c copy assignment operator with const member

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Copy assignment operator.






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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.

Syntax Explanation Implicitly-declared copy assignment operator Deleted implicitly-declared copy assignment operator Trivial copy assignment operator Implicitly-defined copy assignment operator Notes Example Defect reports

[ edit ] Syntax


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)

[ edit ] Explanation

Typical declaration of a copy assignment operator when copy-and-swap idiom can be used.
Typical declaration of a copy assignment operator when copy-and-swap idiom cannot be used (non-swappable type or degraded performance).
Forcing a copy assignment operator to be generated by the compiler.
Avoiding implicit copy assignment.

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.

[ edit ] Implicitly-declared copy assignment operator

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:

each direct base B of T has a copy assignment operator whose parameters are B or const B & or const volatile B & ;
each non-static data member M of T of class type or array of class type has a copy assignment operator whose parameters are M or const M & or const volatile M & .

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.

[ edit ] Deleted implicitly-declared copy assignment operator

A implicitly-declared copy assignment operator for class T is defined as deleted if any of the following is true:

T has a user-declared move constructor;
T has a user-declared move assignment operator.

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:

T has a non-static data member of non-class type (or array thereof) that is const ;
T has a non-static data member of a reference type;
T has a non-static data member or a direct or virtual base class that cannot be copy-assigned (overload resolution for the copy assignment fails, or selects a deleted or inaccessible function);
T is a union-like class , and has a variant member whose corresponding assignment operator is non-trivial.

[ edit ] Trivial copy assignment operator

The copy assignment operator for class T is trivial if all of the following is true:

it is not user-provided (meaning, it is implicitly-defined or defaulted) , , and if it is defaulted, its signature is the same as implicitly-defined (until C++14) ;
T has no virtual member functions;
T has no virtual base classes;
the copy assignment operator selected for every direct base of T is trivial;
the copy assignment operator selected for every non-static class type (or array of class type) member of T is trivial;

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.

[ edit ] Implicitly-defined copy assignment operator

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.

[ edit ] Notes

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 ] Example

[ 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++14	operator=(X&) = default was non-trivial	made trivial

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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 .

[ edit ] Syntax


class_name class_name ( class_name )	(1)	(since C++11)

class_name class_name ( const class_name )	(2)	(since C++11)

class_name class_name ( const class_name ) = default;	(3)	(since C++11)

class_name class_name ( const class_name ) = delete;	(4)	(since C++11)

[ edit ] Explanation

Typical declaration of a copy assignment operator when copy-and-swap idiom can be used
Typical declaration of a copy assignment operator when copy-and-swap idiom cannot be used
Forcing a copy assignment operator to be generated by the compiler
Avoiding implicit copy assignment

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.

[ edit ] Implicitly-declared copy assignment operator

each direct base B of T has a copy assignment operator whose parameters are B or const B& or const volatile B &
each non-static data member M of T of class type or array of class type has a copy assignment operator whose parameters are M or const M& or const volatile M &

[ edit ] Deleted implicitly-declared copy assignment operator

The implicitly-declared or defaulted copy assignment operator for class T is defined as deleted in any of the following is true:

T has a non-static data member that is const
T has a non-static data member of a reference type.
T has a non-static data member that cannot be copy-assigned (has deleted, inaccessible, or ambiguous copy assignment operator)
T has direct or virtual base class that cannot be copy-assigned (has deleted, inaccessible, or ambiguous move assignment operator)
T has a user-declared move constructor
T has a user-declared move assignment operator

[ edit ] Trivial copy assignment operator

The implicitly-declared copy assignment operator for class T is trivial if all of the following is true:

T has no virtual member functions
T has no virtual base classes
The copy assignment operator selected for every direct base of T is trivial
The copy assignment operator selected for every non-static class type (or array of class type) memeber of T is trivial

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.

[ edit ] Implicitly-defined copy assignment operator

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.

[ edit ] Notes

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.

[ edit ] Copy and swap

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

[ edit ] Example

When to use const in C++? Part II: member variables

Just make everything const that you can! That’s the bare minimum you could do for your compiler!

This is a piece of advice, many senior developers tend to repeat to juniors, while so often even the preaching ones - we - fail to follow this rule.

In this series of articles, we discuss about:

In the first episode, we covered const functions and const local variables. Today we’ll speak about the members.

Originally, I didn’t plan this post. I simply wanted to speak about const variables regardless if they have a local scope or if they are members of an object.

Then I saw this tweet from Tina Ulbrich who I met at C++OnSea2020 and I was horrified. Yet another thing in C++, I had no idea about and something I’ve been doing considering that it’s a good practice.

Truth to be told, I didn’t do anything harmful, but that’s only by chance.

Ok, let’s get to it.

Why would you have const members at the first place?

Because you might want to signal that they are immutable, that their values should never change. Some would claim that you have private members for that purpose and you simply should not expose a setter for such members, then there is no need to explicitly make them const .

I get you, you’re right. In an ideal world.

But even if you are a strong believer of the Single Responsibility Principle and small classes, there is a fair chance that others later will change your code, your class will grow, and someone might accidentally change the value inside, plus you haven’t given the compiler a hint for optimization due to immutability.

To me, these are good reasons to make a member const. At least to show the intention.

