So an expression returning a non-const reference is still considered an lvalue. The simplest fix is to simply store the temporary object somewhere, first: Collider c=player. When I discovered this, it seemed odd to me, so I tried. EX: int &var=4; we can change value of reference , but logically it is not possible to change 4. A. C++/SDL "initial value of reference to a non-const must be an lvalue". This section presents an intentionally simplified definition of lvalues and rvalues. The only difference (that I see) is that x2 knows it only has 3 rows, whereas x1 has a dynamic number of rows. Jan 8, 2015 at 8:51. When you pass a pointer by a non- const reference, you are telling the compiler that you are going to modify that. The rest of the article will elaborate on this definition. The core of your question is: can rvalues be bound to non-const lvalue references?. Otherwise, the reference shall be an lvalue reference to a non-volatile const type (i. Oct 10, 2013 at 22:07. C++0x에는 rvalue reference라는 개념이 추가 됩니다. 3. ; T is not reference-related to U. [3] Finally, this temporary variable is used as the value of the initializer. Thus the declaration doesn't have a. nik7. const A& x = 1; //compile x = 2; //error! A&& xxx = 1; //compile A& xx = 1; //does not compile. 3. Non-const reference may only be bound to an lvalue. 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. , cv1 shall be const), or the reference shall be an rvalue reference. The compiler automatically generates a temporary that the reference is bound to. However, A can be converted to an lvalue of type int, and const int is reference-compatible with int, so reference x of type const int can be bound to the conversion result of A(). , temporary) double but a temporary cannot be bound to a non-const reference. rvalue reference 는 rvalue (즉, 상수와 임시객체)도 참조가 가능 하다 점을 빼고는 기존의 참조와 동일합니다. A reference to type “cv1 T1” is initialized by an expression of type “cv2 T2” as follows:I can't be bothered to go looking at that code, but. 4 Why Rvalue cannot bind Lvalue reference? 18 Invalid initialization of non-const reference of type. This rule does not reflect some underlying. You know, just like any other use of const. In summary, after const float & x = true ? a : 2. MS Visual Studio compilers have allowed binding of non- const references to temporary objects but it is not sanctioned by the standard. This won't work. h"` displayPNG("solve. And this is precisely what the compiler is telling you:. " The C++ language doesn't allow you to bind an rvalue to a non-const reference because doing so would allow you to modify the rvalue - which would be impossible if it was a constant and undesirable if it was a temporary. (Binding to a const reference is allowed. The first option can take lvalues because it's an lvalue reference. The implication of a function that takes a non-const reference as an argument is that there is a side-effect applied to the value of that argument. Both const and non-const reference can be binded to a lvalue. The problem is that a non-const lvalue reference cannot bind to a temporary, which is an rvalue. 4. note: A non-const reference may only be bound to an lvalue. It doesn't really matter. What you want is in 40two's answer, but make sure to forward the parameter t. e, the condition. On the contrary, rvalues can be bound to const lvalue references. Since the temporary B that's returned by source () is not. Passing by reference, by a const reference wouldn't cost more than passing by value, especially for templates. Non-explicit constructors have their uses. How to fix depends on what the return type of cleverConfig. This can only bind to a const reference, and then the objec's lifetime will be extended to the lifetime of the const reference it is bound to (hence "binding"). If t returns by lvalue reference, the code does not compile because a rvalue reference cannot bind to it. But a more proper fix is to change the parameter to a const. name. 0 Invalid initialization of non-const reference from a. @acannon828 Okay, but then you'd be modifying the pointer that is internal to World. For reference, the sentence that totally misled me is in [over. E may not have an anonymous union member. Assume a variable name as a label attached to its location in memory. three rules on bit-fields: Rule 1, "A bit-field shall not be a static member. Improve this question. My understanding is that this is largely to avoid breaking several enormous legacy codebases that rely on this "extension. The type of such a reference must be a const qualified lvalue reference or a rvalue references. Your declaration of a is a non-const lvalue reference,. It never makes sense to return a dangling reference, but it's syntactically legal. 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. –Most of the time you don't want a non-const lvalue reference to refer to some temporary object. [ Example: double& rd2 = 2. RVO may explain this particular quark, but you cannot return a reference to something that doesn't exist. void addNeighbour (Element* neighbour); instead of. (1) && attr (optional) declarator. of the Microsoft compiler. This constness can be cast away with a const_cast<>. 