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Default Assignment Operator and References in C++

We have discussed assignment operator overloading for dynamically allocated resources here . In this article, we discussed that when we don’t write our own assignment operator, the compiler creates an assignment operator itself that does shallow copy and thus causes problems. The difference between shallow copy and deep copy becomes visible when the class has pointers as member fields. However, there is no difference when pointers are not used. What happens when we have references in our class and there is no user-defined assignment operator. 

For example, predict the output of the following program,

The compiler doesn’t create default assignment operator in the following cases:

1. Class has a non-static data member of a const type or a reference type.  2. Class has a non-static data member of a type that has an inaccessible copy assignment operator.  3. Class is derived from a base class with an inaccessible copy assignment operator.

When any of the above conditions is true, the user must define the assignment operator. For example, if we add an assignment operator to the above code, the code works fine without any error.

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Explicitly Defaulted and Deleted Functions

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In C++11, defaulted and deleted functions give you explicit control over whether the special member functions are automatically generated. Deleted functions also give you simple language to prevent problematic type promotions from occurring in arguments to functions of all types—special member functions, and normal member functions and nonmember functions—which would otherwise cause an unwanted function call.

Benefits of explicitly defaulted and deleted functions

In C++, the compiler automatically generates the default constructor, copy constructor, copy-assignment operator, and destructor for a type if it doesn't declare its own. These functions are known as the special member functions , and they're what make simple user-defined types in C++ behave like structures do in C. That is, you can create, copy, and destroy them without extra coding effort. C++11 brings move semantics to the language and adds the move constructor and move-assignment operator to the list of special member functions that the compiler can automatically generate.

This is convenient for simple types, but complex types often define one or more of the special member functions themselves, and this can prevent other special member functions from being automatically generated. In practice:

If any constructor is explicitly declared, then no default constructor is automatically generated.

If a virtual destructor is explicitly declared, then no default destructor is automatically generated.

If a move constructor or move-assignment operator is explicitly declared, then:

No copy constructor is automatically generated.

No copy-assignment operator is automatically generated.

If a copy constructor, copy-assignment operator, move constructor, move-assignment operator, or destructor is explicitly declared, then:

No move constructor is automatically generated.

No move-assignment operator is automatically generated.

Additionally, the C++11 standard specifies the following additional rules:

  • If a copy constructor or destructor is explicitly declared, then automatic generation of the copy-assignment operator is deprecated.
  • If a copy-assignment operator or destructor is explicitly declared, then automatic generation of the copy constructor is deprecated.

In both cases, Visual Studio continues to automatically generate the necessary functions implicitly, and doesn't emit a warning by default. Since Visual Studio 2022 version 17.7, C5267 can be enabled to emit a warning.

The consequences of these rules can also leak into object hierarchies. For example, if for any reason a base class fails to have a default constructor that's callable from a deriving class—that is, a public or protected constructor that takes no parameters—then a class that derives from it can't automatically generate its own default constructor.

These rules can complicate the implementation of what should be straight-forward, user-defined types and common C++ idioms—for example, making a user-defined type noncopyable by declaring the copy constructor and copy-assignment operator privately and not defining them.

Before C++11, this code snippet was the idiomatic form of noncopyable types. However, it has several problems:

The copy constructor has to be declared privately to hide it, but because it's declared at all, automatic generation of the default constructor is prevented. You have to explicitly define the default constructor if you want one, even if it does nothing.

Even if the explicitly defined default constructor does nothing, the compiler considers it to be nontrivial. It's less efficient than an automatically generated default constructor and prevents noncopyable from being a true POD type.

Even though the copy constructor and copy-assignment operator are hidden from outside code, the member functions and friends of noncopyable can still see and call them. If they're declared but not defined, calling them causes a linker error.

Although this is a commonly accepted idiom, the intent isn't clear unless you understand all of the rules for automatic generation of the special member functions.

In C++11, the noncopyable idiom can be implemented in a way that is more straightforward.

Notice how the problems with the pre-C++11 idiom are resolved:

Generation of the default constructor is still prevented by declaring the copy constructor, but you can bring it back by explicitly defaulting it.

