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. So the parameter list for a copy constructor consists of an const lvalue reference, like const B& x . png", 560, 120); int x2 = 560 + 54; int x1 = 560; int y1 = 120; int y2 = 291 + 120; const int * xSolv2 = &x2. This program outputs: value = 5 value = 5. Technically, auto is the root of the problem. However, I am. But an rvalue can only be bound to a const reference. You would only need to create such a wrapper if you needed to do things with it that you can't do with C++ objects, such as storing it in an NSArray or. A non-const reference may only be bound to an lvalue[/quote] 1 Reply Last reply Reply Quote 0. error: cannot bind non-const lvalue reference of type 'int&' to an rvalue of type 'int' return std::tie(a. EX: int &var=4; we can change value of reference , but logically it is not possible to change 4. has a class type. 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. s. Consider a function template f that binds a non-const lvalue reference to a deduced non-type template parameter. 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. Sometimes even for the original developer, but definitely for future maintainers. temporary] ( §12. Their very nature implies that the object is transient. C++ only allows non-const binding of an lvalue to a non-const lvalue reference. 5. not an rvalue reference, everything under the sun can be bound by a forwarding reference – Piotr Skotnicki. reference to type 'myclass' could not bind to an rvalue of type 'myclass *'. What "r-value reference for this` does is allow you to add another alternative: void RValueFunc () &&; This allows you to have a function that can only be called if the user calls it through a proper r-value. We should not mix rvalue and lvalue references. 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. 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). 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. 4. rvalue Reference Cannot Bind to a Named lvalue. The problem is that auto and decltype side-step the whole public/private thing, allowing you to create types that you. I am still studying what is the reason in essence in compiler why a non-const reference can not be binded to a rvalue. The foo () function accepts a non-const lvalue reference as an argument, which implies one can modify (read/write) the supplied parameter. rvalue reference versus non-const lvalue. ref], the section on initializers of reference declarations. Follow edited Apr 5, 2021 at 12:41. The int* needs to be converted to void* firstly, which is a temporary object and could be bound to rvalue-reference. Apr 13, 2017 at 13:00. The compiler automatically generates a temporary that the reference is bound to. That's my best guess anyway. : 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. 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. if a. 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. 3. 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. // zcreferencebinding. @MichaelKrelin-hacker: Technically not, you cannot (ever) bind a reference to a value (or compile time constant), the standard is quite explicit as to what actually happens: Otherwise, a temporary of type “cv1 T1” is created and initialized from the initializer expression using the rules for a non-reference copy-initialization (8. Hence, values bound to an rvalue reference can be moved from (not. C++0x에는 rvalue reference라는 개념이 추가 됩니다. 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. An lvalue reference (commonly just called a reference since prior to C++11 there was only one type of reference) acts as an alias for an existing lvalue (such as a variable). 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. — Otherwise, the reference shall be an lvalue reference to a non-volatile const type (i. Notably, types of expressions (i. Mark Forums Read; Quick Links. An entity (such as an object or function) that has. I have looked elsewhere on this site and read similar postings about this error: "initial value of reference to a non-const must be lvalue. You have two options, depending on your intention. However, in VS2010 I seem to be able to do so:. Const reference can be bounded to. Lesley Lai has a blog post on this: “The implication. If t were really an out-parameter, it would be passed by pointer: std::string *t. 0. If C++ allowed you to take literals by non-const reference, then it would either: Have to allow literals to change their meaning dynamically, allowing you to make 1 become 2. C++: rvalue reference converted to non-const lvalue-reference. From the C++20 draft. e. an lvalue, this constructor cannot be used, so the compiler is forced to use. i. It doesn't really matter. 12. 5. A non-const reference may only be bound to an lvalue[/quote] 1 Reply Last reply Reply Quote 0. However, now you've got a temporary A, and that cannot bind to a, which is a non-const lvalue reference. Otherwise, the reference shall be an lvalue reference to a non-volatile const type (i. an identifier) that resolves to a non-type non-static member of X or of a base class of X, is transformed to a member access. That works well with normal variables but uint8Vect_t(dataBlock. 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. 2. . and another 7 more if your interested, all saying about the same thing. E may not have an anonymous union member. I recommend checking how standard library deals with this. The whole idea of forwarding is to accept any value category and preserve it for future calls. Both const and non-const reference can be binded to a lvalue. Non-const lvalue reference to type '_wrap_iter' cannot bind to a value of unrelated type '_wrap_iter' c++;. 19 tricky. 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. We can take the address of an lvalue, but not of an rvalue. g. 5. Anything that is capable of returning a constant expression or value. 上記のようなコードを書いたところ、以下の警告が出た。. 3. 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. That should be a T. e. 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. 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). ; T is not reference-related to U. For reference, the sentence that totally misled me is in [over. r-value causes a warning without the use of std::move. int global_x; void foo (int*& ptr) { ptr = &global_x; } void bar () { int local_x; int * local_ptr = &local_x; foo. The code above is also wrong, because it passes t by non-const reference. 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. (I'll comment on all the answers. Any reference will do. In the second case, fun () returns a non-const lvalue reference, which can bind to another non-const reference, of course. Although the standard formulates it in other words (C++17 standard draft [dcl. And until now we've only touched what already used to happen in C++98. decltype(fun()) b=1; Then, your code initializes a const reference with a prvalue of a different (non-reference-related) type. Lvalue references to const can be bound to. See universal. In the previous lesson ( 12. The const subscript operator returns a const-reference, so the compiler will prevent callers from inadvertently mutating/changing the Fred. If t returns a local variable, then you get a dangling reference, since that variable is gone after the call. 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. An rvalue reference can only bind to an rvalue, which is a candidate for moving. int& func() { int x = 0; return x; } compiles, but it returns a reference to a stack variable that no longer exists. Thus, in case of your variable b: T = int ==> T&& becomes int&& T = int& ==> T&& becomes int. The only way to safely bind an rvalue to an lvalue is either by. end()) is a temporary object and cannot be bound to lvalue reference. Actually for simple types you should prefer to pass by value instead, and let the optimizer worry about providing the best implementation. The problem is that a non-const lvalue reference cannot bind to a temporary, which is an rvalue. The rules were already more complex than "if it has a name it's an lvalue", since you have to consider the references. 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. (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). We don't know which byte should be passed. First of all, an argument to such a reference must have static storage duration and linkage, which your variable cannot have both as it is defined in block-scope. Here you are taking a reference to a uint8Vect_t. 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. 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. Confusion between rvalue references and const lvalue references as parameter. You signed in with another tab or window. C++ does not give that feature to non-const references: A function lvalue; If an rvalue reference or a non-volatile const lvalue reference r to type T is to be initialized by the expression e, and T is reference-compatible with U, reference r can be initialized by expression e and bound directly to e or a base class subobject of e unless T is an inaccessible or ambiguous base class of U. If you want to capture the reference you need to declare a reference. Only const lvalue references (and rvalue references) may be bound to an object accessed through an rvalue expression. int x = 1000; const int &r = x; In this case, its a const reference to a non const variable. C++. That's an exception to the general rule that it is impossible for lvalues to be bound to rvalue. 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. 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. What getPtr () return: std::shared_ptr<int> getPtr (int val) { } is an rvalue reference. Basically, VS will allocate the space somewhere and just let the reference point to it, as if it was a reference-to- const without the constness (or in C++11 an rvalue reference). 6. One way to accomplish this is by overloading on the free parameter with both const and non-const lvalue references. A function lvalue; If an rvalue reference or a nonvolatile const lvalue reference r to type T is to be initialized by the expression e, and T is reference-compatible with U, reference r can be initialized by expression e and bound directly to e or a base class subobject of e unless T is an inaccessible or ambiguous base class of U. 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. A non-const lvalue reference can only bind to non-const lvalues. Thus, in the case where P is const T&& (which is not a forwarding reference), it is transformed to const T and whether or not the argument is an lvalue doesn't affect the type deduction, since value. Same thing can be done with lvalue references to const: const int& x = 10. Fibonacci Series in C++. Cannot bind non-const lvalue reference to an rvalue. double && does not work for lvalues. Your conclusion happens to be correct, but it doesn't follow from your premise. " Rule 2, "A non-const reference shall not be bount to a bit-field". long can be promoted to a long long, and then it gets bound to a const reference. Since the temporary B that's returned by source () is not. The problem is that a non-const lvalue reference cannot bind to a temporary, which is an rvalue. 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. Sometimes even for the original developer, but definitely for future maintainers. Some compilers allow int &r = 5; as an extension, so it makes sense, it's just not allowed by the standard. This operator is trying to return an lvalue reference to a temporary created upon returning from the function: mat2& operator /= ( const GLfloat s. By using the const keyword when declaring an lvalue reference, we tell an lvalue reference to treat the object it is referential when const. the first version essentially returns second of said pair directly. 1. 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. What you probably want is: BYTE *pImage = NULL; x. 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. 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. This rule does not reflect some underlying. Assignment to references, on the other hand, is implicit, so if a is of type int& you simply need to write a=b to make a a reference to b. 3. e. Add a comment. Follow edited Oct 5 at. (1) && attr (optional) declarator. Explanation: const lvalue indicates that the callee wants a read-only view of the object and it does not matter what type of object the caller pass as the argument. Non-const reference may only be bound to an lvalue. Saturday, December 15, 2007 4:49 AM. Non-const reference may only be bound to an lvalue. initial value of reference to non-const must be an lvalue. There is no need for references. Non-const reference may only be bound to an lvalue. int & a=fun(); does not work because a is a non-const reference and fun() is an rvalue expression. It's unclear what you mean by "has". However, lvalue references to const forbid any change to the object and thus you may bind them to an rvalue. The C++ standard does not allow the binding of an anonymous temporary to a reference, although some compilers allow it as an extension. int x; int&& r = x; but also. You're not modifying the given pointer, so just pass it by value instead of by reference. If non-const lvalue references were allowed to refer to rvalues, you would never know if the object referred to was. m. So the temporary value_type () will be bound to val and will persist for the duration of the constructor. 1 invalid initialization of non-const reference of type from an rvalue of type. e. C4239: nonstandard extension used : 'default argument' : conversion from 'QMap<QString,QVariant>' to 'QVariantMap &' A non-const reference may only be bound to an lvalue. ("variable" means object or reference). Improve this question. 7. Then the type of foo would be Foo* const &, which is a reference to const (pointer to non-const Foo), but not const Foo* &, which is a reference to non-const (pointer to const Foo). Non-const reference may only be bound to an lvalue. Non-const reference may only be bound to an lvalue. In the following post: Understanding lvalue/rvalue expression vs object type. A function parameter such as T&& t is known as a forwarding reference. A non-const reference may only be bound to an lvalue? I am debugging MSDN code from, (VS. The temporary int's lifetime will be the same as the const reference. It can appear only on the right-hand side of the assignment operator. The rest of the article will elaborate on this definition. Apr 14 at 22:55. Your declaration of a is a non-const lvalue reference,. 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. 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. Naturally, the same treatment also applies to constructors. reference (such as the B& parameter in the B::B (B&) constructor) can only. Unless an object is created in the read-only section of a program, it is open for modifiction without adverse consequences. "You're not "assigning" to a reference, you're binding to a reference. When you pass a pointer by a non- const reference, you are telling the compiler that you are going to modify that. This means the following. 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(). Since the temporary B that's returned by source () is not. A simple definition. 