3 Answers. Thus you know that you are allowed to manipulate it without damaging other data. This may sound like a silly question, but I was confused about this following behaviour:. it is only accessing the string objects in the array that a points to, so there is no need to pass a by reference, passing it by value will work just fine: void spell(int n, string* a) Live Demo. obj & a1 = bar(); invalid initialization of non-const reference of type ‘obj&’ from an rvalue of type ‘obj’ using g++. It can appear only on the right-hand side of the assignment operator. So naming kInt is not deemed an odr-use as long as it. Rvalue references should be unconditionally cast to rvalues when forwarding them to other functions: void sink (ConcreteType&& ct) // can only be called on rvalues { collection. No, "returning a reference" does not magically extend any lifetime. Constant lvalue references can be bound to all types of values, including non-constant lvalues, constant lvalues. 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. int a = 7. A non-const reference may only be bound to an lvalue[/quote] 1 Reply Last reply Reply Quote 0. e. There are two overloads. 0; // error: not an lvalue and reference not const int i = 2; double& rd3 = i; // error: type mismatch and reference not const —end example] Although not directly related to this case there is another very important difference between const and non-const references. We can take the address of an lvalue, but not of an rvalue. g. There are exceptions, however. Lvalue reference to const. then the reference is bound to the initializer expression lvalue. I can't understand why I have to specify the dynamic type to make it work. e. Testing tools for web developers. This allows you to explicitly move from an lvalue, using move. Understand the design first before you implement. warning C4239: nonstandard extension used : 'initializing' : conversion from 'foo' to 'foo &' A non-const reference may only be bound to an lvalue (note that this remains illegal in C++11) Last edited on Dec 20, 2011 at 2:37am UTC Otherwise, if the reference is lvalue reference to a non-volatile const-qualified type or rvalue reference (since C++11): If target is a non-bit-field rvalue or a function lvalue, and its type is either T or derived from T , equally or less cv-qualified, then the reference is bound to the value of the initializer expression or to its base. e. " followed by a specification of how the result of the conversion is determined. In the case of storing a value, the type can be non-const and the function could modify that object, so the approach is more flexible. In the example above, SomeClass() is not bound to an identifier, so it is an rvalue and can be bound to an rvalue reference -- but not an lvalue reference. @KerrekSB: Binding a temporary to a const reference can cause a copy construction. Visual C++ is non-compliant with the standard in allowing binding of temporaries to non-const lvalue references. There's no difference between a bound rvalue reference and a bound lvalue reference. There's a special rule in C++ template deduction rules which says that when the parameter type is T&& where T is a deduced type, and the argument is an lvalue of type. 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. Allowing non-const references to bind to r-values leads to extremely confusing code. (Binding to a const reference is allowed. (Binding to a const reference is allowed. And const is a constraint imposed by the compiler to the variable that is declared as const. 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. The answer to the question in the title is: yes, the copy-constructor can have a non-const argument. and if you pass it to a function that takes a reference to a non-const - it means that function can change the value. 19 tricky. cpp struct S { }; void f(S&) { } S g() { return S {}; } int main() { S& s = g (); // warning C4239 at /W4 const S& cs = g (); // okay, bound to const ref f (g ()); // Extension: error. Actually for simple types you should prefer to pass by value instead, and let the optimizer worry about providing the best implementation. Binding a reference is always inexpensive,. //. The reference returned from get_value is bound to x which is an l-value, and that's allowed. Expression like a+b will return some constant. Actually the precise reason it doesn't work is not because temporaries cannot be bound to non-const lvalue references, but rather that the initializer of a non-const lvalue reference is subject to certain requirements that char const[N] cannot meet in this case, [dcl. 2) persists until the completion of the full-expression containing the call. That is to say, usage of a reference is syntactically identical to usage of the referent. Reload to refresh your session. obj & a1 = bar(); invalid initialization of non-const reference of type ‘obj&’ from an rvalue of type ‘obj’ using g++. Or, passing it by const reference will also work, since a const lvalue reference can be. (5. " Rule 2, "A non-const reference shall not be bount to a bit-field". In 9. 