Copy Constructor Self Assignment C++ Functions

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


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() { }

C++ Tutorial Operator Overloading II self assignment - 2018 site search:

Self Assignment

Why do we need to protect against the self assignment when we do overloading assignment operator?

But before we dig into the reason why, who would do that kind of silly self assignment?

Window w; w = w; // nobody will do this w[i] = w[j]; // this may happen

Anything can happen. When we write a code, we're not supposed to do make any assumption. So, we need to guard against it.

Here is the code without any protection.

class ScrollBar {}; class Window { ScrollBar *sb; public: Window(ScrollBar *s) : sb(s) {} Window() = default; Window& operator=(const Window&); }; Window& Window::operator=(const Window& rhs) { delete sb; sb = new ScrollBar(*; return *this; } int main() { Window w(new ScrollBar); Window w2(w); }

The code shows a typical overloading assignment overloading.
What could be the flaws in the code?

What could happen if *this and rhs is the same Window instance?

If that's the case, the following line is the problem:

delete sb;

We're deleting since *this and rhs are the same object. After deleting the sb, we're trying to access already deleted object of rhs:

sb = new ScrollBar(*;

We do not want to that happen. The immediate solution is this:

if (this == &rhs;) return *this; delete sb; sb = new ScrollBar(*; return *this;

Still there is a problem.

What if an exception is thrown in the copy constructor after we deleted sb. Then, we're end up having a pointer which is pointing to nothing. So, we need a better code:

Window& Window::operator=(const Window& rhs) { if (this == &rhs;) return *this; ScrollBar *sbOld = sb; sb = new ScrollBar(*; delete sbOld; return *this; }

In this way, the Window object can still holding the ScrollBar even in when there is a thrown exception.

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