But unfortunately, there are some implications.

The first is that classes a const member are not assignable:

MyClassWithConstMember { public: MyClassWithConstMember(int a) : m_a(a) {} private: const int m_a; }; int main() { MyClassWithConstMember o1{666}; MyClassWithConstMember o2{42}; o1 = o2; } /*main.cpp: In function 'int main()': main.cpp:11:8: error: use of deleted function 'MyClassWithConstMember& MyClassWithConstMember::operator=(const MyClassWithConstMember&)' 11 | o1 = o2; | ^~ main.cpp:1:7: note: 'MyClassWithConstMember& MyClassWithConstMember::operator=(const MyClassWithConstMember&)' is implicitly deleted because the default definition would be ill-formed: 1 | class MyClassWithConstMember { | ^~~~~~~~~~~~~~~~~~~~~~ main.cpp:1:7: error: non-static const member 'const int MyClassWithConstMember::m_a', cannot use default assignment operator */

If you think about it, it makes perfect sense. A variable is something you cannot change after initialization. And when you want to assign a new value to an object, thus to its members, it’s not possible anymore.

As such it also makes it impossible to use move semantics, for the same reason.

From the error messages, you can see that the corresponding special functions, such as the assignment operator or the move assignment operator were deleted.

Let’s implement the assignment operator. It will compile, but what the heck would you do?

& operator=(const MyClassWithConstMember&) { // ??? return *this; }

Do you skip assigning to the const members? Not so great, either you depend on that value somewhere, or you should not store the value.

And you cannot assign to a const variable, can you? For a matter of fact, you can…

<utility> #include <iostream> class MyClassWithConstMember { public: MyClassWithConstMember(int a) : m_a(a) {} MyClassWithConstMember& operator=(const MyClassWithConstMember& other) { int* tmp = const_cast<int*>(&m_a); *tmp = other.m_a; std::cout << "copy assignment \n"; return *this; } int getA() {return m_a;} private: const int m_a; }; int main() { MyClassWithConstMember o1{666}; MyClassWithConstMember o2{42}; std::cout << "o1.a: " << o1.getA() << std::endl; std::cout << "o2.a: " << o2.getA() << std::endl; o1 = o2; std::cout << "o1.a: " << o1.getA() << std::endl;

As you cannot cast the constness away from value, you have to turn the member value into a temporary non-const pointer and then you free to rampage.

Is this worth it?

You have your const member, fine. You have the assignment working, fine. Then if anyone comes later and wants to do the same “magic” outside of the special functions, for sure, it would be a red flag in a code review.

Speaking of special functions. Would move semantics work? Well, replace the assignment with this:

= std::move(o2);

You’ll see that it’s still a copy assignment taking place as the the rule of 5 applies. If you implement one special function, you have to implement all of them. The rest is not generated.

In fact, what we have seen is rather dangerous. You think, you have a move and you’re efficient due to having a const member as using move semantics, but in fact, you are using the old copy assignment.

Yet, performance-wise, it seems hard to make a verdict. I ran a couple of tests in QuickBench and there is no significant difference between the above version and the one with non-const member and generated special assignment operator. On low optimization levels (None-O1) it depends on the compiler and its version. With higher optimization levels set there seems to be no difference.

Conclusions

Having const local variables is good. Having const members… It’s not so obvious. We lose the copy assignment and the move semantics as const members cannot be changed anymore.

With “clever” code, we can run a circle around the problem, but then we have to implement all the special functions. For what?

No performance gain. Less readability in the special functions and for slightly higher confidence that nobody will change the value of that member.

Do you think it’s worth it?

Stay tuned, next time we’ll discuss const return types.

If you want to learn more details about How to use const in C++ , check out my book on Leanpub !

When to use const in C++? Part III: return types

When to use const in c++ part iv: parameters.

Don't Make Me Think by Steve Krug

What do you want from a job, that is the question

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Copy assignment operator












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Members


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[ edit ] Syntax


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)

[ edit ] Explanation

Typical declaration of a copy assignment operator when copy-and-swap idiom can be used
Typical declaration of a copy assignment operator when copy-and-swap idiom cannot be used (non-swappable type or degraded performance)
Forcing a copy assignment operator to be generated by the compiler
Avoiding implicit copy assignment

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.

[ edit ] Implicitly-declared copy assignment operator

each direct base B of T has a copy assignment operator whose parameters are B or const B& or const volatile B &
each non-static data member M of T of class type or array of class type has a copy assignment operator whose parameters are M or const M& or const volatile M &

[ edit ] Deleted implicitly-declared copy assignment operator

A implicitly-declared copy assignment operator for class T is defined as deleted if any of the following is true:

T has a user-declared move constructor
T has a user-declared move assignment operator

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:

T has a non-static data member of non-class type (or array thereof) that is const
T has a non-static data member of a reference type.
T has a non-static data member or a direct or virtual base class that cannot be copy-assigned (overload resolution for the copy assignment fails, or selects a deleted or inaccessible function)
T is a union-like class , and has a variant member whose corresponding assignment operator is non-trivial.

