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Move Constructors and Move Assignment Operators (C++)

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This topic describes how to write a move constructor and a move assignment operator for a C++ class. A move constructor enables the resources owned by an rvalue object to be moved into an lvalue without copying. For more information about move semantics, see Rvalue Reference Declarator: && .

This topic builds upon the following C++ class, MemoryBlock , which manages a memory buffer.

The following procedures describe how to write a move constructor and a move assignment operator for the example C++ class.

To create a move constructor for a C++ class

Define an empty constructor method that takes an rvalue reference to the class type as its parameter, as demonstrated in the following example:

In the move constructor, assign the class data members from the source object to the object that is being constructed:

Assign the data members of the source object to default values. This prevents the destructor from freeing resources (such as memory) multiple times:

To create a move assignment operator for a C++ class

Define an empty assignment operator that takes an rvalue reference to the class type as its parameter and returns a reference to the class type, as demonstrated in the following example:

In the move assignment operator, add a conditional statement that performs no operation if you try to assign the object to itself.

In the conditional statement, free any resources (such as memory) from the object that is being assigned to.

The following example frees the _data member from the object that is being assigned to:

Follow steps 2 and 3 in the first procedure to transfer the data members from the source object to the object that is being constructed:

Return a reference to the current object, as shown in the following example:

Example: Complete move constructor and assignment operator

The following example shows the complete move constructor and move assignment operator for the MemoryBlock class:

Example Use move semantics to improve performance

The following example shows how move semantics can improve the performance of your applications. The example adds two elements to a vector object and then inserts a new element between the two existing elements. The vector class uses move semantics to perform the insertion operation efficiently by moving the elements of the vector instead of copying them.

This example produces the following output:

Before Visual Studio 2010, this example produced the following output:

The version of this example that uses move semantics is more efficient than the version that does not use move semantics because it performs fewer copy, memory allocation, and memory deallocation operations.

Robust Programming

To prevent resource leaks, always free resources (such as memory, file handles, and sockets) in the move assignment operator.

To prevent the unrecoverable destruction of resources, properly handle self-assignment in the move assignment operator.

If you provide both a move constructor and a move assignment operator for your class, you can eliminate redundant code by writing the move constructor to call the move assignment operator. The following example shows a revised version of the move constructor that calls the move assignment operator:

The std::move function converts the lvalue other to an rvalue.

Rvalue Reference Declarator: && std::move

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Additional resources

C++ Operator Overloading Guidelines

One of the nice features of C++ is that you can give special meanings to operators, when they are used with user-defined classes. This is called operator overloading . You can implement C++ operator overloads by providing special member-functions on your classes that follow a particular naming convention. For example, to overload the + operator for your class, you would provide a member-function named operator+ on your class.

  • = (assignment operator)
  • + - * (binary arithmetic operators)
  • += -= *= (compound assignment operators)
  • == != (comparison operators)

Here are some guidelines for implementing these operators. These guidelines are very important to follow, so definitely get in the habit early.

Assignment Operator =

The assignment operator has a signature like this: class MyClass { public: ... MyClass & operator=(const MyClass &rhs); ... } MyClass a, b; ... b = a; // Same as b.operator=(a);

Notice that the = operator takes a const-reference to the right hand side of the assignment. The reason for this should be obvious, since we don't want to change that value; we only want to change what's on the left hand side.

  • e = 42 assigns 42 to e , then returns e as the result
  • The value of e is then assigned to d , and then d is returned as the result
  • The value of d is then assigned to c , and then c is returned as the result

Now, in order to support operator chaining, the assignment operator must return some value. The value that should be returned is a reference to the left-hand side of the assignment.

Notice that the returned reference is not declared const . This can be a bit confusing, because it allows you to write crazy stuff like this: MyClass a, b, c; ... (a = b) = c; // What?? At first glance, you might want to prevent situations like this, by having operator= return a const reference. However, statements like this will work with primitive types. And, even worse, some tools actually rely on this behavior. Therefore, it is important to return a non- const reference from your operator= . The rule of thumb is, "If it's good enough for int s, it's good enough for user-defined data-types."

So, for the hypothetical MyClass assignment operator, you would do something like this: // Take a const-reference to the right-hand side of the assignment. // Return a non-const reference to the left-hand side. MyClass& MyClass::operator=(const MyClass &rhs) { ... // Do the assignment operation! return *this; // Return a reference to myself. } Remember, this is a pointer to the object that the member function is being called on. Since a = b is treated as a.operator=(b) , you can see why it makes sense to return the object that the function is called on; object a is the left-hand side.

