Suppose r is an rvalue reference or nonvolatile const lvalue reference to type T, and r is to be initialized by an expression e of type U. A non-const reference may only be bound to an lvalue? (too old to reply) George 15 years ago Hello everyone, I am debugging MSDN code from,. Apr 14 at 22:55. 3 Answers. three rules on bit-fields: Rule 1, "A bit-field shall not be a static member. decltype(fun()) b=1; Then, your code initializes a const reference with a prvalue of a different (non-reference-related) type. When I discovered this, it seemed odd to me, so I tried. Because as_const doesn't take the argument as const reference. Other situations call for other needs, but today we will focus on constant references. There is a special rule in the language that allows binding a const lvalue reference to the rvalue (whether const or not) by extending the lifetime of the rvalue to match the lifetime of the. The binding rules for rvalue references now work differently in one aspect. Data members: Never const. 3. 2. a nonconst reference could only binded to lvalue. reference (such as the B& parameter in the B::B (B&) constructor) can only. 2) persists until the completion of the full-expression containing the call. Or, passing it by const reference will also work, since a const lvalue reference can be. How to fix depends on what the return type of cleverConfig. 7 = a; The compiler / interpreter will work out the right hand side (which may or may not be const), and then put it into the left hand side. 3. Good article to understand both lvalue and rvalue references is C++ Rvalue References Explained. And since that the converted initializer is an xvalue not prvalue, [conv. A temporary has a type, that type can be const, and it can be non-const. 4. Is it for optimization purposes? Take this example:By overloading a function to take a const lvalue reference or an rvalue reference, you can write code that distinguishes between non-modifiable objects (lvalues) and modifiable temporary values. 4 — Lvalue references to const. Thus you know that you are allowed to manipulate it without damaging other data. A non-const reference may only be bound to an lvalue[/quote] 1 Reply Last reply Reply Quote 0. However, an rvalue can be bound to a. E may not have an anonymous union member. )An variable name (which is normally an lvalue) can be moved in a return statement if it names an implicitly movable entity: An implicitly movable entity is a variable of automatic storage duration that is either a non-volatile object or an rvalue reference to a non-volatile object type. if a regular constant can be passed like this: In that example, you have an lvalue reference to const. Take pointers by value -- T const*-- and things are more sane. Early on, when we teach modern C++, we teach that every non-small 1 data should be passed, by default, as constant reference: 1. 806 3 3 gold badges 12 12 silver badges 20 20 bronze badges. thanks in advance, George. Essentially, a mechanism is needed to distinguish between values that can be moved from, and those that cannot be moved from (i. A reference to the container element is obtained from the iterator with the indirection operator: *hand_it. Saturday, December 15, 2007 4:49 AM. So naming kInt is not deemed an odr-use as long as it. C4239 は、以下。. (Binding to a const reference is allowed. What you probably want is: BYTE *pImage = NULL; x. rvalue reference versus non-const lvalue. 2. const reference to non-const object. This means the following is illegal: This is disallowed because it would allow us to modify a const variable ( x) through the non-const reference ( ref ). The temporary int's lifetime will be the same as the const reference. It allows you to do something like swap(a, b), and it will actually swap the values of a and b, instead of having to do swap. You are returning a reference to a local variable. ; T is not reference-related to U. Visual C++ is non-compliant with the standard in allowing binding of temporaries to non-const lvalue references. Non-const reference may only be bound to an lvalue. If /Zc:referenceBinding is specified, the compiler follows section 8. There are several (very constrained) circumstances in which the compiler, with language extensions enabled, will still allow a non-const lvalue reference to bind to an rvalue expression. Const reference can be bounded to. So your reference would be referring to the copy of the pointer which wouldn't be modified if you change the Player object. The standard specifies such behavior in §8. A usual lvalue reference (to a non-const value) won’t do. 3/5, [dcl. Sometimes even for the original developer, but definitely for future maintainers. template <auto N> void f () { auto & n = N; } This works when f is instantiated over class types. The option -qlanglvl=compatrvaluebinding instructs the compiler to allow a non-const or volatile lvalue reference to bind to an. std::is_rvalue_reference<T&&>::value A temporary can only bind to a reference to a prvalue. The conformant behavior does not allow binding a non-const reference to an rvalue. Modified 6 years,. Your declaration of a is a