3. . 1. Their very nature implies that the object is transient. g. Of course, unlike the one in the range-based for loop, this i reference become dangling immediately. If t returns by lvalue reference, the code does not compile because a rvalue reference cannot bind to it. 11. Universal reference is not an actual thing, it just means that we the parameter can have either an lvalue reference and rvalue reference type depending on template instantiation (which depends on the supplied argument at the call site). 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. An lvalue reference is declared using the & operator, for example int& . The foo () function accepts a non-const lvalue reference as an argument, which implies one can modify (read/write) the supplied parameter. C / C++. You have two options, depending on your intention. Writing it gives you the chance to do it wrong (which you already did by. By default, or if /Zc:referenceBinding- is specified, the compiler allows such expressions as a Microsoft extension, but a level 4 warning is issued. yet you can still change the data x by modifying x. i. I agree with the commenter 康桓瑋 that remove_rvalue_reference is a good name for this. I could even (though this is a bit unusual) safely const_cast away the constness of b, since I also hold a non-const reference to the same object. Then you should not have used a forwarding reference. e. 흔히 rvalue reference와 구별하기 위해 기존의 reference를 lvalue reference라고 부릅니다. If the initializer expression. Return by value. Specifically, a const rvalue will prefer to bind to the const rvalue reference rather than the const lvalue reference. So, despite your extra const in your reference type the language still requires it to be bound directly to i. Temporary objects cannot be bound to non-const references; they can only. You can pass lvalues to functions taking rvalues as arguments (tested using a C++ editor). The question about a potential possibility to change a temporary object using a non-const reference. – Joseph Mansfield. 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 ). You can also simplify the return expression, and make the method const, since comparing two objects should not change either of them: bool String::operator< (const String & obj) const { return strcmp (*this, obj) < 0; } although I am not sure strcmp can deal with two. So, when you type const int& ref = 40. rvalue references are marked with two ampersands (&&). Mar 22, 2013 at 18:39. But an rvalue reference can't bind to an lvalue because, as we've said, an rvalue reference refers to a value whose contents it's assumed we don't need to preserve (say, the parameter for a move constructor). an lvalue, this constructor cannot be used, so the compiler is forced to use. You are returning a reference to a local variable. However, the result of that conversion is an rvalue - it is a temporary object. const auto& refInstance = m_map. : if at least one operand is of class type and has a conversion-to-reference operator, the result may be an lvalue designating the object designated by the return value of that operator; and if the designated object is actually a temporary, a dangling reference may result. 25th May 2022, 8:44 AM. is an xvalue, class prvalue, array prvalue or function lvalue and "cv1 T1" is reference-compatible with "cv2 T2", or. error: cannot bind non-const lvalue reference of type 'int&' to an rvalue of type 'int' return std::tie(a. Hot Network Questions Identifying traffic signals for colour blind peopleBut thinking further about it, I think it might be OK :-) Imagine there were three consts (not just two) in const Array &operator=( const Array & ) const; The last const is unacceptable, as it can't even modify itself. C++ : Non-const reference may only be bound to an lvalueTo Access My Live Chat Page, On Google, Search for "hows tech developer connect"As promised, I have a. That is special syntax for a so-called forwarding reference. thanks in advance, George. Rule 3, "Note: if the initializer for a reference of type const T& is. So if the function binds to a rvalue reference, what is seen at the end by the compiler for a certain type T is: std::is_rvalue_reference<T>::value. Note that there is one exception: there can be lvalue const reference binding to an rvalue. rvalues can be residing on read-only memory spaces where changing them might not be allowable and hence the compiler prohibits them. 5. By using the const keyword when declaring an lvalue reference, we tell an lvalue reference to treat the object it is referential when const. Some compilers allow int &r = 5; as an extension, so it makes sense, it's just not allowed by the standard. To produce an xvalue, i. 2 Copy/move constructors [class. 