a non-const reference may only be bound to an lvalue. , you may only want to hold on to a const Bar*, in which case you then can also only pass a const Bar*) Using a const Bar& as parameter type is bound to result in a runtime crash sooner rather than later because:The C++ Standard (2003) indicates that an rvalue can only be bound to a const non-volatile lvalue reference. a non-const reference may only be bound to an lvalue

 
, you may only want to hold on to a const Bar*, in which case you then can also only pass a const Bar*) Using a const Bar& as parameter type is bound to result in a runtime crash sooner rather than later because:The C++ Standard (2003) indicates that an rvalue can only be bound to a const non-volatile lvalue referencea non-const reference may only be bound to an lvalue <samp> The temporary unsigned int could be bound to lvalue-reference to const (i</samp>

However sometimes it is desired to ensure that you can only pass lvalues to a function (this is the case for std::ref for one). ) But there is no way to show me how to solve it;You may modify a non-const object through a non-const reference. 3/5:. Share. In this context, binding an rvalue to the non-const reference behaves the same as if you were binding it to a const reference. So naming kInt is not deemed an odr-use as long as it. for example, to get a reference to the element. The most likely explanation is that the programmer meant to pass by const reference and just forgot the const. 4. ). The number of identifiers must equal the number of non-static data members. For lvalue references, T is deduced to be an lvalue reference, and for rvalue references, T is deduced to be a non-reference. the first version essentially returns second of said pair directly. std::vector<bool> does not return a bool&, but nevertheless this is completely fine: std::vector<bool> x{0,0,0}; x. only call const members of the object, you can not implicitly convert it to non-const, and you cannot perform non-const operations on its members. Ask Question Asked 8 years, 10 months ago. Temporary objects cannot be bound to non-const references; they can only. A reference to type “cv1 T1” is initialized by an expression of type. (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). Visual C++ is non-compliant with the standard in allowing binding of temporaries to non-const lvalue references. // zcreferencebinding. next);. That's my best guess anyway. 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. error: invalid initialization of non-const reference of type 'int&' from an rvalue of type 'int' GCC complains about the reference not being const, namely a constant. temporary] ( §12. The parameter list for a move constructor, however, consists of an rvalue reference, like B&& x. , cv1 shall be const), or the reference shall be an rvalue reference. The option -qlanglvl=compatrvaluebinding instructs the compiler to allow a. 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, an rvalue can be bound to a. Fibonacci Series in C++. – Kerrek SB. Both const and non-const reference can be binded to a lvalue. So in your case, you need to rewrite your. an lvalue, this constructor cannot be used, so the compiler is forced to use. A simple solution is: void foo (MyObject obj) { globalVec. If an rvalue is passed to factory, then an rvalue will be passed to T's constructor with the help of the forward function. reference (such as the B& parameter in the B::B (B&) constructor) can only. Consulting the cppreference documentation for <type_traits>, it appears that there is not such a tool in the standard library. The conversion produces an rvalue (i. Thank you for answering. Overload between rvalue reference and const lvalue reference in template. Constant lvalue references can be bound to all types of values, including non-constant lvalues, constant lvalues. I dont know if its bug in compiler or is it intended. The standard specifies such behavior in §8. In summary, after const float & x = true ? a : 2. The call would bind to f(int&&). (I'll comment on all the answers. Non-compliant compilers might allow a non-const or volatile lvalue reference to be bound to an rvalue. Since C++11, two kinds of references have existed - lvalue and rvalue references. Regarding the second question. Without rvalue expression, we could do only one of the copy assignment/constructor and move assignment/constructor. e. The temporary unsigned int could be bound to lvalue-reference to const (i. 1. If t returns by rvalue reference, you obtain a reference to whatever was returned. A const reference prolongs a lifetime of a temporary object bound to it, so it is destroyed only when the reference goes out of scope. . 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. Thanks. 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. 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. 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. T and U) are never reference types. The Standard says no. int & a=fun (); does not work because a is a non-const reference and fun () is an rvalue expression. g. 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. The simplest fix is to simply store the temporary object somewhere, first: Collider c=player. initial value of reference to non-const must be an lvalue. This sample shows the Microsoft extension that allows a temporary of a user-defined type to be bound to a non-const lvalue reference. (PS the lifetime of the temporary is extended to the lifetime of the reference. 2. The page is trying to say that you can write m. As a reader pointed out, if g() returned const int instead of const T, the output would be different. ; T is not reference-related to U. After some investigation and help from the community, here is the answer:. U is a class type. From the C++20 draft. Apr 14 at 22:55. If U is t’s underlying non-reference type (namely std::remove_reference_t<decltype(t)>), then T. 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. You can disable this behaviour with the /Za (disable language extensions) compiler switch under. find (key);A pointer to non-const is convertible to pointer to const however. A variable is an lvalue, so you are allowed to bind a non const reference to it. s. 124 Non const lvalue references. static_cast<typename remove_reference<T>::type&&> (t) The result of the function call is an rvalue (specifically, an xvalue ), so it can be bound to an rvalue reference where the function argument couldn't. Viewed 3k times. Why can't I bind an Rvalue to a non-const Lvalue reference? C++ does not allow binding Rvalues to non-const Lvalue references because Lvalue references can modify the object they are bound to, and Rvalues. 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. Add a comment. The reference returned from get_value is bound to x which is an l-value, and that's allowed. reference (such as the B& parameter in the B::B (B&) constructor) can only. Overload resolution is usually done in terms of a strict. 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 ). However, C++ makes one exception to this rule and allows const lvalue references to also bind to rvalues. 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. Reference is always constant, you can't change reference. To be standards compliant, you need. The rules were already more complex than "if it has a name it's an lvalue", since you have to consider the references. 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. add (std::move (ct)); } A forwarding reference can bind to both lvalues and rvalues, but. an lvalue, this constructor cannot be used, so the compiler is forced to use. The reference is. Saturday, December 15, 2007 4:49 AM. (Case 1 in the below program). for an lvalue &) and one that is not qualified, the rules are such that they are effectively both qualified and hence ambiguous. A reference is only allowed to bind to a lvalue. Sometimes even for the original developer, but definitely for future maintainers. 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. However, you might need at that returns non-const reference too. Both of g and h are legal and the reference binds directly. Consider also this: the language has no way of knowing that the lvalue reference returned by the iterator's operator * or the vector's operator[] refers to something whose lifetime is bound to that of. A const lvalue reference can be initialized from a bit-field. For example, a const lvalue reference should bind to both lvalue and rvalue arguments, and a non-const lvalue reference should bind to a non-const lvalue, but refuse to bind to rvalues and const lvalues. Given all three functions, this call is ambiguous. 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. 80). 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. g. r-value references are designed to be the subject of a move-constructor or move-assignment. This example is very similar to the previous one, except the temporary object is non-const this time. The non-const reference is converted into a const reference when the print function calls getConstReference. 6 — Pass by const lvalue reference. 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. 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. 1 invalid initialization of non-const reference of type from an rvalue of type. it doesn't say anything else. (2) (since C++11) 1) Lvalue reference declarator: the declaration S& D; declares D as an lvalue reference to the type determined by decl-specifier-seq S. 4 — Lvalue references to const. When I discovered this, it seemed odd to me, so I tried. MSVC has an "extension that allows that. There are exceptions, however. The binding rules for rvalue references now work differently in one. int &a = 5; // error: lvalue cannot be bound to rvalue 5 However, we can bind an rvalue to a const lvalue reference (const reference): const int &a = 5; // Valid In this case, the compiler. Now it makes actually sense to take its address, as it is an lvalue for all intents and purposes. GetImage (iTileId, pulImageSize, a_pImage ); With the method defined as:This change is required by the C++ standard which specifies that a non-const. Improve this question. I've encountered a very weird warning that, although compiles fine on windows, fails to compile for Symbian through CodeWarrior. A simple definition. int* and void* are different types; you can't bind a int* to reference to void* directly. So long as the reference is initially bound to an l-value, everything is fine (so long as you don't use a reference to a stack local variable, of course). 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. The Rvalue refers to a value stored at an address in the memory. What std::string::c_str returns is an rvalue, which can't be bound to an lvalue-reference to non-const (i. 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. C4239: nonstandard extension used : 'default argument' : conversion from 'QMap<QString,QVariant>' to 'QVariantMap &' A non-const reference may only be bound to an lvalue. 1. However, there is a canonical mapping from the. 1 Answer. 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. 4. (5. However, now you've got a temporary A, and that cannot bind to a, which is a non-const lvalue reference. I have fixed these issues and completely understand how/why it gives a warning. It expects an lvalue reference parameter. A const reference could be bound to rvalue, and for this case, a temporary int will be created and initialized from 255. I believe the relevant Standard paragraph is 8. 19 tricky. " I really need some further explanations to solving this: Non-const references cannot bind to rvalues, it's as simple as that. Passing by reference, by a const reference wouldn't cost more than passing by value, especially for templates. 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. const A& x = 1; //compile x = 2; //error! A&& xxx = 1; //compile A& xx = 1; //does not compile. struct S {}; f<S {}> (); // ok. 3. However, getPlayer is returning a copy of that pointer. –And I want to make sure template parameter F&& f only accept a non-const lvalue reference. const int *p; - here it is pointer on const int int const *p; - here it is const pointer on int const int const *p; -. e. and not. So you cannot change the data of x with reference variable r (just acts a read only). Const reference can be bounded to. If you compile with the /Wall flag, you will be given the answer by the compiler itself:. See universal. Return by value. You are returning a reference to a local variable. In the original example , both are xvalues so the ternary operator evaluates to an xvalue. 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. The ability you're talking about is exactly why forwarding references exist: so that the copy/move aspects. 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. Share. Only const lvalue references (and rvalue references) may be bound to an object accessed through an rvalue expression. In the following codes, I have two versions of class A instantiated, one is bound to int and the other to int&. 255 (i. I have to think for a while-_-!. In the second case, fun () returns a non-const lvalue reference, which can bind to another non-const reference, of course. You obviously can't point to a temporary. Accept all cookies Necessary cookies only Customize settings. int const&x = 42; // It's ok. In the above program, because value parameter y is a copy of x, when we increment y, this only affects y. Notably, types of expressions (i. 2nd that, nullptr is the best way to declare the optional parameter. 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& //. lvalue references are marked with one ampersand (&). " 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. You are returning a reference to a local variable. ii. The non-const subscript operator returns a non-const reference, which is your way of telling your callers (and the compiler) that your callers are allowed to modify the Fred object. (Binding to a const reference is allowed. A temporary can only bind to const lvalue references, or rvalue references. Non-const reference may only be bound to an lvalue. The second const is good, as is stops the source item being modified. Calling operator + second time won't be possible because a temporary object can not be passed as reference to a non-const-qualified object. – You may not bind a temporary object with a non-constant lvalue reference. And this is precisely what the compiler is telling you:. – n. To produce an xvalue, i. " I really need some further explanations to solving this: #include "graph1. ref]/5:. rvalue Reference Cannot Bind to a Named lvalue. But a more proper fix is to change the parameter to a const reference:However, you might need at that returns non-const reference too. 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. The type of such a reference must be a const qualified lvalue reference or a rvalue references. However,. Saturday, December 15, 2007 4:49 AM. It never makes sense to return a dangling reference, but it's syntactically legal. So your reference would be referring to the copy of the pointer which wouldn't be modified if you change the Player object. Const reference can be bounded to. -hg. I recently filed a bug against MSVC which relates to this, where the non-standard behavior caused standard-compliant code to fail to compile and/or compile with a deviant behavior. It can appear only on the right-hand side of the assignment operator. rvalues can be residing on read-only memory spaces where changing them might not be allowable and hence the compiler prohibits them. Mark Forums Read; Quick Links. — Otherwise, the reference shall be an lvalue reference to a non-volatile const type (i. 2 Copy/move constructors [class. In such