non-const lvalue reference,. reference to type 'myclass' could not bind to an rvalue of type 'myclass *'. – The outcome is that the code compiles and works when using MSVC, but doesnt on GCC and Clang, with respective errors: GCC: cannot bind non-const lvalue reference of type 'FuncPtr<bool ()>&' to an rvalue of type 'FuncPtr<bool ()>' Clang: no matching constructor for initialization of 'A'. Only const lvalue references (in C++98 and C++11) or rvalue references (in C++11 only) can. Mark Forums Read; Quick Links. My question is, why a non-const reference can not binded to a rvalue? I think the reason is rvalue is not addressable? And we can not change the rvalue through its reference?Warning: "A non-const reference may only be bound to an lvalue" I've encountered a very weird warning that, although compiles fine on windows, fails to. Rule: lvalue, rvalue, const or non-const objects can bind to const lvalue parameters. This rule does not reflect some underlying. Example 5 @relent95 Yes, whether the id-expression refers to a variable of reference or non-reference type doesn't matter because of what you quoted. The Python-side. Saturday, December 15, 2007 4:49 AM. Solution 1: Your problem lies here: The variable is an lvalue reference, that means it's a reference that cannot bind to temporary variables. That is special syntax for a so-called forwarding reference. is an xvalue, class prvalue, array prvalue or function lvalue and "cv1 T1" is reference-compatible with "cv2 T2", or. If an rvalue is passed to factory, then an rvalue will be passed to T's constructor with the help of the forward function. However, since Visual C++ allows this as an extension, how does it work? From what I've gathered, the standard does not allow this since you're getting a reference to a temporary variable, which can cause issues. Since there are some non-modifiable lvalues (so we do not always need to modify values through its reference). (コンパイラは VS2012) warning C4239: nonstandard extension used : 'initializing' : conversion from 'A' to 'A &' A non-const reference may only be bound to an lvalue. As to why using const & or even rvalue && is a bad idea, references are aliases to an object. 21. Let's look at std::vector for example: reference at( size_type pos ); const_reference at( size_type pos ) const; Would you look at that. ref]/5:. You can implement a method and have one "version" for a const object, and one for a non-const object. [ Example: double& rd2 = 2. It's the specific case where changing T& to const T& does more than just ban modifications. Non-compliant compilers might allow a non-const or volatile lvalue reference to be bound to an rvalue. There are better ways to solve your problems. But doesn't work when instantiated over non class types (as I expected)This change is required by the C++ standard which specifies that a non-const. This extends the lifetime of the temporary: base * const &rp = (base*)p; Or bind the reference to an lvalue: base * b = p; base * &rp = b; Share. I get tired of writing a pair of iterators and make a View class. An expression that designates a bit-field (e. — Otherwise, the reference shall be an lvalue reference to a non-volatile const type (i. This approach does not work for two reasons: First, because we modify the source object, we have to pass it as a non-const reference. (After all, there is no actual long long to refer to. If t returns by rvalue reference, you obtain a reference to whatever was returned. Undefined behavior can sometimes look like it's working. To be standards compliant, you need. v = this->v*a. ) Aside from the workaround you already have, if you can change the function to take const QImage& then that would be better. Follow. A non-const reference may only be bound to an lvalue At best, it compiles for reasons of backward compatibility. Remember Me? Forum; FAQ; Calendar; Forum Actions. h(418) : warning C4239: nonstandard extension used : 'argument' : conversion from 'XUTIL::xList<T>::iterator' to. g. A operator*(const A& a) // Return a value, not a reference. Thus, in case of your variable b: T = int ==> T&& becomes int&& T = int& ==> T&& becomes int. 6. And plus more, in this case if I called. An lvalue (locator value) represents an object that occupies some identifiable location in memory (i. I can't understand why I have to specify the dynamic type to make it work. You can either modify the return type of the function from Value* to const Value& , or opt for return *cleverconfig[name]; . Both const and non-const reference can be binded to a lvalue. ii. Reload to refresh your session. Potentially related articles: Overload resolution between object, rvalue reference, const reference; std::begin and R-values; For a STL container C, std::begin(C) and similar access functions including std::data(C) (since C++17) are supposed to have the same behavior of C::begin() and the other corresponding C's methods. Declaring operator + to accept non-const references does not make. e. Share. 4 Why Rvalue cannot bind Lvalue reference? 18 Invalid initialization of non-const reference of type. By the way, don’t return const values from a function, because you make it impossible to use move semantics. I do not quite understand why there is a warning A non-const reference may only be bound to an lvalue and 'B::B (A)' called instead of 'B::B (B &)'? think. But instead removing either reference overload results in ambiguity with f( int ). It reflects the old, not the new. With /W4 you'd see this: warning C4239: nonstandard extension used : 'initializing' : conversion from 'Foo' to 'Foo &' 1> A non-const reference may only be bound to an lvalue Specifically, MSVC 2013 will give a warning of "mysourcefile. Assuming standard data sizes, you have a reference to 2 bytes of data that you're trying to pass into a function that takes a reference to only 1 byte. (The small difference is that the the lambda-expression is converted to a temporary std::function - but that still can't be bound to a non-const reference). Take a look at the swap function signature: swap ( shared_ptr& r ). The advantage of rvalue references over lvalue references is that with rvalue references you know that the object referred to is an rvalue. However, you might need at that returns non-const reference too. initial value of reference to non-const must be an lvalue (emphasis mine). 2 Answers. long can be promoted to a long long, and then it gets bound to a const reference. However, you don't have double && in your code, you have U && for a deduced U. Non-const reference may only be bound to an lvalue. Expression like a+b will return some constant. 3. The whole idea of forwarding is to accept any value category and preserve it for future calls. GetCollider (); platform1. No, "returning a reference" does not magically extend any lifetime. Since the temporary B that's returned by source () is not. A non-const reference can be used to change the value of the variable it is referring to. a. for example, to get a reference to the element. This section presents an intentionally simplified definition of lvalues and rvalues. e. std::tie always expects lvalues for arguments, since its intended purpose is to be used in assignment. std::string&& rref = std::string("hello"); rref has value category lvalue, and it designates a temporary object. Saturday, December 15, 2007 4:49 AM. y()) < std::tie(b. But if you are asking why this doesn't. A reference to type “cv1 T1” is initialized by an expression of type. push_back (std::move (obj)); } If caller passes an lvalue, then there is a copy (into the parameter) and a move (into the vector). r can be bound to the conversion result of e or a base class of e if the following conditions are satisfied. EX: int &var=4; we can change value of reference , but logically it is not possible to change 4. If you are trying to modify the variable 'pImage' inside the method 'GetImage ()' you should either be passing a pointer or a reference to it (not doing both). It allows you to do something like swap(a, b), and it will actually swap the values of a and b, instead of having to do swap. note: A non-const reference may only be bound to an lvalue. error: cannot bind non-const lvalue reference of type 'int&' to an rvalue of type 'int' return std::tie(a. According to the reference collapsing rules, "rvalue reference to rvalue reference collapses to rvalue reference, all other combinations form lvalue reference". Cannot bind non-const lvalue reference to an 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 the first element of the pair is declared as const, you can't bind a non-const rvalue reference (std::string&&) to it. Example 51) Is actually not so arbitrary. In your default constructor, you try to assign a temporary value (the literal 0) to it, and since it's a reference type you can't give it a temporary value. It expects an lvalue reference parameter. You obviously can't point to a temporary. If you want to work with rvalues, perhaps use an rvalue reference. not an rvalue reference, everything under the sun can be bound by a forwarding reference – Piotr Skotnicki. C++/SDL "initial value of reference to a non-const must be an lvalue" 0 non-const lvalue reference to type 'const int *' cannot bind to a value of unrelated type 'int *It is very rarely a good idea to pass a pointer by const &: at best it takes the same overhead, at worst it causes extremely complex pointer reseating logic to surprise readers of your code. including the case where an lvalue is provided, it cannot modify its input (at least not the one bound to the x parameter) - if it did, it would violate the semantics. The const has nothing to do with the lifetime prolongation. By default, or if /Zc:referenceBinding- is specified, the compiler allows such expressions as a Microsoft extension, but a level 4 warning is issued. 3 of the C++11 standard: It doesn't allow expressions that bind a user-defined type temporary to a non-const lvalue reference. Sometimes even for the original developer, but definitely for future maintainers. Thank you. 3. 0. r-value references are designed to be the subject of a move-constructor or move-assignment. an lvalue, this constructor cannot be used, so the compiler is forced to use. That's only proper when the type is const and the context is one where there is automatic lifetime extension of the temporary. The this pointer is defined to be a prvalue, and your function takes an lvalue. So basically, if you have one method that is qualified (e. e. 5. The only way to safely bind an rvalue to an lvalue is either by marking the lvalue as const, or using a mutable rvalue reference && (introduced in C++11 believe?) Alex November 11, 2023 In the previous lesson ( 12. Values are fine: auto refInstance = m_map. e. Binding a reference is always inexpensive,. GetCollider(); platform1. However, int can be implicitly converted to double and this is happening. MS Visual Studio compilers have allowed binding of non- const references to temporary objects but it is not sanctioned by the standard. Properties -> C/C++ -> Language. The best option is to return by copy. Temporary objects cannot be bound to non-const references; they can only. Rule 3, "Note: if the initializer for a reference of type const T& is. clang++ says: " error: non-const lvalue reference to type 'class foo' cannot bind to a temporary of type 'class foo'" Change foo. Alex September 11, 2023. The rules were already more complex than "if it has a name it's an lvalue", since you have to consider the references. 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& //. Fibonacci Series in C++. "You're not "assigning" to a reference, you're binding to a reference. I'll try paraphrasing it: In compiler version 2002, if the compiler had the choice if it would transform an object returned by a function to a non-const-reference or another type, it would have chosen the non-const-reference. A simple definition. 3 -- Lvalue references ), we discussed how an lvalue reference can only bind to a modifiable lvalue. An rvalue reference can only bind to an rvalue, which is a candidate for moving. non-const lvalue reference to type 'int' cannot bind to a. If an rvalue could bind to a non-const lvalue reference, then potentially many modifications could be done that would eventually be discarded (since an rvalue is temporary), this being useless. 3. name. Unfortunately, they may compile with one common compiler, due to language. Because a reference to a non-const value can only bind to a modifiable lvalue (essentially a non. x, a. What this means is that it's technically possible for the function to modify the pointer itself in a way that gets propagated to the caller. All groups and messages. the pointer but not the pointee. The compiler automatically generates a temporary that the reference is bound to. 1/4 of N3337:. The reference in your example is bound to the constructor's argument n, and becomes invalid when the object n is bound to goes out of scope. -hg. However,. Otherwise, the reference shall be an lvalue reference to a non-volatile const type (i. Returning non-const lvalue reference. 3) non-const lvalues can be passed to the parameter. 3 Answers. Hot Network QuestionsNon-const references cannot bind to rvalues, it's as simple as that. This won't work. int&& x = 10; is a declaration and not an expression. The second version is only allowed non- const rvalues because you can't implicitly strip const from the referencee and rvalue references don't allow lvalues to bind to them. m, where a is an lvalue of type struct A {int m: 3;}) is a glvalue expression: it may be used as the left-hand operand. struct Foo{}; { const auto & r = Foo{}; // Foo object not destroyed at semicolon. So obviously it's not portable. If I were to call it with an rvalue, C++ would shout at me. The Standard says no. 1 1 1. In contrast you can bind const references to temporary values as in: std::string const & crs1 = std::string (); However the following is illegal: std::string & rs1 = std::string (); Don't pass int&, it can't be bound to a constant or temporary because those can't be modified - use const int& instead. Overload resolution is usually done in terms of a strict. Even Microsoft engineers like u/STL recommend avoiding this "extension" if I recall correctly. –The pointer returned by the function cannot be bound to a reference. 6 — Pass by const lvalue reference. This is fulfilled by two types being similar, which basically means if they are the same type with the same number of pointers but possibly different cv-qualifiers (e. If P is a forwarding reference and the argument is an lvalue, the type “lvalue reference to A ” is used in place of A for type deduction. The only time that lifetime is extended is when a prvalue (or an xvalue referring to a member of a prvalue) is bound to a reference variable, and the lifetime of the prvalue is extended to that of the variable:. rvalue Reference Cannot Bind to a Named lvalue. C++: rvalue reference converted to non-const lvalue-reference. So you want x to be either an. Once bound, there is no difference in behaviour between an rvalue reference and an lvalue reference. Value categories are applied to expressions, not objects. std::is_rvalue_reference<T&&>::valueA temporary can only bind to a reference to a prvalue. Unless an object is created in the read-only section of a program, it is open for modifiction without adverse consequences. x where s is an object of type struct S { int x:3; };) is an lvalue expression: it may be used on the left hand side of the assignment operator, but its address cannot be taken and a non-const lvalue reference cannot be bound to it. Second, our new version of the copy constructor will just as happily transplant the internals of lvalues: IntVector v1; IntVector v2 (v1); // v1 is no longer. So an expression returning a non-const reference is still considered an lvalue. I do not quite understand why there is a warning A non-const reference may only be bound to an lvalue? A const reference can be bound to: R-value L-value A non-const reference can be bound to: L-value This means that you can do this: int const &x = 5; But you _can't_ do this: int &x = 5;, thus preventing you from trying to modify a. Just like how we don't want the first example to create a temporary int object (a copy of x) and then bind r to that, in the. I recommend checking how standard library deals with this. But a more proper fix is to change the parameter to a const. 1. , cv1 shall be const), or the reference shall be an rvalue reference. . The reference is. const char*&). & attr (optional) declarator. Note that there is one exception: there can be lvalue const reference binding to an rvalue. For sure, string{""} shall have an address somewhere in memory. Non-const reference may only be bound to an lvalue. In other words, in your first example the types actually do match. Only expressions have values. Some compilers allow int &r = 5; as an extension, so it makes sense, it's just not allowed by the standard. Follow edited Apr 5, 2021 at 12:41. Sorted by: 6. The language forbids that sort of binding for various reasons. There are exceptions, however. –Most of the time you don't want a non-const lvalue reference to refer to some temporary object. 3. Actually for simple types you should prefer to. " Rule 2, "A non-const reference shall not be bount to a bit-field". The conversion produces an rvalue (i. Non-const reference may only be bound to an lvalue. C++. Viewed 3k times. If it is not immediately obvious, we can try to check: Therefore, you can opt to change your getPtr ()'s return to a non-const lvalue reference. 2/5 in N4140): A temporary bound to a reference parameter in a function call (5. rvalues can only be bound to const lvalue references. A operator*(const A& a) const { A res; res. The version with const Integer & works as const lvalue references can be bound to both lvalues and rvalues. rvalues can be residing on read-only memory spaces where changing them might not be allowable and hence the compiler prohibits them. Improve this answer. And this is precisely what the compiler is telling you: The Lvalue refers to a modifiable object in c++ that can be either left or right side of the assignment operator. Follow edited Nov 15, 2016 at. In this case, the conversion function is chosen by overload resolution. CheckCollision(0. A simple solution is: void foo (MyObject obj) { globalVec. For non-static member functions, the type of the implicit object parameter is — “lvalue reference to cv X” for functions declared without a ref-qualifier or with the & ref-qualifier — “rvalue reference to cv X” for functions declared with the && ref. However, I am. Generally speaking, when a function takes a parameter by non-const. Passing by reference, by a const reference wouldn't cost more than passing by value, especially for templates. And an rvalue reference is a reference that binds to an rvalue. This operator is trying to return an lvalue reference to a temporary created upon returning from the function: A temporary or an rvalue cannot be changed with a reference to non-const. Reference is always constant, you can't change reference. reference (such as the B& parameter in the B::B (B&) constructor) can only. 5). An lvalue reference is a reference to an object that has a distinct memory address and can be modified. Accept all cookies Necessary cookies only Customize settings. qual] or even [conv. ref], the section on initializers of reference declarations. reference (such as the B& parameter in the B::B (B&) constructor) can only. We can't bind non-const lvalue reference to an rvalue, but it can be bound to the const one. Consider another last example: const int&& r2 = static_cast<int&&>(0); The same wording as above applies: The initializer expression is an rvalue (xvalue) and cv1 T1 (const int) is reference-compatible with cv2 T2 (int). All rvalues are non-const. Now it makes actually sense to take its address, as it is an lvalue for all intents and purposes. 7. It seems perfectly reasonable for the standard to have been that a temporary is created, and dropped at the end of the function's execution (as you currently have to manually do). Don't pass int&, it can't be bound to a constant or temporary because those can't be modified - use const int& instead. You have two options, depending on your intention. When you pass a pointer by a non- const reference, you are telling the compiler that you are going to modify that. Mar 22, 2013 at 18:39. The second const is good, as is stops the source item being modified. However, getPlayer is returning a copy of that pointer. @relent95 Yes, whether the id-expression refers to a variable of reference or non-reference type doesn't matter because of what you quoted. –And I want to make sure template parameter F&& f only accept a non-const lvalue reference. Technically, auto is the root of the problem. A non-const lvalue reference can only bind to non-const lvalues. A non-const reference may only be bound to an lvalue. Lifetime is extended at most once, when first binding to a reference that is not a function parameter, return value, or part of new initialization or parenthesized aggregate initialization and if the expression between the temporary materialization and. The foo () function accepts a non-const lvalue reference as an argument, which implies one can modify (read/write) the supplied parameter. 0f, c); The other similar calls need to be fixed too. Alex September 11, 2023. It's not against the rules in C++ to use a non-const reference but I think it lends to massive confusion and potential bugs. The compiler automatically generates a temporary that the reference is bound to. Fun fact: /W3 is set. Its . Non-const references cannot bind to rvalues, it's as simple as that. 9,096 1 33 54. const int x = 0; int&& r = x; Here, we don't have an exact match in types: the reference wants to bind to an int, but the initializer expression has type const int. Improve this question. non-const reference of type from an rvalue. First of all, I will post the warning I'm getting: xlist. For lvalue references, T is deduced to be an lvalue reference, and for rvalue references, T is deduced to be a non-reference. MS Visual Studio compilers have allowed binding of non- const references to temporary objects but it is not sanctioned by the standard. Both of g and h are legal and the reference binds directly. In general, when Foo isn't a const type your examples should fail to compile. There are exceptions, however. rval] is not applied (i. struct S {}; f<S {}> (); // ok. The page is trying to say that you can write m. Neither the proxy object, nor the converted bool (which is a prvalue) can be bound to a bool& as you try to do in the return statement. Only const lvalue references (and rvalue references) may be bound to an object accessed through an rvalue expression. For non-const references, there is no such extension rule, so the compiler will not allow: bar(std::string("farewell")); because if it did, at the point foo starts, it would only have a reference to the destructed remnants of what was once the farewell string. of the Microsoft compiler. Every non-static data member of E must be a direct member of E or the same base class of E, and must be well-formed in the context of the structured binding when named as e. You are returning a copy of A from test so *c triggers the construction of a copy of c. The rules were already more complex than "if it has a name it's an lvalue", since you have to consider the references. It can take rvalues because it is marked const and rvalues are allowed to bind to const lvalue references. What getPtr () return: std::shared_ptr<int> getPtr (int val) { } is an rvalue reference. A modifiable lvalue is any lvalue expression of complete, non-array type which is not const-qualified, and, if it's a struct/union, has no members that are const-qualified, recursively. You're not modifying the given pointer, so just pass it by value instead of by reference. const int & is a const lvalue reference. An rvalue may be used to initialize a const lvalue [ rvalue] reference, in which case the lifetime of the object identified by the rvalue is extended until the scope of the reference ends. Otherwise, the reference you get behaves more. obj & a1 = bar(); invalid initialization of non-const reference of type ‘obj&’ from an rvalue of type ‘obj’ using g++. The non-const reference is converted into a const reference when the print function calls getConstReference. The compiler automatically generates a temporary that the reference is bound to. 4. Thus the declaration doesn't have a. I agree with the commenter 康桓瑋 that remove_rvalue_reference is a good name for this. GetImage (iTileId, pulImageSize, a_pImage ); With the method defined as: This change is required by the C++ standard which specifies that a non-const. is an xvalue, class prvalue, array prvalue or function lvalue and "cv1 T1" is reference-compatible with "cv2 T2", or. thanks in advance, George For lvalue references, T is deduced to be an lvalue reference, and for rvalue references, T is deduced to be a non-reference. If you are asking why this code doesn't work : const string& val = "hello" string& val = "hello" the answer is you are trying to redeclare the same variable (val) with conflicting definition. So the first fix is to not use the wrong technique here, and accept by an lvalue reference instead:The simple answer is that you are right in essence. But result of such conversion is an rvalue, so your reference to non-const cannot be bound to it. e. Const reference can be bounded to. e. 1.