[ edit ] Trivial copy assignment operator

The copy assignment operator for class T is trivial if all of the following is true:

It is not user-provided (meaning, it is implicitly-defined or defaulted), and if it is defaulted, its signature is the same as implicitly-defined
T has no virtual member functions
T has no virtual base classes
The copy assignment operator selected for every direct base of T is trivial
The copy assignment operator selected for every non-static class type (or array of class type) member of T is trivial

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.

[ edit ] Implicitly-defined copy assignment operator

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.

[ edit ] Notes

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.

[ edit ] Copy and swap

Copy assignment operator can be expressed in terms of copy constructor, destructor, and the swap() member function, if one is provided:

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). However, this approach is not always advisable due to potentially significant overhead: see assignment operator overloading for details.

[ edit ] Example

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C++ Core Guidelines: Rules for Copy and Move

The rules for copy and move are pretty obvious. But before I describe them I have to write about the two remaining rules for constructors. They are about delegating and inheriting constructors.

Here are the two remaining rules:

Constructor Rules

C.51: Use delegating constructors to represent common actions for all constructors of a class

Since C++11, a constructor can delegate its work to another constructor of the same class. This is the modern way in C++ to put common actions for all constructors in one constructor. In C++ code before C++11, you often used for such a job an init function.

The constructors (2) and (3) of the class Degree delegate all its initialization work to the constructor (1), which verifies its arguments. Invoking constructors recursively is undefined behavior.

C.52: Use inheriting constructors to import constructors into a derived class that does not need further explicit initialization

If you can reuse constructors of the base class in the derived class, do it. You violate the DRY (Don’t Repeat Yourself) principle if you don’t do it.

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There is a danger in using inheriting constructors. If your derived class, such as Rec2 has its own members, they are uninitialized (1).

Copy and Move

The chapter starts with a meta-rule. Values types, which behave like an int, should be copyable, but interfaces in class hierarchies are not. The last rule C.67 refers to this meta-rule.

Here are the eight rules:

C.60: Make copy assignment non- virtual , take the parameter by const& , and return by non- const&
C.61: A copy operation should copy
C.62: Make copy assignment safe for self-assignment
C.63: Make move assignment non- virtual , take the parameter by && , and return by non- const&
C.64: A move operation should move and leave its source in a valid state
C.65: Make move assignment safe for self-assignment

C.66: Make move operations noexcept

C.67: a base class should suppress copying, and provide a virtual clone instead, if “copying” is desired.

The first six rules for copy and move consist of 3 similar pairs; therefore, I can explain them together.

C.60 and C.63 state that you should make the copy (move) assignment non-virtual and return a non-const reference. There is a difference in the way you should take the parameter.
Copy assignment should take its parameter by a const lvalue reference (&) because you should not change the source of your assignment
Move assignment should take its parameter by a non-const rvalue reference (&&) because you have to modify the source of your assignment
This is the pattern the assignment operators of the standard template library follow. Here is a simplified look at std::vector.

vector & operator = ( const vector & other ); vector & operator = ( vector && other ); // (since C++11, until C++17) vector & operator = ( vector && other ) noexcept // since C++17)

C.61 and C.64 say a copy (move) operation should copy (move). This is the expected semantic for a = b.

In case of copying, this means that after copying a and b (a = b) a must be the same: (a ==b). Copying can be a deep or shallow. Deep copying means that both objects a and b are afterwards totally independent of each other ( value semantic ). Shallow copying means that both objects a and b share an object afterwards ( reference semantic ). C.64 states the moved-from object should be in a valid state. Most of the times this is the default state of the source. The C++ standard requires, that the moved-from object must be afterwards in an unspecified but valid state.

Copy (move) assignment of the containers of the STL, std::string , and built-in type such as int is safe for self-assignment; therefore, the default generated copy (move) assignment operator is, in this case, safe for self-assignment. The same will hold for an automatically generated copy (move) assignment operator, which uses types that are safe for self-assignment.

class Foo { string s; int i; public: Foo & Foo :: operator = ( const Foo & a){ s = a.s; i = a.i; return * this ; } Foo & Foo :: operator = (Foo && a) noexcept { s = std :: move(a.s); i = a.i; return * this ; } // .... }; The code snippet shows that there is no test for self-assignment such as in the next example necessa r y. Here is the version of the type Foo with redundant (expensive) checks (1) and (2) for self-assignment.

class Foo { string s; int i; public: Foo & Foo :: operator = ( const Foo & a){ if ( this == & a) return * this ; // (1) s = a.s; i = a.i; return * this ; } Foo & Foo :: operator = (Foo && a) noexcept { // (2) if ( this == & a) return * this ; s = std :: move(a.s); i = a.i; return * this ; } // .... };

M ove operations should not throw; therefore, you should declare them as noexcept. You can implement your move constructor and move assignment operators that do not throw.

This is the pattern the move operators of the standard template library follow. Have a look at std::vector.

The last rule C.67 deserves more attention.

The main reason for this rule is that slicing is not possible. Slicing is one of these phenomena in C++, and my colleagues always warned me. There also exists an article on Wikipedia about object slicing .

Slicing will happen when an object of a derived class is copied to an object of a base class.

In this scenario, the copy operations of the base class are used; therefore, only the base part of d is copied.