But, the member function needs to return a reference to the object, not a pointer to the object. So, it returns *this , which returns what this points at (i.e. the object), not the pointer itself. (In C++, instances are turned into references, and vice versa, pretty much automatically, so even though *this is an instance, C++ implicitly converts it into a reference to the instance.)

Now, one more very important point about the assignment operator:

YOU MUST CHECK FOR SELF-ASSIGNMENT!

This is especially important when your class does its own memory allocation. Here is why: The typical sequence of operations within an assignment operator is usually something like this: MyClass& MyClass::operator=(const MyClass &rhs) { // 1. Deallocate any memory that MyClass is using internally // 2. Allocate some memory to hold the contents of rhs // 3. Copy the values from rhs into this instance // 4. Return *this } Now, what happens when you do something like this: MyClass mc; ... mc = mc; // BLAMMO. You can hopefully see that this would wreak havoc on your program. Because mc is on the left-hand side and on the right-hand side, the first thing that happens is that mc releases any memory it holds internally. But, this is where the values were going to be copied from, since mc is also on the right-hand side! So, you can see that this completely messes up the rest of the assignment operator's internals.

The easy way to avoid this is to CHECK FOR SELF-ASSIGNMENT. There are many ways to answer the question, "Are these two instances the same?" But, for our purposes, just compare the two objects' addresses. If they are the same, then don't do assignment. If they are different, then do the assignment.

So, the correct and safe version of the MyClass assignment operator would be this: MyClass& MyClass::operator=(const MyClass &rhs) { // Check for self-assignment! if (this == &rhs) // Same object? return *this; // Yes, so skip assignment, and just return *this. ... // Deallocate, allocate new space, copy values... return *this; } Or, you can simplify this a bit by doing: MyClass& MyClass::operator=(const MyClass &rhs) { // Only do assignment if RHS is a different object from this. if (this != &rhs) { ... // Deallocate, allocate new space, copy values... } return *this; } Remember that in the comparison, this is a pointer to the object being called, and &rhs is a pointer to the object being passed in as the argument. So, you can see that we avoid the dangers of self-assignment with this check.

  • Take a const-reference for the argument (the right-hand side of the assignment).
  • Return a reference to the left-hand side, to support safe and reasonable operator chaining. (Do this by returning *this .)
  • Check for self-assignment, by comparing the pointers ( this to &rhs ).

Compound Assignment Operators += -= *=

I discuss these before the arithmetic operators for a very specific reason, but we will get to that in a moment. The important point is that these are destructive operators, because they update or replace the values on the left-hand side of the assignment. So, you write: MyClass a, b; ... a += b; // Same as a.operator+=(b) In this case, the values within a are modified by the += operator.

How those values are modified isn't very important - obviously, what MyClass represents will dictate what these operators mean.

The member function signature for such an operator should be like this: MyClass & MyClass::operator+=(const MyClass &rhs) { ... } We have already covered the reason why rhs is a const-reference. And, the implementation of such an operation should also be straightforward.

But, you will notice that the operator returns a MyClass -reference, and a non-const one at that. This is so you can do things like this: MyClass mc; ... (mc += 5) += 3;

Don't ask me why somebody would want to do this, but just like the normal assignment operator, this is allowed by the primitive data types. Our user-defined datatypes should match the same general characteristics of the primitive data types when it comes to operators, to make sure that everything works as expected.

This is very straightforward to do. Just write your compound assignment operator implementation, and return *this at the end, just like for the regular assignment operator. So, you would end up with something like this: MyClass & MyClass::operator+=(const MyClass &rhs) { ... // Do the compound assignment work. return *this; }

As one last note, in general you should beware of self-assignment with compound assignment operators as well. Fortunately, none of the C++ track's labs require you to worry about this, but you should always give it some thought when you are working on your own classes.

Binary Arithmetic Operators + - *

The binary arithmetic operators are interesting because they don't modify either operand - they actually return a new value from the two arguments. You might think this is going to be an annoying bit of extra work, but here is the secret:

Define your binary arithmetic operators using your compound assignment operators.

There, I just saved you a bunch of time on your homeworks.

So, you have implemented your += operator, and now you want to implement the + operator. The function signature should be like this: // Add this instance's value to other, and return a new instance // with the result. const MyClass MyClass::operator+(const MyClass &other) const { MyClass result = *this; // Make a copy of myself. Same as MyClass result(*this); result += other; // Use += to add other to the copy. return result; // All done! } Simple!