2), an xvalue if T is an rvalue reference to object type, and a prvalue otherwise. In other words, in your first example the types actually do match. reference (such as the B& parameter in the B::B (B&) constructor) can only. I have fixed these issues and completely understand how/why it gives a warning. 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. Since there are some non-modifiable lvalues (so we do not always need to modify values through its reference). It looks like well formed code with defined behavior to me. 5The Lvalue refers to a modifiable object in c++ that can be either left or right side of the assignment operator. 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). thanks in advance, George. All rvalues are non-const. In the second case, fun () returns a non-const lvalue reference, which can bind to another non-const reference, of course. –The pointer returned by the function cannot be bound to a reference. 12. 2) x is a variable of non-reference type that is usable in constant expressions and has no mutable subobjects, and E is an element of the set of potential results of an expression of non-volatile-qualified non-class type to which the lvalue-to-rvalue conversion is applied, or. move simply returns an rvalue reference to its argument, equivalent to. This program outputs: value = 5 value = 5. The solution depends on the value of return type in cleverConfig. The advantage of rvalue references over lvalue references is that with rvalue references you know that the object referred to is an rvalue. Although the standard formulates it in other words (C++17 standard draft [dcl. That's not it. Non-const reference may only be bound to an lvalue. 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. Without the function, you are essentially writing: int x = 10; // x is an l-value int &get_x = x; // just a variable instead of a function get_x = 20; // assignment is ok By float&, he means he wants to take a reference to a float. There are better ways to solve your problems. Value categories are applied to expressions, not objects. x, a. C4239 は、以下。. warning C4239: nonstandard extension used: 'default argument': conversion from 'std::shared_ptr' to 'std::shared_ptr &'. could be an AI. "A reference to type 'cv1 T1' is initialized" refers to the variable that is being initialized, not to the expression in its initializer. So the temporary value_type () will be bound to val and will persist for the duration of the constructor. Non-compliant compilers might allow a non-const or volatile lvalue reference to be bound to an rvalue. Some similar case give me the reason: The C++ standard does not allow the binding of an anonymous temporary to a reference, although some compilers allow it as an extension. And the this pointer is a const pointer, so the instance cannot be changed. The reference returned from get_value is bound to x which is an l-value, and that's allowed. 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. rvalue Reference Cannot Bind to a Named lvalue. A function parameter such as T&& t is known as a forwarding reference. for an lvalue &) and one that is not qualified, the rules are such that they are effectively both qualified and hence ambiguous. ;, x is an lvalue denoting a float whose value is the result of converting a to double and back. I believe the relevant Standard paragraph is 8. An expression that designates a bit field (e. But in your case the operands are different category (123 is a prvalue, a is an lvalue). e. Both const and non-const reference can be binded to a lvalue. 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. CheckCollision (0. In the above code, getStr(); on line 12 is an rvalue since it returns a temporary object as well as the right-hand side of the expression on line 13. For example, when passing things by value, or else with things like A a; B b = a;. )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. A temporary is a prvalue whilst a reference is a lvalue. 3/5:. It's unclear what you mean by "has". C++: Variable that is passed by const referance changes value. A non-const reference may only be bound to an lvalue? I am debugging MSDN code from, (VS. We can't bind rvalue reference to an lvalue also. std::is_rvalue_reference<T&&>::valueA temporary can only bind to a reference to a prvalue. If you want to capture the reference you need to declare a reference. What I have seen however is that you can bind an rvalue to an rvalue reference and since a named rvalue reference is inherently an lvalue, you can bind it to an lvalue reference. This rule covers not only cases such as. C++ only allows non-const binding of an lvalue to a non-const lvalue reference. @YueZhou Function lvalues may be bound to rvalue references. 