From the object-oriented perspective, an instance of Derived is-a an instance of Base. That means whenever you need an instance of Base you can use an instance of Derived. But you have to be careful. If you take the instance of Base by copy (value-semantic), you will only get the base parts of an instance of Derived.

The cure that the guidelines suggest is that the base class should suppress copying but provide a virtual clone member function if copying is desired. Here is an example from the guidelines.

The clone method returns the newly created object in a std::unique_ptr; therefore, the ownership goes to the caller. Such a clone method is better known as a factory method. A factory method is one of the creational patterns from the book: Design Pattern: Elements of Reusable Object-Oriented Software.

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C++11 and C++14
Concurrency with Modern C++
Design Pattern and Architectural Pattern with C++
Embedded Programming with Modern C++
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Clean Code with Modern C++

Online Seminars (German)

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Mobil: +49 176 5506 5086
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Mentoring Page: www.ModernesCpp.org

Modernes C++ Mentoring,

c copy assignment operator with const member

22.3 — Move constructors and move assignment

Windows Programming
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General C++ Programming
assignment constructor for const member

assignment constructor for const member

std; Foo { : m_val; : Foo( n=0): m_val(n) {} ~Foo() {} Foo( Foo &f): m_val(f.m_val) {} Foo& =( Foo &f): m_val(f.m_val) { ( ==&f) * ; * ; } GetVal() { m_val; } }; main() { Foo f(10); Foo g(f); cout << g.GetVal() << endl; 0; }

ans = 42; n = 77; ans = n; Foo f1(42); Foo f2(77); f1 = f2;

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C++ At Work

Copy Constructors, Assignment Operators, and More

Paul DiLascia

Code download available at: CAtWork0509.exe (276 KB) Browse the Code Online

Q I have a simple C++ problem. I want my copy constructor and assignment operator to do the same thing. Can you tell me the best way to accomplish this?

A At first glance this seems like a simple question with a simple answer: just write a copy constructor that calls operator=.

Or, alternatively, write a common copy method and call it from both your copy constructor and operator=, like so:

This code works fine for many classes, but there's more here than meets the eye. In particular, what happens if your class contains instances of other classes as members? To find out, I wrote the test program in Figure 1 . It has a main class, CMainClass, which contains an instance of another class, CMember. Both classes have a copy constructor and assignment operator, with the copy constructor for CMainClass calling operator= as in the first snippet. The code is sprinkled with printf statements to show which methods are called when. To exercise the constructors, cctest first creates an instance of CMainClass using the default ctor, then creates another instance using the copy constructor:

Figure 1 Copy Constructors and Assignment Operators

If you compile and run cctest, you'll see the following printf messages when cctest constructs obj2:

The member object m_obj got initialized twice! First by the default constructor, and again via assignment. Hey, what's going on?

In C++, assignment and copy construction are different because the copy constructor initializes uninitialized memory, whereas assignment starts with an existing initialized object. If your class contains instances of other classes as data members, the copy constructor must first construct these data members before it calls operator=. The result is that these members get initialized twice, as cctest shows. Got it? It's the same thing that happens with the default constructor when you initialize members using assignment instead of initializers. For example:

As opposed to:

Using assignment, m_obj is initialized twice; with the initializer syntax, only once. So, what's the solution to avoid extra initializations during copy construction? While it goes against your instinct to reuse code, this is one situation where it's best to implement your copy constructor and assignment operator separately, even if they do the same thing. Calling operator= from your copy constructor will certainly work, but it's not the most efficient implementation. My observation about initializers suggests a better way:

Now the main copy ctor calls the member object's copy ctor using an initializer, and m_obj is initialized just once by its copy ctor. In general, copy ctors should invoke the copy ctors of their members. Likewise for assignment. And, I may as well add, the same goes for base classes: your derived copy ctor and assignment operators should invoke the corresponding base class methods. Of course, there are always times when you may want to do something different because you know how your code works—but what I've described are the general rules, which are to be broken only when you have a compelling reason. If you have common tasks to perform after the basic objects have been initialized, you can put them in a common initialization method and call it from your constructors and operator=.

Q Can you tell me how to call a Visual C++® class from C#, and what syntax I need to use for this?

Sunil Peddi

Q I have an application that is written in both C# (the GUI) and in classic C++ (some business logic). Now I need to call from a DLL written in C++ a function (or a method) in a DLL written in Visual C++ .NET. This one calls another DLL written in C#. The Visual C++ .NET DLL acts like a proxy. Is this possible? I was able to use LoadLibrary to call a function present in the Visual C++ .NET DLL, and I can receive a return value, but when I try to pass some parameters to the function in the Visual C++ .NET DLL, I get the following error:

How can I resolve this problem?

Giuseppe Dattilo

A I get a lot of questions about interoperability between the Microsoft® .NET Framework and native C++, so I don't mind revisiting this well-covered topic yet again. There are two directions you can go: calling the Framework from C++ or calling C++ from the Framework. I won't go into COM interop here as that's a separate issue best saved for another day.