Actually, this explicitly spells out all of the steps, and if you want, you can combine them all into a single statement, like so: // Add this instance's value to other, and return a new instance // with the result. const MyClass MyClass::operator+(const MyClass &other) const { return MyClass(*this) += other; } This creates an unnamed instance of MyClass , which is a copy of *this . Then, the += operator is called on the temporary value, and then returns it.

If that last statement doesn't make sense to you yet, then stick with the other way, which spells out all of the steps. But, if you understand exactly what is going on, then you can use that approach.

You will notice that the + operator returns a const instance, not a const reference. This is so that people can't write strange statements like this: MyClass a, b, c; ... (a + b) = c; // Wuh...? This statement would basically do nothing, but if the + operator returns a non- const value, it will compile! So, we want to return a const instance, so that such madness will not even be allowed to compile.

  • Implement the compound assignment operators from scratch, and then define the binary arithmetic operators in terms of the corresponding compound assignment operators.
  • Return a const instance, to prevent worthless and confusing assignment operations that shouldn't be allowed.

Comparison Operators == and !=

The comparison operators are very simple. Define == first, using a function signature like this: bool MyClass::operator==(const MyClass &other) const { ... // Compare the values, and return a bool result. } The internals are very obvious and straightforward, and the bool return-value is also very obvious.

The important point here is that the != operator can also be defined in terms of the == operator, and you should do this to save effort. You can do something like this: bool MyClass::operator!=(const MyClass &other) const { return !(*this == other); } That way you get to reuse the hard work you did on implementing your == operator. Also, your code is far less likely to exhibit inconsistencies between == and != , since one is implemented in terms of the other.

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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:

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?

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…

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:

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 !

Further Reading

When to use const in c++ part i: functions and local 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 th...

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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21.12 — Overloading the assignment operator

The copy assignment operator (operator=) is used to copy values from one object to another already existing object .

Related content

As of C++11, C++ also supports “Move assignment”. We discuss move assignment in lesson 22.3 -- Move constructors and move assignment .

Copy assignment vs Copy constructor

The purpose of the copy constructor and the copy assignment operator are almost equivalent -- both copy one object to another. However, the copy constructor initializes new objects, whereas the assignment operator replaces the contents of existing objects.

The difference between the copy constructor and the copy assignment operator causes a lot of confusion for new programmers, but it’s really not all that difficult. Summarizing:

  • If a new object has to be created before the copying can occur, the copy constructor is used (note: this includes passing or returning objects by value).
  • If a new object does not have to be created before the copying can occur, the assignment operator is used.

Overloading the assignment operator

Overloading the copy assignment operator (operator=) is fairly straightforward, with one specific caveat that we’ll get to. The copy assignment operator must be overloaded as a member function.

This prints:

This should all be pretty straightforward by now. Our overloaded operator= returns *this, so that we can chain multiple assignments together:

Issues due to self-assignment

Here’s where things start to get a little more interesting. C++ allows self-assignment:

This will call f1.operator=(f1), and under the simplistic implementation above, all of the members will be assigned to themselves. In this particular example, the self-assignment causes each member to be assigned to itself, which has no overall impact, other than wasting time. In most cases, a self-assignment doesn’t need to do anything at all!

However, in cases where an assignment operator needs to dynamically assign memory, self-assignment can actually be dangerous:

First, run the program as it is. You’ll see that the program prints “Alex” as it should.

Now run the following program:

You’ll probably get garbage output. What happened?

Consider what happens in the overloaded operator= when the implicit object AND the passed in parameter (str) are both variable alex. In this case, m_data is the same as str.m_data. The first thing that happens is that the function checks to see if the implicit object already has a string. If so, it needs to delete it, so we don’t end up with a memory leak. In this case, m_data is allocated, so the function deletes m_data. But because str is the same as *this, the string that we wanted to copy has been deleted and m_data (and str.m_data) are dangling.

Later on, we allocate new memory to m_data (and str.m_data). So when we subsequently copy the data from str.m_data into m_data, we’re copying garbage, because str.m_data was never initialized.

Detecting and handling self-assignment

Fortunately, we can detect when self-assignment occurs. Here’s an updated implementation of our overloaded operator= for the MyString class:

By checking if the address of our implicit object is the same as the address of the object being passed in as a parameter, we can have our assignment operator just return immediately without doing any other work.

Because this is just a pointer comparison, it should be fast, and does not require operator== to be overloaded.