Explicitly defaulted special member functions are still considered trivial, so there's no performance penalty, and noncopyable isn't prevented from being a true POD type.

The copy constructor and copy-assignment operator are public but deleted. It's a compile-time error to define or call a deleted function.

The intent is clear to anyone who understands =default and =delete . You don't have to understand the rules for automatic generation of special member functions.

Similar idioms exist for making user-defined types that are nonmovable, that can only be dynamically allocated, or that can't be dynamically allocated. Each of these idioms have pre-C++11 implementations that suffer similar problems, and that are similarly resolved in C++11 by implementing them in terms of defaulted and deleted special member functions.

Explicitly defaulted functions

You can default any of the special member functions—to explicitly state that the special member function uses the default implementation, to define the special member function with a nonpublic access qualifier, or to reinstate a special member function whose automatic generation was prevented by other circumstances.

You default a special member function by declaring it as in this example:

Notice that you can default a special member function outside the body of a class as long as it's inlinable.

Because of the performance benefits of trivial special member functions, we recommend that you prefer automatically generated special member functions over empty function bodies when you want the default behavior. You can do this either by explicitly defaulting the special member function, or by not declaring it (and also not declaring other special member functions that would prevent it from being automatically generated.)

Deleted functions

You can delete special member functions and normal member functions and nonmember functions to prevent them from being defined or called. Deleting of special member functions provides a cleaner way of preventing the compiler from generating special member functions that you don't want. The function must be deleted as it's declared; it can't be deleted afterwards in the way that a function can be declared and then later defaulted.

Deleting of normal member function or nonmember functions prevents problematic type promotions from causing an unintended function to be called. This works because deleted functions still participate in overload resolution and provide a better match than the function that could be called after the types are promoted. The function call resolves to the more-specific—but deleted—function and causes a compiler error.

Notice in the preceding sample that calling call_with_true_double_only by using a float argument would cause a compiler error, but calling call_with_true_double_only by using an int argument wouldn't; in the int case, the argument will be promoted from int to double and successfully call the double version of the function, even though that might not be what you intend. To ensure that any call to this function by using a non-double argument causes a compiler error, you can declare a template version of the deleted function.

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Operator overloading.

Customizes the C++ operators for operands of user-defined types.

[ edit ] Syntax

Overloaded operators are functions with special function names:

[ edit ] Overloaded operators

When an operator appears in an expression , and at least one of its operands has a class type or an enumeration type , then overload resolution is used to determine the user-defined function to be called among all the functions whose signatures match the following:

Note: for overloading co_await , (since C++20) user-defined conversion functions , user-defined literals , allocation and deallocation see their respective articles.

Overloaded operators (but not the built-in operators) can be called using function notation:

[ edit ] Restrictions

  • The operators :: (scope resolution), . (member access), .* (member access through pointer to member), and ?: (ternary conditional) cannot be overloaded.
  • New operators such as ** , <> , or &| cannot be created.
  • It is not possible to change the precedence, grouping, or number of operands of operators.
  • The overload of operator -> must either return a raw pointer, or return an object (by reference or by value) for which operator -> is in turn overloaded.
  • The overloads of operators && and || lose short-circuit evaluation.

[ edit ] Canonical implementations

Besides the restrictions above, the language puts no other constraints on what the overloaded operators do, or on the return type (it does not participate in overload resolution), but in general, overloaded operators are expected to behave as similar as possible to the built-in operators: operator + is expected to add, rather than multiply its arguments, operator = is expected to assign, etc. The related operators are expected to behave similarly ( operator + and operator + = do the same addition-like operation). The return types are limited by the expressions in which the operator is expected to be used: for example, assignment operators return by reference to make it possible to write a = b = c = d , because the built-in operators allow that.

Commonly overloaded operators have the following typical, canonical forms: [1]

[ edit ] Assignment operator

The assignment operator ( operator = ) has special properties: see copy assignment and move assignment for details.