cases: [1] First, implicit type conversion to T is applied if necessary. The compiler preventing this is a way of catching these kinds of errors. It's just that non-const lvalue references can't bind to rvalues, so the can never be used that way. There is no need for references. According to the language specifications, you are allowed to bind a const lvalue to an rvalue. In this case, returning a non-const lvalue reference compiles because x is an lvalue (just one whose lifetime is about to end). A temporary is a prvalue whilst a reference is a lvalue. non-const lvalue reference to type 'int' cannot bind to a. What you want is in 40two's answer, but make sure to forward the parameter t. e. 3. C++ initial value of reference to non-const must be an lvalue and I'm sure I have done everything right. 3 Answers. The rules about reference binding are that a non-const lvalue reference may only bind to an lvalue expression. int & a=fun(); does not work because a is a non-const reference and fun() is an rvalue expression. 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. Case 3: binding to data members. Non-const reference may only be bound to an lvalue. Non-explicit constructors have their uses. thus, this is legal: string&& s = foo (); // extends lifetime as before s += "bar"; baz (std::move (s)); // move the temporary into the baz function. a. An rvalue reference can only bind to an rvalue, which is a candidate for moving. 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. obj in f is an lvalue expression, and will therefore be treated as such. Note also that if you simply use CList<DATA>, the second template argument ARG_TYPE is correctly deduced to be const DATA& by default, as per CList template declaration (TYPE = DATA, ARG_TYPE = const DATA&): template<class TYPE, class ARG_TYPE = const TYPE&> class CList : public CObjectT& data; There's your problem. According to the reference collapsing rules, "rvalue reference to rvalue reference collapses to rvalue reference, all other combinations form lvalue reference". " Rule 2, "A non-const reference shall not be bount to a bit-field". That is to say, usage of a reference is syntactically identical to usage of the referent. Hence, C++ does not permit a non-const reference to a const variable. 5. template <auto N> void f () { auto & n = N; } This works when f is instantiated over class types. C++: rvalue reference converted to non-const lvalue-reference. Non-const lvalue reference to type 'Common::XYZCallbackInterface' cannot bind to a temporary of type 'Common::XYZCallbackInterface *'. It doesn't really matter. Non-const reference may only be bound to an lvalue. 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. Your declaration of a is a non-const lvalue reference,. match. However, this is deceptive, because it may or may not be an rvalue reference depending on the type of T. There are exceptions, however. const char*&). A non-const reference must be bound to lvalue (i. You signed out in another tab or window. The rules were already more complex than "if it has a name it's an lvalue", since you have to consider the references. rvalue reference 는 rvalue (즉, 상수와 임시객체)도 참조가 가능 하다 점을 빼고는 기존의 참조와 동일합니다. e. Hot Network QuestionsNon-const references cannot bind to rvalues, it's as simple as that. Now an lvalue reference is a reference that binds to an lvalue. If you used a reference to const, it would extend the lifetime of the temporary result of the implicit conversion: const int * const &j = i;The iterator object itself refers to an element of the container. There is no implicit conversion as suggested in the title, the reference binds directly to the. , cv1 shall be const), or the reference shall be an rvalue reference. e. g. e. 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. You normally point to some spot in memory where you stored a value of interest. 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. 2 Answers. – Vlad from Moscow. it is explained that an lvalue is when you can take its address. Expression like a+b will return some constant. The temporary int's lifetime will be the same as the const reference. The reference returned from get_value is bound to x which is an l-value, and that's allowed. In the second case, fun() returns a non-const lvalue reference, which can bind to another non-const reference, of course. could be an AI. 2. rvalue reference versus non-const lvalue. 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. Unfortunately, they may compile with one common compiler, due to language. A non-const reference may only be bound to an lvalue[/quote] 1 Reply Last reply Reply Quote 0. 3. This means the following. ) Aside from the workaround you already have, if you can change the function to take const QImage& then that would be better. e. E may not have an anonymous union member. You switched accounts on another tab or window. Non-const lvalue reference to type '_wrap_iter' cannot bind to a value of unrelated type '_wrap_iter' c++;. e. . This constness can be cast away with a const_cast<>. 