1 Answer. ) Thus the return type is also int&. 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. 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. ii. Alex September 11, 2023. , cv1 shall be const), or the reference shall be an rvalue reference. The core of your question is: can rvalues be bound to non-const lvalue references?. constexpr T& value() &; constexpr const T & value() const &; constexpr T&& value() &&; constexpr const T&& value() const &&; What is the point of returning a const rvalue reference? The only reason I can think of is to enable the compiler to help catch undefined behavior in (really really weird) cases like the followingA non-const reference may only be bound to an lvalue[/quote] Reply Quote 0. The call would bind to f(int&&). Consulting the cppreference documentation for <type_traits>, it appears that there is not such a tool in the standard library. The reference returned from get_value is bound to x which is an l-value, and that's allowed. In the previous lesson ( 12. Properties -> C/C++ -> Language. Otherwise, the reference shall be an lvalue reference to a non-volatile const type (i. The advantage of rvalue references over lvalue references is that with rvalue references you know that the object referred to is an rvalue. In 9. Named variables are lvalues. The solution depends on the value of return type in cleverConfig. 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. ref/6] ). . Value categories pertain to expressions, not objects. Reference is always constant, you can't change reference. std::is_rvalue_reference<T&&>::value A temporary can only bind to a reference to a prvalue. Share. You know, just like any other use of const. The binding rules for rvalue references now work differently in one. r-value:-. 10 is a prvalue expression. unsigned int&). –The pointer returned by the function cannot be bound to a reference. If t returns by rvalue reference, you obtain a reference to whatever was returned. 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. The default is -qlanglvl. Can someone given an example of a "non-const lvalue reference"? I need to pass an object to a routine where the object's state will be modified, after the routine has completed I expect to use the object with the modified state. Some older compilers couldn't support the latter in proper way. [2] Then, the resulting value is placed in a temporary variable of type T. The linked page uses the words "rvalue" and "lvalue" incorrectly . If you need different semantics, you would require explicit specialization of template. Const reference can be bounded to. Share. And const is a constraint imposed by the compiler to the variable that is declared as const. decltype (fun ()) b=1;Syntax: void foo (std::string& str); // non-constant lvalue reference overload. 5. and if you pass it to a function that takes a reference to a non-const - it means that function can change the value. Since the constructor in your example only takes lvalues, you can only pass lvalues into the factory function. It reflects the old, not the new. But a is an lvalue expression because it refers to an object's name . For lvalue-references (that is, the type T&) there isn't. The option -qlanglvl=compatrvaluebinding instructs the compiler to allow a non-const or volatile lvalue reference to bind to an. Thank you. Would you explain why you need a non-const reference that cannot bind to non-const objects?. This way, if the user passes in a U as an lvalue, it will be passed as U&, and if the user passes in a U as an rvalue, it will be passed as U&&. 3. Reload to refresh your session. Create_moneys () is a function that takes a mutable reference to a pointer. [ Example: double& rd2 = 2. r can be bound to the conversion result of e or a base class of e if the following conditions are satisfied. thanks in advance, George. " In other words, at that point the value is pretty much like any other local. -hg. One const and the other non-const. Rule 3, "Note: if the initializer for a reference of type const T& is. Specifically, a const rvalue will prefer to bind to the const rvalue reference rather than the const lvalue reference. You can disable this behaviour with the /Za (disable language extensions) compiler switch under. 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. There are exceptions, however. initial value of reference to non-const must be an lvalue. Of course, unlike the one in the range-based for loop, this i reference become dangling immediately. But if you are asking why this doesn't. Am getting cannot bind non-const lvalue reference of type ‘Type&’ to an rvalue of type 'Type' The function returns a pointer, which you are trying to bind to a reference. An rvalue reference can only bind to non-const rvalues. Yes, some times it is very convenient to be able to locally modify a pass-by-value argument to a function. Apparently, the Standard agrees. ctor] A non-template constructor for class X is a copy constructor if its first parameter is of type X&, const X&, volatile X& or const volatile X&, and either there are. Generally speaking, when a function takes a parameter by non-const. So how to solve that. 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. operator[] is - either change the return type of the function from Value* to const Value&, or return *cleverconfig[name]; With the option -qinfo=por specified, when the compiler chooses such a binding, the following informational message is emitted. This may sound like a silly question, but I was confused about this following behaviour:. Non. So, despite your extra const in your reference type the language still requires it to be bound directly to i. Calling operator + second time won't be possible because a temporary object can not be passed as reference to a non-const-qualified object. 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. y()) < std::tie(b. By the way, don’t return const values from a function, because you make it impossible to use move semantics. Case 3: binding to data members. 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 ). find (key);A pointer to non-const is convertible to pointer to const however. Expression like a+b will return some constant. e. Improve this question. Thus the declaration doesn't have a. So an expression returning a non-const reference is still considered an lvalue. Constructor by the definition does not have a return value. 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& //. 3. However, this is deceptive, because it may or may not be an rvalue reference depending on the type of T. Non-const reference may only be bound to an lvalue. 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. Take a look at the swap function signature: swap ( shared_ptr& r ). 3 of the C++11 standard: It doesn't allow expressions that bind a user-defined type temporary to a non-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. T may resolve to different types of reference, but the type trait don't know about references. 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. In general, when Foo isn't a const type your examples should fail to compile. "non-const lvalue reference to type 'QByteArray' cannot bind to a temporary of type 'QByteArray'". There are exceptions, however. It is unusual to use references to iterators. That is to say, usage of a reference is syntactically identical to usage of the referent. But that doesn't make sense. 15. Regarding the second question. Follow edited Nov 15, 2016 at. You can't. rvalue reference 는 rvalue (즉, 상수와 임시객체)도 참조가 가능 하다 점을 빼고는 기존의 참조와 동일합니다. int f( int ); int f( int && ); int f( int const & ); int q = f( 3 ); Removing f( int ) causes both Clang and GCC to prefer the rvalue reference over the lvalue reference. GetCollider(); platform1. Fibonacci Series in C++. Once bound, there is no difference in behaviour between an rvalue reference and 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. 1. Creating a const reference does not need to be created from a lvalue variable, because if it is created from a non-lvalue variable, it creates a. Early on, when we teach modern C++, we teach that every non-small 1 data should be passed, by default, as constant reference: 1. If the initializer expression. Use a const reference, which can be bound to rvalues. GetImage (iTileId, pulImageSize, a_pImage ); With the method defined as: This change is required by the C++ standard which specifies that a non-const. //. c++; Share. C4239 は、以下。. They can bind to const lvalue-references because then a promise has been made. The number of identifiers must equal the number of non-static data members. There are better ways to solve your problems. You must handle the case. const int & is a const lvalue reference. As I understand it, the compiler has to create an implicit read-only object so that ri3 can be a reference to it; note that &ri3 yields a valid address. 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. 71. It matches arguments of any value category, making t an lvalue reference if the supplied argument was an lvalue or an rvalue reference if the supplied argument was an rvalue. 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. According to the reference collapsing rules, "rvalue reference to rvalue reference collapses to rvalue reference, all other combinations form lvalue reference". doesn't that mean that an rvalue ref is an lvalue. A rvalue can be used as const T&, however, the compiler complains about binding a non-const lvalue to a rvalue. In the second case, fun() returns a non-const lvalue reference, which can bind to another non-const reference, of course. Overload resolution is usually done in terms of a strict. copy. The compiler automatically generates a temporary that the reference is bound to. To reduce template instantiation overhead, I would recommend a more direct implementation:will result in output: Constructor called 42. ningaman151 November 23, 2019, 7:39pm 8. a. In function 'int main()': Line 15: error: invalid initialization of non-const reference of type 'std::string&' from a temporary of type 'std::string' compilation terminated due to -Wfatal-errors.