Let's start with the easiest one first: calling the Framework from C++. The simplest and easiest way to call the Framework from your C++ program is to use the Managed Extensions. These Microsoft-specific C++ language extensions are designed to make calling the Framework as easy as including a couple of files and then using the classes as if they were written in C++. Here's a very simple C++ program that calls the Framework's Console class:

To use the Managed Extensions, all you need to do is import <mscorlib.dll> and whatever .NET assemblies contain the classes you plan to use. Don't forget to compile with /clr:

Your C++ code can use managed classes more or less as if they were ordinary C++ classes. For example, you can create Framework objects with operator new, and access them using C++ pointer syntax, as shown in the following:

Here, the String s is declared as pointer-to-String because String::Format returns a new String object.

The "Hello, world" and date/time programs seem childishly simple—and they are—but just remember that however complex your program is, however many .NET assemblies and classes you use, the basic idea is the same: use <mscorlib.dll> and whatever other assemblies you need, then create managed objects with new, and use pointer syntax to access them.

So much for calling the Framework from C++. What about going the other way, calling C++ from the Framework? Here the road forks into two options, depending on whether you want to call extern C functions or C++ class member functions. Again, I'll take the simpler case first: calling C functions from .NET. The easiest thing to do here is use P/Invoke. With P/Invoke, you declare the external functions as static methods of a class, using the DllImport attribute to specify that the function lives in an external DLL. In C# it looks like this:

This tells the compiler that MessageBox is a function in user32.dll that takes an IntPtr (HWND), two Strings, and an int. You can then call it from your C# program like so:

Of course, you don't need P/Invoke for MessageBox since the .NET Framework already has a MessageBox class, but there are plenty of API functions that aren't supported directly by the Framework, and then you need P/Invoke. And, of course, you can use P/Invoke to call C functions in your own DLLs. I've used C# in the example, but P/Invoke works with any .NET-based language like Visual Basic® .NET or JScript®.NET. The names are the same, only the syntax is different.

Note that I used IntPtr to declare the HWND. I could have got away with int, but you should always use IntPtr for any kind of handle such as HWND, HANDLE, or HDC. IntPtr will default to 32 or 64 bits depending on the platform, so you never have to worry about the size of the handle.

DllImport has various modifiers you can use to specify details about the imported function. In this example, CharSet=CharSet.Auto tells the Framework to pass Strings as Unicode or Ansi, depending on the target operating system. Another little-known modifier you can use is CallingConvention. Recall that in C, there are different calling conventions, which are the rules that specify how the compiler should pass arguments and return values from one function to another across the stack. The default CallingConvention for DllImport is CallingConvention.Winapi. This is actually a pseudo-convention that uses the default convention for the target platform; for example, StdCall (in which the callee cleans the stack) on Windows® platforms and CDecl (in which the caller cleans the stack) on Windows CE .NET. CDecl is also used with varargs functions like printf.

The calling convention is where Giuseppe ran into trouble. C++ uses yet a third calling convention: thiscall. With this convention, the compiler uses the hardware register ECX to pass the "this" pointer to class member functions that don't have variable arguments. Without knowing the exact details of Giuseppe's program, it sounds from the error message that he's trying to call a C++ member function that expects thiscall from a C# program that's using StdCall—oops!

Aside from calling conventions, another interoperability issue when calling C++ methods from the Framework is linkage: C and C++ use different forms of linkage because C++ requires name-mangling to support function overloading. That's why you have to use extern "C" when you declare C functions in C++ programs: so the compiler won't mangle the name. In Windows, the entire windows.h file (now winuser.h) is enclosed in extern "C" brackets.

While there may be a way to call C++ member functions in a DLL directly using P/Invoke and DllImport with the exact mangled names and CallingConvention=ThisCall, it's not something to attempt if you're in your right mind. The proper way to call C++ classes from managed code—option number two—is to wrap your C++ classes in managed wrappers. Wrapping can be tedious if you have lots of classes, but it's really the only way to go. Say you have a C++ class CWidget and you want to wrap it so .NET clients can use it. The basic formula looks something like this:

The pattern is the same for any class. You write a managed (__gc) class that holds a pointer to the native class, you write a constructor and destructor that allocate and destroy the instance, and you write wrapper methods that call the corresponding native C++ member functions. You don't have to wrap all the member functions, only the ones you want to expose to the managed world.

Figure 2 shows a simple but concrete example in full detail. CPerson is a class that holds the name of a person, with member functions GetName and SetName to change the name. Figure 3 shows the managed wrapper for CPerson. In the example, I converted Get/SetName to a property, so .NET-based programmers can use the property syntax. In C#, using it looks like this:

Figure 3 Managed Person Class

Figure 2 Native CPerson Class

Using properties is purely a matter of style; I could equally well have exposed two methods, GetName and SetName, as in the native class. But properties feel more like .NET. The wrapper class is an assembly like any other, but one that links with the native DLL. This is one of the cool benefits of the Managed Extensions: You can link directly with native C/C++ code. If you download and compile the source for my CPerson example, you'll see that the makefile generates two separate DLLs: person.dll implements a normal native DLL and mperson.dll is the managed assembly that wraps it. There are also two test programs: testcpp.exe, a native C++ program that calls the native person.dll and testcs.exe, which is written in C# and calls the managed wrapper mperson.dll (which in turn calls the native person.dll).

Figure 4** Interop Highway **

I've used a very simple example to highlight the fact that there are fundamentally only a few main highways across the border between the managed and native worlds (see Figure 4 ). If your C++ classes are at all complex, the biggest interop problem you'll encounter is converting parameters between native and managed types, a process called marshaling. The Managed Extensions do an admirable job of making this as painless as possible (for example, automatically converting primitive types and Strings), but there are times where you have to know something about what you're doing.

For example, you can't pass the address of a managed object or subobject to a native function without pinning it first. That's because managed objects live in the managed heap, which the garbage collector is free to rearrange. If the garbage collector moves an object, it can update all the managed references to that object—but it knows nothing of raw native pointers that live outside the managed world. That's what __pin is for; it tells the garbage collector: don't move this object. For strings, the Framework has a special function PtrToStringChars that returns a pinned pointer to the native characters. (Incidentally, for those curious-minded souls, PtrToStringChars is the only function as of this date defined in <vcclr.h>. Figure 5 shows the code.) I used PtrToStringChars in MPerson to set the Name (see Figure 3 ).

Figure 5 PtrToStringChars

Pinning isn't the only interop problem you'll encounter. Other problems arise if you have to deal with arrays, references, structs, and callbacks, or access a subobject within an object. This is where some of the more advanced techniques come in, such as StructLayout, boxing, __value types, and so on. You also need special code to handle exceptions (native or managed) and callbacks/delegates. But don't let these interop details obscure the big picture. First decide which way you're calling (from managed to native or the other way around), and if you're calling from managed to native, whether to use P/Invoke or a wrapper.

In Visual Studio® 2005 (which some of you may already have as beta bits), the Managed Extensions have been renamed and upgraded to something called C++/CLI. Think of the C++/CLI as Managed Extensions Version 2, or What the Managed Extensions Should Have Been. The changes are mostly a matter of syntax, though there are some important semantic changes, too. In general C++/CLI is designed to highlight rather than blur the distinction between managed and native objects. Using pointer syntax for managed objects was a clever and elegant idea, but in the end perhaps a little too clever because it obscures important differences between managed and native objects. C++/CLI introduces the key notion of handles for managed objects, so instead of using C pointer syntax for managed objects, the CLI uses ^ (hat):

As you no doubt noticed, there's also a gcnew operator to clarify when you're allocating objects on the managed heap as opposed to the native one. This has the added benefit that gcnew doesn't collide with C++ new, which can be overloaded or even redefined as a macro. C++/CLI has many other cool features designed to make interoperability as straightforward and intuitive as possible.

Send your questions and comments for Paul to [email protected] .

Paul DiLascia is a freelance software consultant and Web/UI designer-at-large. He is the author of Windows ++: Writing Reusable Windows Code in C ++ (Addison-Wesley, 1992). In his spare time, Paul develops PixieLib, an MFC class library available from his Web site, www.dilascia.com .

Additional resources

Copy assignment operator






(C++20)





(C++20)




(C++11)
(C++11)
(C++11)
(C++17)

General




Members


pointer







(C++11)
specifier
specifier

Special member functions


(C++11)

(C++11)

Inheritance




(C++11)
(C++11)


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)

Explanation

Typical declaration of a copy assignment operator when copy-and-swap idiom can be used.
Typical declaration of a copy assignment operator when copy-and-swap idiom cannot be used (non-swappable type or degraded performance).
Forcing a copy assignment operator to be generated by the compiler.
Avoiding implicit copy assignment.

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.

Implicitly-declared copy assignment operator

each direct base B of T has a copy assignment operator whose parameters are B or const B & or const volatile B & ;
each non-static data member M of T of class type or array of class type has a copy assignment operator whose parameters are M or const M & or const volatile M & .

Deleted implicitly-declared copy assignment operator

A implicitly-declared copy assignment operator for class T is defined as deleted if any of the following is true:

T has a user-declared move constructor;
T has a user-declared move assignment operator.

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:

T has a non-static data member of non-class type (or array thereof) that is const ;
T has a non-static data member of a reference type;
T has a non-static data member or a direct or virtual base class that cannot be copy-assigned (overload resolution for the copy assignment fails, or selects a deleted or inaccessible function);
T is a union-like class , and has a variant member whose corresponding assignment operator is non-trivial.

Trivial copy assignment operator

The copy assignment operator for class T is trivial if all of the following is true:

it is not user-provided (meaning, it is implicitly-defined or defaulted) , , and if it is defaulted, its signature is the same as implicitly-defined (until C++14) ;
T has no virtual member functions;
T has no virtual base classes;
the copy assignment operator selected for every direct base of T is trivial;
the copy assignment operator selected for every non-static class type (or array of class type) member of T is trivial;

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.

Implicitly-defined copy assignment operator

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.

See assignment operator overloading for additional detail on the expected behavior of a user-defined copy-assignment operator.

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++14	operator=(X&) = default was non-trivial	made trivial

C++ Classes and Objects
C++ Polymorphism
C++ Inheritance
C++ Abstraction
C++ Encapsulation
C++ OOPs Interview Questions
C++ OOPs MCQ
C++ Interview Questions
C++ Function Overloading
C++ Programs
C++ Preprocessor
C++ Templates

The Rule of Five in C++

The “ Rule of Five” is a guideline for efficient and bug-free programming in C++. The Rule of Five states that,

If any of the below functions is defined for a class, then it is better to define all of them.

