Overload Assignment Operator C++ Struct Constructor

Quick Answer:

It depends on what do you want to achieve.

Long, Extended, Boring Answer:

You hit the nail.

I usually dislike that "C++" allows "Struct (s)" allows to declare methods. Preferably, I use explicit "Class (es)" for methods required and P.O.D. "Struct (s)" for only fields.

Yet, I agree that some basic simple operations, like:

  • assign initial values ("constructor")
  • make a copy of a structure ("copy constructor)
  • assign values to an existing structure ("overload assign operator")

Are required, and, in those circumstances, methods for structures, make sense.

Suggestion

Another potential solution is to use P.O.D. structures, but, still conceptually treat them as classes and objects.

Wrap those declarations in a namespace, and, add global functions, for the most important actions.

The code declaration could be similar to this:

The code that applies the solution, could be something like this:

This alternative approach is better when a huge memory allocation of data is required, or interacting with other low-level shared libraries.

This approach, with some changes, is applied in Game Development.

Extra

Personally, I consider a syntax extension for "C++", or even, a new "C++" based P.L. that solves this issue:

Cheers.

answered Nov 11 '14 at 18:45

The latest version of this topic can be found at Copy Constructors and Copy Assignment Operators (C++).

NOTE]

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. Therefore,

    causes the value of to be copied to .

  • Initialization: Initialization occurs when a new object is declared, when arguments are passed to functions by value, or when values are returned from functions by value.

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 a second handle to FILE1.DAT." You must attach appropriate copying semantics to each class, as follows.

  • By using the assignment operator together with a reference to the class type as the return type and the parameter that is passed by reference—for example .

  • By using the copy constructor. For more information about the copy constructor, see Rules for Declaring Constructors.

If you do not declare a copy constructor, the compiler generates a member-wise copy constructor for you. If you do not declare a copy assignment operator, the compiler generates a member-wise copy assignment operator for you. Declaring a copy constructor does not suppress the compiler-generated copy assignment operator, nor vice versa. If you implement either one, we recommend that you also implement the other one so that the meaning of the code is clear.

Member-wise assignment is covered in more detail in (NOTINBUILD) Memberwise Assignment and Initialization.

The copy constructor takes an argument of type class-name&, where class-name is the name of the class for which the constructor is defined. For example:

Make the type of the copy constructor's argument const class-name& whenever possible. This prevents the copy constructor from accidentally changing the object from which it is copying. It also enables copying 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 constclass-name&. In such a case, the compiler-generated copy constructor's argument is also const.

When the argument type to the copy constructor is not const, initialization by copying a const object generates an error. The reverse is not true: If the argument is const, you can initialize by copying an object that is not const.

Compiler-generated assignment operators follow the same pattern with regard to const. They take a single argument of type class-name& unless the assignment operators in all base and member classes take arguments of type constclass-name&. In this case, the class's generated assignment operator takes a const argument.

When virtual base classes are initialized by copy constructors, compiler-generated or user-defined, they are initialized only once: at the point when they are constructed.

The implications are similar to those of the copy constructor. When the argument type is not const, assignment from a const object generates an error. The reverse is not true: If a const value is assigned to a value that is not const, the assignment succeeds.

For more information about overloaded assignment operators, see Assignment.

Special Member Functions

TextFile a, b; a.Open( "FILE1.DAT" ); b.Open( "FILE2.DAT" ); b = a;
// spec1_copying_class_objects.cpp class Window { public: Window( const Window& ); // Declare copy constructor. // ... }; int main() { }
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