_cast. According to the rule of forwarding reference, when an lvalue is passed to add, the template type argument Element will be deduced as SomeClass&. enum type init and assignment must be enum inside，so enum type can't is lvalue。. Forwarding references are very greedy, and if you don't pass in the exact same type (including. 2 Answers. Each expression in C (an operator with its arguments, a function call, a constant, a variable name, etc) is characterized by two independent properties: a type and a value category . 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. fstream file{"filename"}; print_stream(file);I would like to write a variadic template function that accepts rvalues and lvalue references. rvalue references are sausage-making devices added later after nobody could find a. 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. The quote doesn't say anything about the result of &, which in fact is an rvalue. b is just an alternative name to the memory assigned to the variable a. The rvalue reference is bound to the temporary materialized from the prvalue conversion of arr. First the compiler performs an implicit array-to-pointer conversion for "abc", so the type of "abc" becomes const char*. As long as no const is involved, the expression T() is a modifiable rvalue, to be more precise. An example of an rvalue would be a literal constant – something like ’8′, or ’3. It's just that type of that lvalue is "rvalue reference to Key ". 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. 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.