It includes the following functions of a class:

Copy Constructor
Copy Assignment Operator
Move Constructor
Move Assignment Operator

The Rule of Big Five is an extension of the Rule of Three to include move semantics. The Rule of Three, consists of a destructor, copy constructor, and, copy assignment operator, use all these three functions when you are dealing with dynamically allocated resources Whereas The Rule of Five includes two more functions i.e. move constructor and move assignment operator .

Need for Rule of Five in C++

In C++ “The Rule of Big Five” is needed to properly manage the resources and efficiently handle objects. Following are the important reasons why we should use the Rule of the Big Five:

Assume that you have some dynamically allocated resources in the object and you copy them using shallow copy. This will lead to problems like segmentation faults when the source object is destroyed. Similar thing will happen with copy assignment, move constructor and move assignment operator.
By implementing Rule of Big Five properly we can make sure that there are no resource leaks. This can be achieved by ensuring that all the dynamically allocated memory or any other resources are released appropriately.

Let’s discuss each of the five special member functions that a class should define in detail

1. Destructor

The Destructor is used for removing/freeing up all the resources that an object has taken throughout its lifetime.With the help of this destructor, we make sure that any resources taken by objects are properly released before the object is no longer in scope.

2. Copy Constructor

Copy Constructor is used to make a new object by copying an existing object. Copy constructor is invoked when we use it to pass an object by value or when we make a copy explicitly. mainly we use copy constructor to replicate an already existing object.

3. Copy Assignment Operator

Copy Assignment Operator is a special type of function that takes care of assigning the data of one object to another object. It gets called when you use this assignment operator (=) between objects.

4. Move Constructor

Move Constructor is one of the member functions that is used to transfer the ownership of resources from one object to another. This job can easily be done by this move constructor by using a temporary object called rvalue

Explanation: In the above syntax, “noexcept” is used to indicate that the given function, like the move constructor in this case, does not throw any kind of exceptions. The move constructor is specifically designed to handle temporary objects and it requires an rvalue reference (ClassName&& other) as a parameter.

5. Move Assignment Operator

The Move Assignment Operator is comparable to the Move Constructor. It is used when an existing object is assigned the value of an rvalue. It is activated when you use the assignment operator (=) to assign the data of a temporary object(value) to an existing object.

The below example demonstrates the use of all five member functions: Destructor, Copy Constructor, Copy Assignment Operator, Move Constructor, and Move Assignment Operator.

In C++ “The Rule of Big Five” is used to ensure that the memory is properly handled i.e. memory allocation and deallocation is done properly and also resource management. It is an extension of Rule of Three. This rule says that we should use or try to use all the five functions (Destructor, Copy Constructor, Copy Assignment Operator, Move Constructor, and Move Assignment Operator) even if initially we require only one so that we can achieve application optimization, less bugs, efficient code and manage the resources efficiently.

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When to use const class member variables
Therefore, when a class has const member variables and requires object assignment, a custom assignment operator is necessary. This custom operator should handle the assignment of non-const members as needed. As for the copy constructor, the default one provided by the compiler does handle const member variables correctly. It initializes new ...
Copy assignment operator
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.
21.12
21.12 — Overloading the assignment operator. Alex July 22, 2024. The copy assignment operator (operator=) is used to copy values from one object to another already existing object. As of C++11, C++ also supports "Move assignment". We discuss move assignment in lesson 22.3 -- Move constructors and move assignment.
Copy Constructor vs Assignment Operator in C++
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. 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 ...
Copy assignment operator
Copy assignment operator. 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.
When to use const in C++? Part II: member variables
3. 4. MyClassWithConstMember& operator=(const MyClassWithConstMember&) { // ??? return *this; } Do you skip assigning to the const members? Not so great, either you depend on that value somewhere, or you should not store the value. And you cannot assign to a const variable, can you? For a matter of fact, you can….
Copy assignment operator
Copy assignment operator. 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.
PDF Copy Constructors and Assignment Operators
Unless you specify otherwise, C++ will automatically provide objects a basic copy constructor and assignment operator that simply invoke the copy constructors and assignment operators of all the class's data members. In many cases, this is exactly what you want. For example, consider the following class: class MyClass {public: /* Omitted ...
C++ Core Guidelines: Rules for Copy and Move
Here are the eight rules: C.60: Make copy assignment non- virtual, take the parameter by const&, and return by non- const&. C.61: A copy operation should copy. C.62: Make copy assignment safe for self-assignment. C.63: Make move assignment non- virtual, take the parameter by &&, and return by non- const&.
Assignment operators
Correct behavior. CWG 1527. C++11. for assignments to class type objects, the right operand could be an initializer list only when the assignment is defined by a user-defined assignment operator. removed user-defined assignment constraint. CWG 1538. C++11. E1 ={E2} was equivalent to E1 = T(E2) (T is the type of E1), this introduced a C-style cast.
Copy constructors
each non-static data member M of T of class type or array of class type has a copy constructor whose parameters are of type const M& or constvolatile M&. Otherwise, the implicitly-declared copy constructor is T::T(T&). Due to these rules, the implicitly-declared copy constructor cannot bind to a volatile lvalue argument.
22.3
C++11 defines two new functions in service of move semantics: a move constructor, and a move assignment operator. Whereas the goal of the copy constructor and copy assignment is to make a copy of one object to another, the goal of the move constructor and move assignment is to move ownership of the resources from one object to another (which is typically much less expensive than making a copy).
C++ Assignment Operator Overloading
Overloading assignment operator in C++ copies all values of one object to another object. Only a non-static member function should be used to overload the assignment operator. In C++, the compiler automatically provides a default assignment operator for classes. This operator performs a shallow copy of each member of the class from one object ...
Special members
The copy assignment operator is an overload of operator= which takes a value or reference of the class itself as parameter. The return value is generally a reference to *this ... Or even better, since its string member is not constant, it could re-utilize the same string object: 1 2 3 4:
Copy Constructor in C++
Copy Constructor in C++. A copy constructor is a type of constructor that initializes an object using another object of the same class. In simple terms, a constructor which creates an object by initializing it with an object of the same class, which has been created previously is known as a copy constructor. The process of initializing members ...
Copy assignment operator
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 .
The rule of three/five/zero
Rule of three. If a class requires a user-defined destructor, a user-defined copy constructor, or a user-defined copy assignment operator, it almost certainly requires all three.. Because C++ copies and copy-assigns objects of user-defined types in various situations (passing/returning by value, manipulating a container, etc), these special member functions will be called, if accessible, and ...
assignment constructor for const member
Once the constant member is assigned a value, it's value remains constant. Wazzak. Last edited on . Cubbi. A const member simply cannot be assigned to, and non-static const member makes the entire class non-assignable* ... * technically, you could write a copy-assignment operator that doesn't attempt to modify m_val and that will compile, but ...
C++ at Work: Copy Constructors, Assignment Operators, and More
In C++, assignment and copy construction are different because the copy constructor initializes uninitialized memory, whereas assignment starts with an existing initialized object. If your class contains instances of other classes as data members, the copy constructor must first construct these data members before it calls operator=.
Why copy constructor argument should be const in C++?
A copy constructor is a member function that initializes an object using another object of the same class. The Copy constructor is called mainly when a new object is created from an existing object, as a copy of the existing object. In C++, a Copy Constructor may be called for the following cases: 1) When an object of the class is returned by value
Copy assignment operators (C++ only)
Copy assignment operators (C++ only) The copy assignment operator lets you create a new object from an existing one by initialization. A copy assignment operator of a class A is a nonstatic non-template member function that has one of the following forms: A::operator= (A) A::operator= (A&) A::operator= (const A&)
Copy assignment operator
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)
Mastering the Rule of 5 in C++: Best Practices and Pitfalls
It ensures proper copying of data members and resources. Example: class MyClass {public: MyClass& operator=(const MyClass& other) { /* Copy assignment implementation */ }}; Move Assignment Operator: A move assignment operator efficiently transfers resources from one object to another, typically using the assignment operator (=). Example:
The Rule of Five in C++
The "Rule of Five" is a guideline for efficient and bug-free programming in C++. The Rule of Five states that, If any of the below functions is defined for a class, then it is better to define all of them. It includes the following functions of a class: Destructor. Copy Constructor.

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[ edit ] Syntax

[ edit ] Explanation

[ edit ] Implicitly-declared copy assignment operator

[ edit ] Deleted implicitly-declared copy assignment operator

[ edit ] Trivial copy assignment operator

[ edit ] Implicitly-defined copy assignment operator

[ edit ] Notes

[ edit ] Example

Copy assignment operator

[ edit ] Syntax

[ edit ] Explanation

[ edit ] Implicitly-declared copy assignment operator

[ edit ] Deleted implicitly-declared copy assignment operator

[ edit ] Trivial copy assignment operator

[ edit ] Implicitly-defined copy assignment operator

[ edit ] Notes

[ edit ] Copy and swap

[ edit ] Example

When to use const in C++? Part II: member variables

Conclusions

Further Reading

When to use const in C++? Part III: return types

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Copy assignment operator

[ edit ] Syntax

[ edit ] Explanation

[ edit ] Implicitly-declared copy assignment operator

[ edit ] Deleted implicitly-declared copy assignment operator

[ edit ] Trivial copy assignment operator

[ edit ] Implicitly-defined copy assignment operator

[ edit ] Notes

[ edit ] Copy and swap

[ edit ] Example

C++ Core Guidelines: Rules for Copy and Move

Constructor Rules

C.51: Use delegating constructors to represent common actions for all constructors of a class

C.52: Use inheriting constructors to import constructors into a derived class that does not need further explicit initialization

Modernes C++ Mentoring

Copy and Move

C.66: Make move operations noexcept

Standard Seminars (English/German)

Online Seminars (German)

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22.3 — Move constructors and move assignment

assignment constructor for const member

Additional resources

Copy assignment operator

Explanation

Implicitly-declared copy assignment operator

Deleted implicitly-declared copy assignment operator

Trivial copy assignment operator

Implicitly-defined copy assignment operator

Defect reports

The Rule of Five in C++

Need for Rule of Five in C++

1. Destructor

2. Copy Constructor

3. Copy Assignment Operator

4. Move Constructor

5. Move Assignment Operator

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