The canonical copy-assignment operator is expected to be safe on self-assignment , and to return the lhs by reference:

In those situations where copy assignment cannot benefit from resource reuse (it does not manage a heap-allocated array and does not have a (possibly transitive) member that does, such as a member std::vector or std::string ), there is a popular convenient shorthand: the copy-and-swap assignment operator, which takes its parameter by value (thus working as both copy- and move-assignment depending on the value category of the argument), swaps with the parameter, and lets the destructor clean it up.

This form automatically provides strong exception guarantee , but prohibits resource reuse.

[ edit ] Stream extraction and insertion

The overloads of operator>> and operator<< that take a std:: istream & or std:: ostream & as the left hand argument are known as insertion and extraction operators. Since they take the user-defined type as the right argument ( b in a @ b ), they must be implemented as non-members.

These operators are sometimes implemented as friend functions .

[ edit ] Function call operator

When a user-defined class overloads the function call operator, operator ( ) , it becomes a FunctionObject type.

An object of such a type can be used in a function call expression:

Many standard algorithms, from std:: sort to std:: accumulate accept FunctionObject s to customize behavior. There are no particularly notable canonical forms of operator ( ) , but to illustrate the usage:

[ edit ] Increment and decrement

When the postfix increment or decrement operator appears in an expression, the corresponding user-defined function ( operator ++ or operator -- ) is called with an integer argument 0 . Typically, it is implemented as T operator ++ ( int ) or T operator -- ( int ) , where the argument is ignored. The postfix increment and decrement operators are usually implemented in terms of the prefix versions:

Although the canonical implementations of the prefix increment and decrement operators return by reference, as with any operator overload, the return type is user-defined; for example the overloads of these operators for std::atomic return by value.

[ edit ] Binary arithmetic operators

Binary operators are typically implemented as non-members to maintain symmetry (for example, when adding a complex number and an integer, if operator+ is a member function of the complex type, then only complex + integer would compile, and not integer + complex ). Since for every binary arithmetic operator there exists a corresponding compound assignment operator, canonical forms of binary operators are implemented in terms of their compound assignments:

[ edit ] Comparison operators

Standard algorithms such as std:: sort and containers such as std:: set expect operator < to be defined, by default, for the user-provided types, and expect it to implement strict weak ordering (thus satisfying the Compare requirements). An idiomatic way to implement strict weak ordering for a structure is to use lexicographical comparison provided by std::tie :

Typically, once operator < is provided, the other relational operators are implemented in terms of operator < .

Likewise, the inequality operator is typically implemented in terms of operator == :

When three-way comparison (such as std::memcmp or std::string::compare ) is provided, all six two-way comparison operators may be expressed through that:

[ edit ] Array subscript operator

User-defined classes that provide array-like access that allows both reading and writing typically define two overloads for operator [ ] : const and non-const variants:

If the value type is known to be a scalar type, the const variant should return by value.

Where direct access to the elements of the container is not wanted or not possible or distinguishing between lvalue c [ i ] = v ; and rvalue v = c [ i ] ; usage, operator [ ] may return a proxy. See for example std::bitset::operator[] .

Because a subscript operator can only take one subscript until C++23, to provide multidimensional array access semantics, e.g. to implement a 3D array access a [ i ] [ j ] [ k ] = x ; , operator [ ] has to return a reference to a 2D plane, which has to have its own operator [ ] which returns a reference to a 1D row, which has to have operator [ ] which returns a reference to the element. To avoid this complexity, some libraries opt for overloading operator ( ) instead, so that 3D access expressions have the Fortran-like syntax a ( i, j, k ) = x ; .

[ edit ] Bitwise arithmetic operators

User-defined classes and enumerations that implement the requirements of BitmaskType are required to overload the bitwise arithmetic operators operator & , operator | , operator ^ , operator~ , operator & = , operator | = , and operator ^ = , and may optionally overload the shift operators operator << operator >> , operator >>= , and operator <<= . The canonical implementations usually follow the pattern for binary arithmetic operators described above.

[ edit ] Boolean negation operator

[ edit ] rarely overloaded operators.