When not to handle self-assignment

Typically the self-assignment check is skipped for copy constructors. Because the object being copy constructed is newly created, the only case where the newly created object can be equal to the object being copied is when you try to initialize a newly defined object with itself:

In such cases, your compiler should warn you that c is an uninitialized variable.

Second, the self-assignment check may be omitted in classes that can naturally handle self-assignment. Consider this Fraction class assignment operator that has a self-assignment guard:

If the self-assignment guard did not exist, this function would still operate correctly during a self-assignment (because all of the operations done by the function can handle self-assignment properly).

Because self-assignment is a rare event, some prominent C++ gurus recommend omitting the self-assignment guard even in classes that would benefit from it. We do not recommend this, as we believe it’s a better practice to code defensively and then selectively optimize later.

The copy and swap idiom

A better way to handle self-assignment issues is via what’s called the copy and swap idiom. There’s a great writeup of how this idiom works on Stack Overflow .

The implicit copy assignment operator

Unlike other operators, the compiler will provide an implicit public copy assignment operator for your class if you do not provide a user-defined one. This assignment operator does memberwise assignment (which is essentially the same as the memberwise initialization that default copy constructors do).

Just like other constructors and operators, you can prevent assignments from being made by making your copy assignment operator private or using the delete keyword:

Note that if your class has const members, the compiler will instead define the implicit operator= as deleted. This is because const members can’t be assigned, so the compiler will assume your class should not be assignable.

If you want a class with const members to be assignable (for all members that aren’t const), you will need to explicitly overload operator= and manually assign each non-const member.

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Assignment operators

Assignment operators modify the value of the object.

Explanation

copy assignment operator replaces the contents of the object a with a copy of the contents of b ( b is not modified). For class types, this is a special member function, described in copy assignment operator .

move assignment operator replaces the contents of the object a with the contents of b , avoiding copying if possible ( b may be modified). For class types, this is a special member function, described in move assignment operator . (since C++11)

For non-class types, copy and move assignment are indistinguishable and are referred to as direct assignment .

compound assignment operators replace the contents of the object a with the result of a binary operation between the previous value of a and the value of b .

Builtin direct assignment

The direct assignment expressions have the form

For the built-in operator, lhs may have any non-const scalar type and rhs must be implicitly convertible to the type of lhs .

The direct assignment operator expects a modifiable lvalue as its left operand and an rvalue expression or a braced-init-list (since C++11) as its right operand, and returns an lvalue identifying the left operand after modification.

For non-class types, the right operand is first implicitly converted to the cv-unqualified type of the left operand, and then its value is copied into the object identified by left operand.

When the left operand has reference type, the assignment operator modifies the referred-to object.

If the left and the right operands identify overlapping objects, the behavior is undefined (unless the overlap is exact and the type is the same)

In overload resolution against user-defined operators , for every type T , the following function signatures participate in overload resolution:

For every enumeration or pointer to member type T , optionally volatile-qualified, the following function signature participates in overload resolution:

For every pair A1 and A2, where A1 is an arithmetic type (optionally volatile-qualified) and A2 is a promoted arithmetic type, the following function signature participates in overload resolution:

Builtin compound assignment

The compound assignment expressions have the form

The behavior of every builtin compound-assignment expression E1 op = E2 (where E1 is a modifiable lvalue expression and E2 is an rvalue expression or a braced-init-list (since C++11) ) is exactly the same as the behavior of the expression E1 = E1 op E2 , except that the expression E1 is evaluated only once and that it behaves as a single operation with respect to indeterminately-sequenced function calls (e.g. in f ( a + = b, g ( ) ) , the += is either not started at all or is completed as seen from inside g ( ) ).

In overload resolution against user-defined operators , for every pair A1 and A2, where A1 is an arithmetic type (optionally volatile-qualified) and A2 is a promoted arithmetic type, the following function signatures participate in overload resolution:

For every pair I1 and I2, where I1 is an integral type (optionally volatile-qualified) and I2 is a promoted integral type, the following function signatures participate in overload resolution:

For every optionally cv-qualified object type T , the following function signatures participate in overload resolution:

Operator precedence

Operator overloading

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Const member functions in C++

Constant member functions are those functions that are denied permission to change the values of the data members of their class. To make a member function constant, the keyword const is appended to the function prototype and also to the function definition header.