1. struct Foo{}; { const auto & r = Foo{}; // Foo object not destroyed at semicolon. Of course since methods can be called on rvalue (and thus prvalue) and those methods may return a reference to the objects they were called on we can easily bypass the silly (1) a reference is only allowed to bind to a lvalue restriction. bind to an lvalue. My understanding is that this is largely to avoid breaking several enormous legacy codebases that rely on this "extension. void foo(int& x)) and then complaining that you can't call foo(5). Const reference can be bounded to. 2. Such one reference is called an lvalue reference to a constant true (sometimes called a reference to konst or a const. Change the declaration of the function GetStart like: Point & GetStart(); Also at least the function GetEnd should be changed like: Point & GetEnd(); You could overload the functions for constant and non-constant objects:It looks like we are actually able to bind temporary object to non-const reference, but only if this object. 5. And until now we've only touched what already used to happen in C++98. obj in f is an lvalue expression, and will therefore be treated as such. 3/5, [dcl. Note that for const auto& foo, const is qualified on the auto part, i. 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. 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. An rvalue reference can only bind to an rvalue, which is a candidate for moving. an lvalue that refers to. Notes: A non-const or volatile lvalue reference cannot be bound to anrvalue of a built-in type. It got me quite curious. Sounds like you actually want getPlayer to return a reference too and then to. This means the following is illegal: int main() { const int x { 5 }; int& ref { x }; return 0; } This sample shows the Microsoft extension that allows a temporary of a user-defined type to be bound to a non-const lvalue reference. A temporary has a type, that type can be const, and it can be non-const. The default is -qlanglvl. Since the constructor in your example only takes lvalues, you can only pass lvalues into the factory function. There is no need for references. i have a player class in which i have a function to return a SDL_Rect for the dimensions and the position of my player object: SDL_Rect Player::pos () { return SDL_Rect { mPosX, mPosY, PLAYER_WIDTH, PLAYER_HEIGHT }; } i get the error: "initial value of. 4 — Lvalue references to const. int&& x = 10; is a declaration and not an expression. 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. Viewed 3k times. A reference (of any kind) is just an alias for the referenced object. 5) rvalues can be passed to the parameter. 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. You cannot do that with a non-member function that accepts an lvalue reference. That should be a T. a. Share. Non-const reference may only be bound to an lvalue. Reload to refresh your session. It's just that type of that lvalue is "rvalue reference to Key ". The problem is that auto and decltype side-step the whole public/private thing, allowing you to create types that you. That's an exception to the general rule that it is impossible for lvalues to be bound to rvalue. In the following copy-initialization contexts, a move. You obviously can't point to a temporary. cannot bind non-const lvalue reference of type to an rvalue of type 0 Implementation of the decorator class in C++ using a member reference to the decorated object not working as expected 12. So if this is in the type Object:So we have a reference being initialized by an xvalue of type const foo. Don't pass int&, it can't be bound to a constant or temporary because those can't be modified - use const int& instead. C++ prohibits passing a temporary object as a non-const reference parameter. It cannot be done with lvalue references to non-const since they cannot be bound to rvalues. Remember Me? Forum; FAQ; Calendar; Forum Actions. decltype(fun()) b=1; Then, your code initializes a const reference with a prvalue of a different (non-reference-related) type. warning C4239: nonstandard extension used: 'initializing': conversion from 'A' to 'A &' note: A non-const reference may only be bound to an lvalue warning C4239: nonstandard extension used: 'initializing': conversion from 'A' to 'A &' note: A non-const reference may only be bound to an lvalue 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. operator[] is - either change the return type of the function from Value* to const Value&, or return *cleverconfig[name];The C++ Standard (2003) indicates that an rvalue can only be bound to a const non-volatile lvalue reference. Without rvalue expression, we could do only one of the copy assignment/constructor and move assignment/constructor. Since you cannot access the result of that side-effect if you are passing a temporary, then I would conclude that you're very likely doing something wrong. 1. Actor actor = get_actor_ref_from_ped (PLAYER::PLAYER_PED_ID ()); Is going to make a copy of the value returned from the function as it calls the copy constructor. It's the specific case where changing T& to const T& does more than just ban modifications. For example inc(1). Sorted by: 6. There is no implicit conversion as suggested in the title, the reference binds directly to the. e. C++ initial value of reference to non-const must be an lvalue and I'm sure I have done everything right. png", 560, 120); int x2 = 560 + 54; int x1 = 560; int y1 = 120; int y2 = 291 + 120; const int * xSolv2 = &x2. 