The following operators are rarely overloaded:

  • The address-of operator, operator & . If the unary & is applied to an lvalue of incomplete type and the complete type declares an overloaded operator & , it is unspecified whether the operator has the built-in meaning or the operator function is called. Because this operator may be overloaded, generic libraries use std::addressof to obtain addresses of objects of user-defined types. The best known example of a canonical overloaded operator& is the Microsoft class CComPtrBase . An example of this operator's use in EDSL can be found in boost.spirit .
  • The boolean logic operators, operator && and operator || . Unlike the built-in versions, the overloads cannot implement short-circuit evaluation. Also unlike the built-in versions, they do not sequence their left operand before the right one. (until C++17) In the standard library, these operators are only overloaded for std::valarray .
  • The comma operator, operator, . Unlike the built-in version, the overloads do not sequence their left operand before the right one. (until C++17) Because this operator may be overloaded, generic libraries use expressions such as a, void ( ) ,b instead of a,b to sequence execution of expressions of user-defined types. The boost library uses operator, in boost.assign , boost.spirit , and other libraries. The database access library SOCI also overloads operator, .
  • The member access through pointer to member operator - > * . There are no specific downsides to overloading this operator, but it is rarely used in practice. It was suggested that it could be part of a smart pointer interface , and in fact is used in that capacity by actors in boost.phoenix . It is more common in EDSLs such as cpp.react .

[ edit ] Notes

[ edit ] example, [ edit ] defect reports.

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

[ edit ] See also

  • Operator precedence
  • Alternative operator syntax
  • Argument-dependent lookup

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Derived classes still have a default assignment operator

If a class does not declare an explicit assignment operator ( operator=() ), a default operator will be created by the C++ compiler which will copy over each of the data members of the class. A class can (and often they do) override this behaviour by declaring the assignment operator. If a class derives from a base class with a non-default assignment operator, but does not have an assignment operator itself , the C++ compiler will still create a default operator which copies the members of the derived (but not the base) class.

This above may seem like an obvious statement to experienced C++ users, but a recent question came up which showed to me that less experienced people can be tripped up by this issue, so here is a quick review of the above.

A class without an assignment operator, e.g.,

will have a default assignment operator defined by the compiler that is functionally similar to:

Declaring a assignment operator overrides this behaviour, e.g.,

in this case the operator does not copy the members of the class.

If a class is derived from such a class with non-default operator, but itself does not declare this operator, e.g.,

then a default operator will created by the C++ compiler that will copy over the members of the derived class only. In the above case this means y will be copied on assignment while x will not be copied.

Here is the complete program illustrating this:

IMAGES

  1. Assignment Operators in C

    c declare default assignment operator

  2. [100% Working Code]

    c declare default assignment operator

  3. C# Assignment Operator

    c declare default assignment operator

  4. C++ : c++ default assignment operator

    c declare default assignment operator

  5. Assignment Operators in C » PREP INSTA

    c declare default assignment operator

  6. C programming +=

    c declare default assignment operator

VIDEO

  1. How to Define a Class in C#

  2. Assignment Operator in C Programming

  3. Augmented assignment operators in C

  4. Assignment Operator in C Programming

  5. "Banks Are About To Declare Default & Property Values Will Plunge As Much As 90%"

  6. C Programming Tutorial

COMMENTS

  1. Copy assignment operator

    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 operatoIf no user-defined copy assignment operators are provided for a class type, the compiler will always declare one as an inline public member of the class.

  2. c++

    The C++17 standard states: 15.8.2 Copy/move assignment operator [class.copy.assign]. 10 A copy/move assignment operator for a class X that is defaulted and not defined as deleted is implicitly defined when it is odr-used (6.2) (e.g., when it is selected by overload resolution to assign to an object of its class type) or when it is explicitly defaulted after its first declaration.

  3. Default Assignment Operator and References in C++

    The compiler doesn't create default assignment operator in the following cases: 1. Class has a non-static data member of a const type or a reference type. 2. Class has a non-static data member of a type that has an inaccessible copy assignment operator. 3.

  4. Assignment operators

    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. it is equivalent to E1 = T{E2}

  5. Assignment operator (C++)

    In the C++ programming language, the assignment operator, =, is the operator used for assignment. Like most other operators in C++, it can be overloaded . The copy assignment operator, often just called the "assignment operator", is a special case of assignment operator where the source (right-hand side) and destination (left-hand side) are of ...