5). const unsigned int&), (and its lifetime is extended to the lifetime of the reference,) but can't be bound to lvalue-reference to non-const (i. 4. test (const std::string& a): a is const lvalue reference and like before I have lvalue and rvalue. The make_range function doesn't use that constructor. The option -qlanglvl=compatrvaluebinding instructs the compiler to allow a non-const or volatile lvalue reference to bind to an. To handle other value categories, one may use std::forward_as_tuple:. You have two options, depending on your intention. h"` displayPNG("solve. 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. 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 expected12. Thus, in case of your variable b: T = int ==> T&& becomes int&& T = int& ==> T&& becomes int. The basic idea behind references is that lvalue references bind to lvalues, and rvalue references bind to rvalues; the reference thus bound henceforth refers to the value it was bound to. at(0) = false; The reaons is that x. Suppose r is an rvalue reference or non-volatile const lvalue reference to type T, and r is to be initialized by an expression e of type U. In the case of built-in types, the result is a prvalue, so a temporary (of type const int) is always created from this prvalue and bound to x. 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. C / C++. A temporary or an rvalue cannot be changed with a reference to non-const. reference to type 'myclass' could not bind to an rvalue of type 'myclass *'. 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. Naturally, the same treatment also applies to constructors. I don't get why the make_range function doesn't work unless I remove the View (View<C>& r) constructor. However, since a reference acts identically to the object being referenced, when using pass by reference, any changes made to the reference parameter will affect the argument: #include <iostream. In the previous lesson ( 12. The following example shows the function g, which is overloaded to take an lvalue reference and an rvalue. -1. 0. at returns a proxy (of type std::vector<bool>::reference) that allows you to write the element. Improve this question. r can be bound to the conversion result of e or a base class of e if the following conditions are satisfied. However, lvalue references to const forbid any change to the object and thus you may bind them to an rvalue. a copy would be needed). g. There are two aspects to the const in C++: logical constness: When you create a variable and point a const pointer or reference to it, the compiler simply checks that you don't modify the variable via the const pointer or reference, directly or indirectly. In the previous lesson ( 12. void checkMe (shared_ptr<string>& param = shared_ptr<string> ()); This MSDN article says it is a /W4 warning. 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. 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. Follow edited Nov 15, 2016 at. My understanding is that this is largely to avoid breaking several enormous legacy codebases that rely on this "extension. The only difference (that I see) is that x2 knows it only has 3 rows, whereas x1 has a dynamic number of rows. 21. 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. Since the constructor in your example only takes lvalues, you can only pass lvalues into the factory function. e. Actually for simple types you should prefer to. reference (such as the B& parameter in the B::B (B&) constructor) can only. Thus, the standard allows all types. m. 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,. Actually the Standard say so: 8. 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. And the lvalue-reference to const could bind to. Rvalues (including xvalues) can be bound to const lvalue references so that you can pass a temporary to a function with such a parameter:With pointers, you can mostly correctly use const and non const versions, whatever is more appropriate (i. 0f, c); The other similar calls need to be fixed too. 71. In 9. This may sound like a silly question, but I was confused about this following behaviour:. May 4, 2013 at 16:38. 2. Of course, unlike the one in the range-based for loop, this i reference become dangling immediately. In other words, in your first example the types actually do match. I agree with the commenter 康桓瑋 that remove_rvalue_reference is a good name for this. e. Ok, so, I already know that returning a local variable as reference will cause undefined behavior when we try to use it and that we can create a non-const reference to only form a lvalue variable. C++ prohibits passing a temporary object as a non-const reference parameter. Confusion between rvalue references and const lvalue references as parameter. Therefore, if we make a reference parameter const, then it will be able to bind to any type of argument:I suppose I'd think of it along the lines of, in C++: If I have a mutable lvalue reference a and const lvalue reference b to the same object, I can always mutate b by mutating a. However, when you use a const reference to a non-const object, you are asking the compiler to not let you modify the object through that particular. e. i. decltype (fun ()) b=1;Syntax: void foo (std::string& str); // non-constant lvalue reference overload. It can take rvalues because it is marked const and rvalues are allowed to bind to const lvalue references. Because as_const doesn't take the argument as const reference. 3. This rule covers not only cases such as.