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A copy assignment operator is a non-template non-static member function with the name operator = that can be called with an argument of the same class type and copies the content of the argument without mutating the argument.
Syntax Explanation Implicitly-declared copy assignment operator Implicitly-defined copy assignment operator Deleted copy assignment operator Trivial copy assignment operator Eligible copy assignment operator Notes Example Defect reports See also |
For the formal copy assignment operator syntax, see function declaration . The syntax list below only demonstrates a subset of all valid copy assignment operator syntaxes.
return-type parameter-list | (1) | ||||||||
return-type parameter-list function-body | (2) | ||||||||
return-type parameter-list-no-default | (3) | (since C++11) | |||||||
return-type parameter-list | (4) | (since C++11) | |||||||
return-type class-name parameter-list function-body | (5) | ||||||||
return-type class-name parameter-list-no-default | (6) | (since C++11) | |||||||
class-name | - | the class whose copy assignment operator is being declared, the class type is given as in the descriptions below |
parameter-list | - | a of only one parameter, which is of type , , const T&, volatile T& or const volatile T& |
parameter-list-no-default | - | a of only one parameter, which is of type , , const T&, volatile T& or const volatile T& and does not have a default argument |
function-body | - | the of the copy assignment operator |
return-type | - | any type, but is favored in order to allow chaining asssignments |
The copy assignment operator is called whenever selected by overload resolution , e.g. when an object appears on the left side of an assignment expression.
If no user-defined copy assignment operators are provided for a class type, the compiler will always declare one as an inline public member of the class. This implicitly-declared copy assignment operator has the form T & T :: operator = ( const T & ) if all of the following is true:
Otherwise the implicitly-declared copy assignment operator is declared as T & T :: operator = ( T & ) .
Due to these rules, the implicitly-declared copy assignment operator cannot bind to a volatile lvalue argument.
A class can have multiple copy assignment operators, e.g. both T & T :: operator = ( T & ) and T & T :: operator = ( T ) . If some user-defined copy assignment operators are present, the user may still force the generation of the implicitly declared copy assignment operator with the keyword default . (since C++11)
The implicitly-declared (or defaulted on its first declaration) copy assignment operator has an exception specification as described in dynamic exception specification (until C++17) noexcept specification (since C++17)
Because the copy assignment operator is always declared for any class, the base class assignment operator is always hidden. If a using-declaration is used to bring in the assignment operator from the base class, and its argument type could be the same as the argument type of the implicit assignment operator of the derived class, the using-declaration is also hidden by the implicit declaration.
If the implicitly-declared copy assignment operator is neither deleted nor trivial, it is defined (that is, a function body is generated and compiled) by the compiler if odr-used or needed for constant evaluation (since C++14) . For union types, the implicitly-defined copy assignment copies the object representation (as by std::memmove ). For non-union class types, the operator performs member-wise copy assignment of the object's direct bases and non-static data members, in their initialization order, using built-in assignment for the scalars, memberwise copy-assignment for arrays, and copy assignment operator for class types (called non-virtually).
The implicitly-defined copy assignment operator for a class is if is a , and that is of class type (or array thereof), the assignment operator selected to copy that member is a constexpr function. | (since C++14) (until C++23) |
The implicitly-defined copy assignment operator for a class is . | (since C++23) |
The generation of the implicitly-defined copy assignment operator is deprecated if has a user-declared destructor or user-declared copy constructor. | (since C++11) |
An implicitly-declared or explicitly-defaulted (since C++11) copy assignment operator for class T is undefined (until C++11) defined as deleted (since C++11) if any of the following conditions is satisfied:
The implicitly-declared copy assignment operator for class is defined as deleted if declares a or . | (since C++11) |
The copy assignment operator for class T is trivial if all of the following is true:
A trivial copy assignment operator makes a copy of the object representation as if by std::memmove . All data types compatible with the C language (POD types) are trivially copy-assignable.