Like member functions and member function arguments, the objects of a class can also be declared as const. An object declared as const cannot be modified and hence, can invoke only const member functions as these functions ensure not to modify the object. A const object can be created by prefixing the const keyword to the object declaration. Any attempt to change the data member of const objects results in a compile-time error.

The const member function can be defined in three ways:

1. For function declaration within a class.

2. for function definition within the class declaration., 3. for function definition outside the class., important points.

  • When a function is declared as const, it can be called on any type of object, const object as well as non-const objects.
  • Whenever an object is declared as const, it needs to be initialized at the time of declaration. however, the object initialization while declaring is possible only with the help of constructors.
  • A function becomes const when the const keyword is used in the function’s declaration. The idea of const functions is not to allow them to modify the object on which they are called.
  • It is recommended practice to make as many functions const as possible so that accidental changes to objects are avoided.

Examples of Const Member Functions

The below C++ program demonstrates that data members can be updated in a member function that is not constant.

The below C++ program demonstrates that data cannot be updated in a Constant member function.

The below C++ code demonstrates how to define constant member functions outside the class definition and showcases the usage of a constant member function to set and retrieve the value of a private member variable.

The below C++ program demonstrates that const functions can be called by non-const objects.

When a function is declared as const, it can be called on any type of object. Non-const functions can only be called by non-const objects.

For example, the following program has compiler errors.

Let’s look at another example:

FAQs on Const Member Functions

Q1. can const objects of a class call non-const member functions.

No,an object declared as const cannot be modified and hence, can invoke only const member functions as these functions ensure not to modify the object.

Q2. Can non-const objects of a class call const member function?

When a function is declared as const, it can be called on any type of object.

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cppreference.com

Non-static member functions.

A non-static member function is a function that is declared in a member specification of a class without a static or friend specifier (see static member functions and friend declaration for the effect of those keywords).

Constructors , destructors , and conversion functions use special syntaxes for their declarations. The rules described in this page may not apply to these functions. See their respective pages for details.

An implicit object member function is a non-static member function without an explicit object parameter (prior to C++23, this was the only kind of non-static member function, and hence referred to as "non-static member function" in the literature).

[ edit ] Explanation

Any function declarations are allowed, with additional syntax elements that are only available for non-static member functions: pure-specifiers , cv-qualifiers , ref-qualifiers, final and override specifiers (since C++11) , and member initialization lists .

A non-static member function of class X may be called

Calling a non-static member function of class X on an object that is not of type X , or of a type derived from X invokes undefined behavior.

Within the body of a non-static member function of X , any id-expression e (e.g. an identifier) that resolves to a non-type non-static member of X or of a base class of X , is transformed to a member access expression ( * this ) . e (unless it's already a part of a member access expression). This does not occur in template definition context, so a name may have to be prefixed with this - > explicitly to become dependent .

Within the body of a non-static member function of X , any unqualified-id that resolves to a static member, an enumerator or a nested type of X or of a base class of X , is transformed to the corresponding qualified-id:

[ edit ] Member functions with cv-qualifiers

An implicit object member function can be declared with a cv-qualifier sequence ( const , volatile , or a combination of const and volatile ), this sequence appears after the parameter list in the function declaration . Functions with different cv-qualifier sequences (or no sequence) have different types and so may overload each other.

In the body of a function with a cv-qualifier sequence, * this is cv-qualified, e.g. in a member function with const qualifier, only other member functions with const qualifier may be called normally. A member function without const qualifier may still be called if const_cast is applied or through an access path that does not involve this .

[ edit ] Virtual and pure virtual functions

A non-static member function may be declared virtual or pure virtual . See virtual functions and abstract classes for details.

[ edit ] Special member functions

Some member functions are special : under certain circumstances they are defined by the compiler even if not defined by the user. They are:

  • Default constructor
  • Copy constructor
  • Copy assignment operator
  • Destructor (until C++20) Prospective destructor (since C++20)

Special member functions along with the comparison operators (since C++20) are the only functions that can be defaulted , that is, defined using = default instead of the function body (see their pages for details).

[ edit ] Notes

[ edit ] example, [ edit ] defect reports.

The following behavior-changing defect reports were applied retroactively to previously published C++ standards.

[ edit ] See also

  • non-static data members
  • static data members
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    assignment constructor for const member assignment constructor for const member Nov 22, 2011 at 1:15pm csxpeng (4) Hi all, is there a good way to define a assignment constructor for const members? Below code doesnot compile for function Foo& operator= (const Foo &f), Edit & run on cpp.sh Nov 22, 2011 at 1:18pm closed account ( zb0S216C)

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