1. For example, the argument might be a reference to a node of a linked list, and within the function you may want to traverse the list, so you will want to be doing node = * (node. Only local const references prolong the lifespan. cpp(10): warning C4239: nonstandard extension used : 'argument' : conversion from '<type1>' to '<type2>' 1> A non-const reference may only be bound to an lvalue" only on warning level /W4 or above. a. The reference is. But that doesn't make sense. The option -qlanglvl=compatrvaluebinding instructs the compiler to allow a non-const or volatile lvalue reference to bind to an. Calling a non-static member function of class X on an object that is not of type X, or of a type derived from X invokes undefined behavior. 2: the reference shall be an lvalue reference to a non-volatile const type (i. Otherwise, the reference you get behaves more. (Case 1 in the below program). A temporary can only bind to const lvalue references, or rvalue references. std::vector<bool> is special from all other std::vector specializations. Improve this answer. -hg. There are two overloads. To reduce template instantiation overhead, I would recommend a more direct implementation:will result in output: Constructor called 42. There's no reason to make it a reference. And since that the converted initializer is an xvalue not prvalue, [conv. 3 Answers. , int and const int are similar, int* const ** volatile and volatile int** const * are similar, and crucially int* and. 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. 124 Non const lvalue references. VS2008 is not too bad as at least it gives a compile warning: warning C4239: nonstandard extension used : 'initializing' : conversion from std::string to std::string & A non-const reference may only be bound to an lvalue A non-const reference may only be bound to an lvalue. "non-const lvalue reference to type 'QByteArray' cannot bind to a temporary of type 'QByteArray'". v = this->v*a. 1. 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. – Vlad from Moscow. C4239: nonstandard extension used : 'default argument' : conversion from 'QMap<QString,QVariant>' to 'QVariantMap &' A non-const reference may only be bound to an lvalue. Saturday, December 15, 2007 4:49 AM. 2/5 in N4140): A temporary bound to a reference parameter in a function call (5. rvalues are defined by exclusion, by saying that every expression is. Some older compilers couldn't support the latter in proper way. A rvalue can be used as const T&, however, the compiler complains about binding a non-const lvalue to a rvalue. const int & is a const lvalue reference. 3. Now an lvalue reference is a reference that binds to an lvalue. The first variant returns a reference to the actual value associated with the key test, whereas the second one returns a reference to the map element, which is a pair<const key_type, mapped_type>, i. Furthermore, we don't know if somefunc2 modifies the referenced byte, and if it does then we don't know what should happen to the other byte. 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. 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). If you compile with the /Wall flag, you will be given the answer by the compiler itself:. It can appear only on the right-hand side of the assignment operator. 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. Sometimes even for the original developer, but definitely for future maintainers. In this case, returning a non-const lvalue reference compiles because x is an lvalue (just one whose lifetime is about to end). , temporary) double but a temporary cannot be bound to a non-const reference. The language forbids that sort of binding for various reasons. GetImage (iTileId, pulImageSize, a_pImage ); With the method defined as: This change is required by the C++ standard which specifies that a non-const. Expect the programmer to take extra care to modify values only via those references which do not refer to literals, and invoke undefined behaviour if you get it wrong. warning C4239: nonstandard extension used : 'initializing' : conversion from 'foo' to 'foo &' A non-const reference may only be bound to an lvalue (note that this remains illegal in C++11) Last edited on. If t returns by rvalue reference, you obtain a reference to whatever was returned. The standard specifies such behavior in §8. 1 1 1. 1 Answer. — Otherwise, the reference shall be an lvalue reference to a non-volatile const type (i. it doesn't say anything else. In the second case, fun() returns a non-const lvalue reference, which can bind to another non-const reference, of course. If t returns a local variable, then you get a dangling reference, since that variable is gone after the call. next);. Alex September 11, 2023. However, I am.