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A declaration of a class/struct or union may appear within another class. Such declaration declares a nested class .
The name of the nested class exists in the scope of the enclosing class, and name lookup from a member function of a nested class visits the scope of the enclosing class after examining the scope of the nested class. Like any member of its enclosing class, the nested class has access to all names (private, protected, etc) to which the enclosing class has access, but it is otherwise independent and has no special access to the this pointer of the enclosing class. Declarations in a nested class can use any members of the enclosing class, following the usual usage rules for the non-static members.
Friend functions defined within a nested class have no special access to the members of the enclosing class even if lookup from the body of a member function that is defined within a nested class can find the private members of the enclosing class.
Out-of-class definitions of the members of a nested class appear in the namespace of the enclosing class:
Nested classes can be forward-declared and later defined, either within the same enclosing class body, or outside of it:
Nested class declarations obey member access specifiers, a private member class cannot be named outside the scope of the enclosing class, although objects of that class may be manipulated:
The following behavior-changing defect reports were applied retroactively to previously published C++ standards.
DR | Applied to | Behavior as published | Correct behavior |
---|---|---|---|
C++98 | the members of a nested class cannot access the enclosing class and its friends | they have the same access rights as other members of the enclosing class (also resolves CWG issues #8 and #10) |
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Starting in C++11, two kinds of assignment are supported in the language: copy assignment and move assignment . In this article "assignment" means copy assignment unless explicitly stated otherwise. For information about move assignment, see Move Constructors and Move Assignment Operators (C++) .
Both the assignment operation and the initialization operation cause objects to be copied.
Assignment : When one object's value is assigned to another object, the first object is copied to the second object. So, this code copies the value of b into a :
Initialization : Initialization occurs when you declare a new object, when you pass function arguments by value, or when you return by value from a function.
You can define the semantics of "copy" for objects of class type. For example, consider this code:
The preceding code could mean "copy the contents of FILE1.DAT to FILE2.DAT" or it could mean "ignore FILE2.DAT and make b a second handle to FILE1.DAT." You must attach appropriate copying semantics to each class, as follows:
Use an assignment operator operator= that returns a reference to the class type and takes one parameter that's passed by const reference—for example ClassName& operator=(const ClassName& x); .
Use the copy constructor.
If you don't declare a copy constructor, the compiler generates a member-wise copy constructor for you. Similarly, if you don't declare a copy assignment operator, the compiler generates a member-wise copy assignment operator for you. Declaring a copy constructor doesn't suppress the compiler-generated copy assignment operator, and vice-versa. If you implement either one, we recommend that you implement the other one, too. When you implement both, the meaning of the code is clear.
The copy constructor takes an argument of type ClassName& , where ClassName is the name of the class. For example:
Make the type of the copy constructor's argument const ClassName& whenever possible. This prevents the copy constructor from accidentally changing the copied object. It also lets you copy from const objects.
Compiler-generated copy constructors, like user-defined copy constructors, have a single argument of type "reference to class-name ." An exception is when all base classes and member classes have copy constructors declared as taking a single argument of type const class-name & . In such a case, the compiler-generated copy constructor's argument is also const .
When the argument type to the copy constructor isn't const , initialization by copying a const object generates an error. The reverse isn't true: If the argument is const , you can initialize by copying an object that's not const .
Compiler-generated assignment operators follow the same pattern for const . They take a single argument of type ClassName& unless the assignment operators in all base and member classes take arguments of type const ClassName& . In this case, the generated assignment operator for the class takes a const argument.
When virtual base classes are initialized by copy constructors, whether compiler-generated or user-defined, they're initialized only once: at the point when they are constructed.
The implications are similar to the copy constructor. When the argument type isn't const , assignment from a const object generates an error. The reverse isn't true: If a const value is assigned to a value that's not const , the assignment succeeds.
For more information about overloaded assignment operators, see Assignment .
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What is the difference between inside and outside class overloading?
First consider this example using your class (reformatted to a style I prefer):
The conventional way to do this is therefore to pass the argument by reference, and to make the method const :
If you define the comparison outside the class, like this:
… then
If you choose to place these non-member operators inline inside the class definition, via the friend mechanism (so that they're found via ADL lookup), then you lose the first bullet point's advantage, but you then have all the code relevant for use of the class, within the class definition:
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24. First: the two different ways are really "overload as a member" and "overload as a non-member", and the latter has two different ways to write it (as-friend-inside class definition and outside class definition). Calling them "inside class" and "outside class" is going to confuse you. Overloads for +=, +, -=, -, etc. have a special pattern:
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 .
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}
Assignment Operators in C are used to assign values to the variables. They come under the category of binary operators as they require two operands to operate upon. The left side operand is called a variable and the right side operand is the value. The value on the right side of the "=" is assigned to the variable on the left side of "=".
Code language:C++(cpp) The = assignment operator is called a simple assignment operator. It assigns the value of the left operand to the right operand. Besides the simple assignment operator, C supports compound assignment operators. A compound assignment operator performs the operation specified by the additional operator and then assigns the ...
Simple assignment operator. Assigns values from right side operands to left side operand. C = A + B will assign the value of A + B to C. +=. Add AND assignment operator. It adds the right operand to the left operand and assign the result to the left operand. C += A is equivalent to C = C + A. -=.
5,6) Definition of a copy assignment operator outside of class definition (the class must contain a declaration (1) ). 6) The copy assignment operator is explicitly-defaulted. 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.
Overloading assignments (C++ only) You overload the assignment operator, operator=, with a nonstatic member function that has only one parameter. You cannot declare an overloaded assignment operator that is a nonmember function. The following example shows how you can overload the assignment operator for a particular class: struct X {. int data;
I have a header file which contains a class with an assignment operator. Now I want to define my function inside my .cpp file. My header file contains: Car &operator=(const Car &other); Now in my cpp file I would like to do something like: Car::Car &operator=(const Car &other) {. } Unfortunately this doesn't seem to be the right syntax.
The assignment operator that affects the instance of the class still can go in the class, but the assignment operator that affects the primitive can only go into the global namespace.<BR><BR>For ...
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. The following set of operators is commonly overloaded for user-defined classes:
The assignment operator,"=", is the operator used for Assignment. It copies the right value into the left value. Assignment Operators are predefined to operate only on built-in Data types. Assignment operator overloading is binary operator overloading. Overloading assignment operator in C++ copies all values of one object to another object.
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: Expression. As member function. As non-member function.
It allows you to provide an intuitive interface to users of your class, plus makes it possible for templates to work equally well with classes and built-in/intrinsic types. Operator overloading allows C/C++ operators to have user-defined meanings on user-defined types (classes). Overloaded operators are syntactic sugar for function calls: class ...
I read that and understand how to write an overload function like that using a member function inside a class, but I need to be able to write a regular function outside of a class that can do the same thing.
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 ...
Explanation. The name of the nested class exists in the scope of the enclosing class, and name lookup from a member function of a nested class visits the scope of the enclosing class after examining the scope of the nested class. Like any member of its enclosing class, the nested class has access to all names (private, protected, etc) to which ...
Use an assignment operator operator= that returns a reference to the class type and takes one parameter that's passed by const reference—for example ClassName& operator=(const ClassName& x);. Use the copy constructor. If you don't declare a copy constructor, the compiler generates a member-wise copy constructor for you.
auto operator==( int const u, B const& v ) -> bool { return true; // Whatever. } If you choose to place these non-member operators inline inside the class definition, via the friend mechanism (so that they're found via ADL lookup), then you lose the first bullet point's advantage, but you then have all the code relevant for use of the class ...