A copy assignment operator is eligible if it is either user-declared or both implicitly-declared and definable. | (until C++11) |
A copy assignment operator is eligible if it is not deleted. | (since C++11) (until C++20) |
A copy assignment operator is eligible if all following conditions are satisfied: (if any) are satisfied. than any other copy assignment operator. | (since C++20) |
Triviality of eligible copy assignment operators determines whether the class is a trivially copyable type .
If both copy and move assignment operators are provided, overload resolution selects the move assignment if the argument is an rvalue (either a prvalue such as a nameless temporary or an xvalue such as the result of std::move ), and selects the copy assignment if the argument is an lvalue (named object or a function/operator returning lvalue reference). If only the copy assignment is provided, all argument categories select it (as long as it takes its argument by value or as reference to const, since rvalues can bind to const references), which makes copy assignment the fallback for move assignment, when move is unavailable.
It is unspecified whether virtual base class subobjects that are accessible through more than one path in the inheritance lattice, are assigned more than once by the implicitly-defined copy assignment operator (same applies to move assignment ).
See assignment operator overloading for additional detail on the expected behavior of a user-defined copy-assignment operator.
[ edit ] defect reports.
The following behavior-changing defect reports were applied retroactively to previously published C++ standards.
DR | Applied to | Behavior as published | Correct behavior |
---|---|---|---|
C++98 | the conditions where implicitly-declared copy assignment operators are undefined did not consider multi-dimensional array types | consider these types | |
C++11 | a volatile subobject made defaulted copy assignment operators non-trivial ( ) | triviality not affected | |
C++11 | operator=(X&) = default was non-trivial | made trivial | |
C++11 | a defaulted copy assignment operator for class was not defined as deleted if is abstract and has non-copy-assignable direct virtual base classes | the operator is defined as deleted in this case | |
C++20 | a copy assignment operator was not eligible if there is another copy assignment operator which is more constrained but does not satisfy its associated constraints | it can be eligible in this case |
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A copy assignment operator of class T is a non-template non-static member function with the name operator = that takes exactly one parameter of type T , T & , const T & , volatile T & , or const volatile T & . For a type to be CopyAssignable , it must have a public copy assignment operator.
class_name class_name ( class_name ) | (1) | ||||||||
class_name class_name ( const class_name ) | (2) | ||||||||
class_name class_name ( const class_name ) = default; | (3) | (since C++11) | |||||||
class_name class_name ( const class_name ) = delete; | (4) | (since C++11) | |||||||
The copy assignment operator is called whenever selected by overload resolution , e.g. when an object appears on the left side of an assignment expression.
If no user-defined copy assignment operators are provided for a class type ( struct , class , or union ), the compiler will always declare one as an inline public member of the class. This implicitly-declared copy assignment operator has the form T & T :: operator = ( const T & ) if all of the following is true:
Otherwise the implicitly-declared copy assignment operator is declared as T & T :: operator = ( T & ) . (Note that due to these rules, the implicitly-declared copy assignment operator cannot bind to a volatile lvalue argument.)
A class can have multiple copy assignment operators, e.g. both T & T :: operator = ( const T & ) and T & T :: operator = ( T ) . If some user-defined copy assignment operators are present, the user may still force the generation of the implicitly declared copy assignment operator with the keyword default . (since C++11)
The implicitly-declared (or defaulted on its first declaration) copy assignment operator has an exception specification as described in dynamic exception specification (until C++17) exception specification (since C++17)
Because the copy assignment operator is always declared for any class, the base class assignment operator is always hidden. If a using-declaration is used to bring in the assignment operator from the base class, and its argument type could be the same as the argument type of the implicit assignment operator of the derived class, the using-declaration is also hidden by the implicit declaration.
A implicitly-declared copy assignment operator for class T is defined as deleted if any of the following is true:
Otherwise, it is defined as defaulted.
A defaulted copy assignment operator for class T is defined as deleted if any of the following is true:
The copy assignment operator for class T is trivial if all of the following is true:
has no non-static data members of -qualified type. | (since C++14) |
A trivial copy assignment operator makes a copy of the object representation as if by std::memmove . All data types compatible with the C language (POD types) are trivially copy-assignable.
If the implicitly-declared copy assignment operator is neither deleted nor trivial, it is defined (that is, a function body is generated and compiled) by the compiler if odr-used . For union types, the implicitly-defined copy assignment copies the object representation (as by std::memmove ). For non-union class types ( class and struct ), the operator performs member-wise copy assignment of the object's bases and non-static members, in their initialization order, using built-in assignment for the scalars and copy assignment operator for class types.
The generation of the implicitly-defined copy assignment operator is deprecated (since C++11) if T has a user-declared destructor or user-declared copy constructor.
If both copy and move assignment operators are provided, overload resolution selects the move assignment if the argument is an rvalue (either a prvalue such as a nameless temporary or an xvalue such as the result of std::move ), and selects the copy assignment if the argument is an lvalue (named object or a function/operator returning lvalue reference). If only the copy assignment is provided, all argument categories select it (as long as it takes its argument by value or as reference to const, since rvalues can bind to const references), which makes copy assignment the fallback for move assignment, when move is unavailable.
It is unspecified whether virtual base class subobjects that are accessible through more than one path in the inheritance lattice, are assigned more than once by the implicitly-defined copy assignment operator (same applies to move assignment ).
See assignment operator overloading for additional detail on the expected behavior of a user-defined copy-assignment operator.
The following behavior-changing defect reports were applied retroactively to previously published C++ standards.
DR | Applied to | Behavior as published | Correct behavior |
---|---|---|---|
C++14 | operator=(X&) = default was non-trivial | made trivial |
Copy constructor and Assignment operator are similar as they are both used to initialize one object using another object. But, there are some basic differences between them:
Copy constructor | Assignment operator |
---|---|
It is called when a new object is created from an existing object, as a copy of the existing object | This operator is called when an already initialized object is assigned a new value from another existing object. |
It creates a separate memory block for the new object. | It does not automatically create a separate memory block or new memory space. However, if the class involves dynamic memory management, the assignment operator must first release the existing memory on the left-hand side and then allocate new memory as needed to copy the data from the right-hand side. |
It is an overloaded constructor. | It is a bitwise operator. |
C++ compiler implicitly provides a copy constructor, if no copy constructor is defined in the class. | A bitwise copy gets created, if the Assignment operator is not overloaded. |
className(const className &obj) { // body } |
className obj1, obj2; obj2 = obj1; |
Consider the following C++ program.
Explanation: Here, t2 = t1; calls the assignment operator , same as t2.operator=(t1); and Test t3 = t1; calls the copy constructor , same as Test t3(t1);
Must Read: When is a Copy Constructor Called in C++?
Similar reads.
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Starting in C++11, two kinds of assignment are supported in the language: copy assignment and move assignment . In this article "assignment" means copy assignment unless explicitly stated otherwise. For information about move assignment, see Move Constructors and Move Assignment Operators (C++) .
Both the assignment operation and the initialization operation cause objects to be copied.
Assignment : When one object's value is assigned to another object, the first object is copied to the second object. So, this code copies the value of b into a :
Initialization : Initialization occurs when you declare a new object, when you pass function arguments by value, or when you return by value from a function.
You can define the semantics of "copy" for objects of class type. For example, consider this code:
The preceding code could mean "copy the contents of FILE1.DAT to FILE2.DAT" or it could mean "ignore FILE2.DAT and make b a second handle to FILE1.DAT." You must attach appropriate copying semantics to each class, as follows:
Use an assignment operator operator= that returns a reference to the class type and takes one parameter that's passed by const reference—for example ClassName& operator=(const ClassName& x); .
Use the copy constructor.
If you don't declare a copy constructor, the compiler generates a member-wise copy constructor for you. Similarly, if you don't declare a copy assignment operator, the compiler generates a member-wise copy assignment operator for you. Declaring a copy constructor doesn't suppress the compiler-generated copy assignment operator, and vice-versa. If you implement either one, we recommend that you implement the other one, too. When you implement both, the meaning of the code is clear.
The copy constructor takes an argument of type ClassName& , where ClassName is the name of the class. For example:
Make the type of the copy constructor's argument const ClassName& whenever possible. This prevents the copy constructor from accidentally changing the copied object. It also lets you copy from const objects.
Compiler-generated copy constructors, like user-defined copy constructors, have a single argument of type "reference to class-name ." An exception is when all base classes and member classes have copy constructors declared as taking a single argument of type const class-name & . In such a case, the compiler-generated copy constructor's argument is also const .
When the argument type to the copy constructor isn't const , initialization by copying a const object generates an error. The reverse isn't true: If the argument is const , you can initialize by copying an object that's not const .
Compiler-generated assignment operators follow the same pattern for const . They take a single argument of type ClassName& unless the assignment operators in all base and member classes take arguments of type const ClassName& . In this case, the generated assignment operator for the class takes a const argument.
When virtual base classes are initialized by copy constructors, whether compiler-generated or user-defined, they're initialized only once: at the point when they are constructed.
The implications are similar to the copy constructor. When the argument type isn't const , assignment from a const object generates an error. The reverse isn't true: If a const value is assigned to a value that's not const , the assignment succeeds.
For more information about overloaded assignment operators, see Assignment .
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A copy assignment operator of class T is a non-template non-static member function with the name operator = that takes exactly one parameter of type T , T & , const T & , volatile T & , or const volatile T & . A type with a public copy assignment operator is CopyAssignable .
Syntax Explanation Implicitly-declared copy assignment operator Deleted implicitly-declared copy assignment operator Trivial copy assignment operator Implicitly-defined copy assignment operator Notes Copy and swap Example |
class_name class_name ( class_name ) | (1) | ||||||||
class_name class_name ( const class_name ) | (2) | ||||||||
class_name class_name ( const class_name ) = default; | (3) | (since C++11) | |||||||
class_name class_name ( const class_name ) = delete; | (4) | (since C++11) | |||||||
The copy assignment operator is called whenever selected by overload resolution , e.g. when an object appears on the left side of an assignment expression.
If no user-defined copy assignment operators are provided for a class type ( struct , class , or union ), the compiler will always declare one as an inline public member of the class. This implicitly-declared copy assignment operator has the form T & T :: operator = ( const T & ) if all of the following is true:
Otherwise the implicitly-declared copy assignment operator is declared as T & T :: operator = ( T & ) . (Note that due to these rules, the implicitly-declared copy assignment operator cannot bind to a volatile lvalue argument)
A class can have multiple copy assignment operators, e.g. both T & T :: operator = ( const T & ) and T & T :: operator = ( T ) . If some user-defined copy assignment operators are present, the user may still force the generation of the implicitly declared copy assignment operator with the keyword default . (since C++11)
Because the copy assignment operator is always declared for any class, the base class assignment operator is always hidden. If a using-declaration is used to bring in the assignment operator from the base class, and its argument type could be the same as the argument type of the implicit assignment operator of the derived class, the using-declaration is also hidden by the implicit declaration.
The implicitly-declared or defaulted copy assignment operator for class T is defined as deleted in any of the following is true:
The copy assignment operator for class T is trivial if all of the following is true:
has no non-static data members of -qualified type | (since C++14) |
A trivial copy assignment operator makes a copy of the object representation as if by std::memmove . All data types compatible with the C language (POD types) are trivially copy-assignable.
If the implicitly-declared copy assignment operator is not deleted or trivial, it is defined (that is, a function body is generated and compiled) by the compiler. For union types, the implicitly-defined copy assignment copies the object representation (as by std::memmove ). For non-union class types ( class and struct ), the operator performs member-wise copy assignment of the object's bases and non-static members, in their initialization order, using, using built-in assignment for the scalars and copy assignment operator for class types.
The generation of the implicitly-defined copy assignment operator is deprecated (since C++11) if T has a user-declared destructor or user-declared copy constructor.
If both copy and move assignment operators are provided, overload resolution selects the move assignment if the argument is an rvalue (either prvalue such as a nameless temporary or xvalue such as the result of std::move ), and selects the copy assignment if the argument is lvalue (named object or a function/operator returning lvalue reference). If only the copy assignment is provided, all argument categories select it (as long as it takes its argument by value or as reference to const, since rvalues can bind to const references), which makes copy assignment the fallback for move assignment, when move is unavailable.
Copy assignment operator can be expressed in terms of copy constructor, destructor, and the swap() member function, if one is provided:
T & T :: operator = ( T arg ) { // copy/move constructor is called to construct arg swap ( arg ) ; // resources exchanged between *this and arg return * this ; } // destructor is called to release the resources formerly held by *this
For non-throwing swap(), this form provides strong exception guarantee . For rvalue arguments, this form automatically invokes the move constructor, and is sometimes referred to as "unifying assignment operator" (as in, both copy and move).
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Salim Khan and Javed Akhtar pictured together.
Salim-Javed , who were regarded as significant and powerful as the leading actors of their era, transformed the status of writers in Hindi cinema. The legendary pair wrote 24 films together, 22 of which were successful at the box office. Even though screenwriters and other technicians in a movie production usually do not have as much influence as the leading star, there was a time when Salim-Javed used to charge more money than the biggest of stars.
In a conversation with NDTV, Salim Khan said that he and Javed Akhtar were insistent on being paid well because the script remains the most important part of the film. He said, “ Ek film ke andar aisa hua ki humne actor se zyada paisa liya…Producer ko maine poocha kitna diya 12 lakh humne kahaan humein 12.5 lakh. [There was a film where we took more money. We asked the producer ‘How much are you pacing the lead actor, he said ‘Rs 12 lakh'. We said, ‘We will take Rs 12.5 lakh'. He agreed]"
The iconic duo didn't think they would ask for more money than the actors, but they wanted to better the situation for writers. Salim Khan said, “ Aisa lagta that ki bohut jatti hain..jaise kikoi value nahin hain [We ere not happy with the situation. It seemed as if writers had no value].”
Then Salim-Javed did something unheard of. Zanjeer, the iconic Amitabh Bachchan movie, did not have credits of the writers. What did they do? “ Humne ek painter ko bola ki purra Bombay ke posters mein Salim-Javed likh do. Sako jhatka laga. Log bole aap aise na karo, naamdenge hum. [We hired a painter and asked him to put our name in posters across the city. People were shocked],” said Salim Khan.
Salim Khan was also asked about the industry's reaction to their asking price. To which he said, " Bolte the pagal hai, ye pagalpane ki baat karte hain. Humne kaha theek hai mat do paise agar script chahiye to is see kam m nahi milegi. Hum kahani sunane se pehle hi price bta dete they to vo bhi jhatka kha jate the. Jis zamane m mushkil see Rs 10,000 milte the writers ko, hum lakho maangte the. [They called us crazy, used to say we talk in madness. We said okay don't give us money but if you need the script we will not go any lower than this. We used to tell our price even before telling the story, they used to get so shocked. In the time, when the writers were not getting even Rs 10,000, we used to ask for lakhs of money]"
Angry Young Men, a docu-series based on Salim-Javed's life, also touches upon this anecdote. It showed how at the height of their fame, Salim-Javed received payments of Rs 21 lakh for each film—more than even Amitabh Bachchan, who was the highest-paid actor in India at the time. The show also explores Salim Khan and Javed Akhtar's personal lives.
Salim-Javed delivered numerous Bollywood hits, including Haathi Mere Saathi , Seeta Aur Geeta , Yaadon Ki Baaraat , Deewaar , Sholay , Trishul , Don , and many more.
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In C++, the concept of returning reference from the copy assignment operator is unclear to me. Why can't the copy assignment operator return a copy of the new object? In addition, if I have class A , and the following:
The operator= is defined as follows:
A bit of clarification as to why it's preferable to return by reference for operator= versus return by value --- as the chain a = b = c will work fine if a value is returned.
If you return a reference, minimal work is done. The values from one object are copied to another object.
However, if you return by value for operator= , you will call a constructor AND destructor EACH time that the assignment operator is called!!
In sum, there is nothing gained by returning by value, but a lot to lose.
( Note : This isn't meant to address the advantages of having the assignment operator return an lvalue. Read the other posts for why that might be preferable)
Strictly speaking, the result of a copy assignment operator doesn't need to return a reference, though to mimic the default behavior the C++ compiler uses, it should return a non-const reference to the object that is assigned to (an implicitly generated copy assignment operator will return a non-const reference - C++03: 12.8/10). I've seen a fair bit of code that returns void from copy assignment overloads, and I can't recall when that caused a serious problem. Returning void will prevent users from 'assignment chaining' ( a = b = c; ), and will prevent using the result of an assignment in a test expression, for example. While that kind of code is by no means unheard of, I also don't think it's particularly common - especially for non-primitive types (unless the interface for a class intends for these kinds of tests, such as for iostreams).
I'm not recommending that you do this, just pointing out that it's permitted and that it doesn't seem to cause a whole lot of problems.
These other SO questions are related (probably not quite dupes) that have information/opinions that might be of interest to you.
When you overload operator= , you can write it to return whatever type you want. If you want to badly enough, you can overload X::operator= to return (for example) an instance of some completely different class Y or Z . This is generally highly inadvisable though.
In particular, you usually want to support chaining of operator= just like C does. For example:
That being the case, you usually want to return an lvalue or rvalue of the type being assigned to. That only leaves the question of whether to return a reference to X, a const reference to X, or an X (by value).
Returning a const reference to X is generally a poor idea. In particular, a const reference is allowed to bind to a temporary object. The lifetime of the temporary is extended to the lifetime of the reference to which it's bound--but not recursively to the lifetime of whatever that might be assigned to. This makes it easy to return a dangling reference--the const reference binds to a temporary object. That object's lifetime is extended to the lifetime of the reference (which ends at the end of the function). By the time the function returns, the lifetime of the reference and temporary have ended, so what's assigned is a dangling reference.
Of course, returning a non-const reference doesn't provide complete protection against this, but at least makes you work a little harder at it. You can still (for example) define some local, and return a reference to it (but most compilers can and will warn about this too).
Returning a value instead of a reference has both theoretical and practical problems. On the theoretical side, you have a basic disconnect between = normally means and what it means in this case. In particular, where assignment normally means "take this existing source and assign its value to this existing destination", it starts to mean something more like "take this existing source, create a copy of it, and assign that value to this existing destination."
From a practical viewpoint, especially before rvalue references were invented, that could have a significant impact on performance--creating an entire new object in the course of copying A to B was unexpected and often quite slow. If, for example, I had a small vector, and assigned it to a larger vector, I'd expect that to take, at most, time to copy elements of the small vector plus a (little) fixed overhead to adjust the size of the destination vector. If that instead involved two copies, one from source to temp, another from temp to destination, and (worse) a dynamic allocation for the temporary vector, my expectation about the complexity of the operation would be entirely destroyed. For a small vector, the time for the dynamic allocation could easily be many times higher than the time to copy the elements.
The only other option (added in C++11) would be to return an rvalue reference. This could easily lead to unexpected results--a chained assignment like a=b=c; could destroy the contents of b and/or c , which would be quite unexpected.
That leaves returning a normal reference (not a reference to const, nor an rvalue reference) as the only option that (reasonably) dependably produces what most people normally want.
It's partly because returning a reference to self is faster than returning by value, but in addition, it's to allow the original semantics that exist in primitive types.
operator= can be defined to return whatever you want. You need to be more specific as to what the problem actually is; I suspect that you have the copy constructor use operator= internally and that causes a stack overflow, as the copy constructor calls operator= which must use the copy constructor to return A by value ad infinitum.
There is no core language requirement on the result type of a user-defined operator= , but the standard library does have such a requirement:
C++98 §23.1/3:
” The type of objects stored in these components must meet the requirements of CopyConstructible types (20.1.3), and the additional requirements of Assignable types.
C++98 §23.1/4:
” In Table 64, T is the type used to instantiate the container, t is a value of T , and u is a value of (possibly const ) T .
Returning a copy by value would still support assignment chaining like a = b = c = 42; , because the assignment operator is right-associative, i.e. this is parsed as a = (b = (c = 42)); . But returning a copy would prohibit meaningless constructions like (a = b) = 666; . For a small class returning a copy could conceivably be most efficient, while for a larger class returning by reference will generally be most efficient (and a copy, prohibitively inefficient).
Until I learned about the standard library requirement I used to let operator= return void , for efficiency and to avoid the absurdity of supporting side-effect based bad code.
With C++11 there is additionally the requirement of T& result type for default -ing the assignment operator, because
C++11 §8.4.2/1:
” A function that is explicitly defaulted shall […] have the same declared function type (except for possibly differing ref-qualifiers and except that in the case of a copy constructor or copy assignment operator, the parameter type may be “reference to non-const T ”, where T is the name of the member function’s class) as if it had been implicitly declared
I guess, because user defined object should behave like builtin types. For example:
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Copy assignment operator
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A user-declared copy assignment operator X::operator= is a non-static non-template member function of class X with exactly one parameter of type X, X&, const X&, volatile X& or const volatile X&. So for example: struct X {. int a; // an assignment operator which is not a copy assignment operator. X &operator=(int rhs) { a = rhs; return *this; }
the copy assignment operator selected for every non-static class type (or array of class type) member of T is trivial; T has no non-static data members of volatile-qualified type. (since C++14) A trivial copy assignment operator makes a copy of the object representation as if by std::memmove. All data types compatible with the C language (POD ...
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Copy Constructor vs Assignment Operator in C
Assignment: When one object's value is assigned to another object, the first object is copied to the second object.So, this code copies the value of b into a:. Point a, b; ... a = b; Initialization: Initialization occurs when you declare a new object, when you pass function arguments by value, or when you return by value from a function.. You can define the semantics of "copy" for objects of ...
The copy assignment operator selected for every non-static class type (or array of class type) memeber of T is trivial T has no non-static data members of volatile-qualified type (since C++14) A trivial copy assignment operator makes a copy of the object representation as if by std::memmove. All data types compatible with the C language (POD ...
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"We used to tell our price even before telling the story, they used to get so shocked. In the time, when the writers were not getting even Rs 10,000, we used to ask for lakhs of money," Salim Khan ...
7. Test::operator=(const Test&) is not a constructor. It's a copy assignment operator. A constructor creates a new object; an assignment operator modifies an existing object. - Pete Becker. Jun 7, 2019 at 21:14. 4. @SteliosPapamichail Seems your teacher is not too familiar with C++ concepts. - john.
C++98 §23.1/4: " In Table 64, T is the type used to instantiate the container, t is a value of T, and u is a value of (possibly const) T. Returning a copy by value would still support assignment chaining like a = b = c = 42;, because the assignment operator is right-associative, i.e. this is parsed as a = (b = (c = 42));.