  6. Explicitly Defaulted and Deleted Functions

    In C++, the compiler automatically generates the default constructor, copy constructor, copy-assignment operator, and destructor for a type if it doesn't declare its own. These functions are known as the special member functions, and they're what make simple user-defined types in C++ behave like structures do in C. That is, you can create, copy ...

  7. Everything You Need To Know About The Copy Assignment Operator In C++

    The Copy Assignment Operator in a class is a non-template non-static member function that is declared with the operator=. When you create a class or a type that is copy assignable (that you can copy with the = operator symbol), it must have a public copy assignment operator. Here is a simple syntax for the typical declaration of a copy ...

  8. What Is A Forced (Default) Copy Assignment Operator In Modern C++

    The forced copy assignment operator forces a copy assignment operator to be generated by the compiler and it is default in any class declarations. This means you don't need to declare it as above, let's give examples without using it. Let's give a simple C++ example to copy assignment operator with default option, here is a simple class.

  9. operator overloading

    In those situations where copy assignment cannot benefit from resource reuse (it does not manage a heap-allocated array and does not have a (possibly transitive) member that does, such as a member std::vector or std::string), there is a popular convenient shorthand: the copy-and-swap assignment operator, which takes its parameter by value (thus working as both copy- and move-assignment ...

  10. Derived classes still have a default assignment operator

    If a class does not declare an explicit assignment operator ( operator= () ), a default operator will be created by the C++ compiler which will copy over each of the data members of the class. A class can (and often they do) override this behaviour by declaring the assignment operator. If a class derives from a base class with a non-default ...

  11. c++

    Suppose I declare two person variables: person a; person b; where a.name = "George", a.ssn = 1, and b.name = "Fred" and b.ssn = 2. ... The default assignment operator in C++ uses Memberwise Assignment to copy the values. That is it effectively assigns all members to each other. In this case that would cause b to have the same values as a.

  12. compiler errors

    1 Answer. Sorted by: 4. From the C++11 draft, section 12.8: The implicitly-defined copy/move assignment operator for a non-union class X performs memberwise copy/move assignment of its subobjects. The direct base classes of X are assigned first, in the order of their declaration in the base-specifier-list, and then the immediate non-static data ...

  13. What is Implicitly-declared Copy Assignment Operator In C++?

    In modern C++, the compiler declares an inline copy assignment operator as a public member of the class. This is called as Implicitly-declared copy assignment operator. This operator may have two different forms, with const T& and T& as below, and, These may depends on the parameters used in assignment operator.

  14. C++ default assignment operator from different type

    When you do. b = true; The compiler sees that it can convert true to a MyBoolean using. MyBoolean(bool value = false):value(value) {}; Since you get one user provided conversion the compiler constructs a temporary MyBoolean and then assigns that to b using the default copy assignment operator. answered Jul 31, 2017 at 13:32.

  15. Non-static const member, can't use default assignment operator

    Please note that this is no GNU specific problem here. The ISO C++ standard requires that T has an assignment operator (see section 23.2.4.3). I just showed on the example of GNU's STL implementation where this can lead to.

  16. c++

    The copy assignment operator, often just called the "assignment operator", is a special case of assignment operator where the source (right-hand side) and destination (left-hand side) are of the same class type. It is one of the special member functions, which means that a default version of it is generated automatically by the compiler if the ...

  17. c++

    Well, you have no choice. s& operator =(const s& m) {. return *new(this) s(m); } Undefined behaviour. There's a reason why pretty much nobody uses const member variables, and it's because of this. There's nothing you can do about it. const member variables simply cannot be used in types you want to be assignable.

  18. C++ assignment operator during declaration

    When initializing an object at definition, it's not using assignment even when the = syntax is used. When you do: Hero zak = foo; it's equivalent to: Hero zak(foo); Which is copy-initialization and as such invokes the copy-constructor. The problem with. Hero bar = "HelloWorld"; is that it's equivalent to: