assignment operator on copy constructor

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Copy constructors, assignment operators, and exception safe assignment

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Copy assignment operator.

A copy assignment operator is a non-template non-static member function with the name operator = that can be called with an argument of the same class type and copies the content of the argument without mutating the argument.

[ edit ] Syntax

For the formal copy assignment operator syntax, see function declaration . The syntax list below only demonstrates a subset of all valid copy assignment operator syntaxes.

[ edit ] Explanation

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.

[ edit ] Implicitly-declared copy assignment operator

If 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. This implicitly-declared copy assignment operator has the form T & T :: operator = ( const T & ) if all of the following is true:

• each direct base B of T has a copy assignment operator whose parameters are B or const B & or const volatile B & ;
• each non-static data member M of T of class type or array of class type has a copy assignment operator whose parameters are M or const M & or const volatile M & .

Otherwise the implicitly-declared copy assignment operator is declared as T & T :: operator = ( T & ) .

Due to these rules, the implicitly-declared copy assignment operator cannot bind to a volatile lvalue argument.

A class can have multiple copy assignment operators, e.g. both T & T :: operator = ( T & ) and T & T :: operator = ( T ) . If some user-defined copy assignment operators are present, the user may still force the generation of the implicitly declared copy assignment operator with the keyword default . (since C++11)

The implicitly-declared (or defaulted on its first declaration) copy assignment operator has an exception specification as described in dynamic exception specification (until C++17) noexcept specification (since C++17)

Because the copy assignment operator is always declared for any class, the base class assignment operator is always hidden. If a using-declaration is used to bring in the assignment operator from the base class, and its argument type could be the same as the argument type of the implicit assignment operator of the derived class, the using-declaration is also hidden by the implicit declaration.

[ edit ] Implicitly-defined copy assignment operator

If the implicitly-declared copy assignment operator is neither deleted nor trivial, it is defined (that is, a function body is generated and compiled) by the compiler if odr-used or needed for constant evaluation (since C++14) . For union types, the implicitly-defined copy assignment copies the object representation (as by std::memmove ). For non-union class types, the operator performs member-wise copy assignment of the object's direct bases and non-static data members, in their initialization order, using built-in assignment for the scalars, memberwise copy-assignment for arrays, and copy assignment operator for class types (called non-virtually).

[ edit ] Deleted copy assignment operator

An implicitly-declared or explicitly-defaulted (since C++11) copy assignment operator for class T is undefined (until C++11) defined as deleted (since C++11) if any of the following conditions is satisfied:

• T has a non-static data member of a const-qualified non-class type (or possibly multi-dimensional array thereof).
• T has a non-static data member of a reference type.
• T has a potentially constructed subobject of class type M (or possibly multi-dimensional array thereof) such that the overload resolution as applied to find M 's copy assignment operator
• does not result in a usable candidate, or
• in the case of the subobject being a variant member , selects a non-trivial function.

[ edit ] Trivial copy assignment operator

The copy assignment operator for class T is trivial if all of the following is true:

• it is not user-provided (meaning, it is implicitly-defined or defaulted);
• T has no virtual member functions;
• T has no virtual base classes;
• the copy assignment operator selected for every direct base of T is trivial;
• the copy assignment operator selected for every non-static class type (or array of class type) member of T is trivial.

A trivial copy assignment operator makes a copy of the object representation as if by std::memmove . All data types compatible with the C language (POD types) are trivially copy-assignable.

[ edit ] Eligible copy assignment operator

Triviality of eligible copy assignment operators determines whether the class is a trivially copyable type .

[ edit ] Notes

If both copy and move assignment operators are provided, overload resolution selects the move assignment if the argument is an rvalue (either a prvalue such as a nameless temporary or an xvalue such as the result of std::move ), and selects the copy assignment if the argument is an lvalue (named object or a function/operator returning lvalue reference). If only the copy assignment is provided, all argument categories select it (as long as it takes its argument by value or as reference to const, since rvalues can bind to const references), which makes copy assignment the fallback for move assignment, when move is unavailable.

It is unspecified whether virtual base class subobjects that are accessible through more than one path in the inheritance lattice, are assigned more than once by the implicitly-defined copy assignment operator (same applies to move assignment ).

See assignment operator overloading for additional detail on the expected behavior of a user-defined copy-assignment operator.

[ edit ] Example

[ edit ] defect reports.

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

[ edit ] See also

• converting constructor
• copy constructor
• copy elision
• default constructor
• aggregate initialization
• constant initialization
• copy initialization
• default initialization
• direct initialization
• initializer list
• list initialization
• reference initialization
• value initialization
• zero initialization
• move assignment
• move constructor
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Copy constructor vs assignment operator in C++

The Copy constructor and the assignment operators are used to initializing one object to another object. The main difference between them is that the copy constructor creates a separate memory block for the new object. But the assignment operator does not make new memory space. It uses the reference variable to point to the previous memory block.

Copy Constructor (Syntax)

Assignment operator (syntax).

Let us see the detailed differences between Copy constructor and Assignment Operator.

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C++ At Work

Copy Constructors, Assignment Operators, and More

Paul DiLascia

Code download available at: CAtWork0509.exe (276 KB) Browse the Code Online

Q I have a simple C++ problem. I want my copy constructor and assignment operator to do the same thing. Can you tell me the best way to accomplish this?

A At first glance this seems like a simple question with a simple answer: just write a copy constructor that calls operator=.

Or, alternatively, write a common copy method and call it from both your copy constructor and operator=, like so:

This code works fine for many classes, but there's more here than meets the eye. In particular, what happens if your class contains instances of other classes as members? To find out, I wrote the test program in Figure 1 . It has a main class, CMainClass, which contains an instance of another class, CMember. Both classes have a copy constructor and assignment operator, with the copy constructor for CMainClass calling operator= as in the first snippet. The code is sprinkled with printf statements to show which methods are called when. To exercise the constructors, cctest first creates an instance of CMainClass using the default ctor, then creates another instance using the copy constructor:

Figure 1 Copy Constructors and Assignment Operators

If you compile and run cctest, you'll see the following printf messages when cctest constructs obj2:

The member object m_obj got initialized twice! First by the default constructor, and again via assignment. Hey, what's going on?

In C++, assignment and copy construction are different because the copy constructor initializes uninitialized memory, whereas assignment starts with an existing initialized object. If your class contains instances of other classes as data members, the copy constructor must first construct these data members before it calls operator=. The result is that these members get initialized twice, as cctest shows. Got it? It's the same thing that happens with the default constructor when you initialize members using assignment instead of initializers. For example:

As opposed to:

Using assignment, m_obj is initialized twice; with the initializer syntax, only once. So, what's the solution to avoid extra initializations during copy construction? While it goes against your instinct to reuse code, this is one situation where it's best to implement your copy constructor and assignment operator separately, even if they do the same thing. Calling operator= from your copy constructor will certainly work, but it's not the most efficient implementation. My observation about initializers suggests a better way:

Now the main copy ctor calls the member object's copy ctor using an initializer, and m_obj is initialized just once by its copy ctor. In general, copy ctors should invoke the copy ctors of their members. Likewise for assignment. And, I may as well add, the same goes for base classes: your derived copy ctor and assignment operators should invoke the corresponding base class methods. Of course, there are always times when you may want to do something different because you know how your code works—but what I've described are the general rules, which are to be broken only when you have a compelling reason. If you have common tasks to perform after the basic objects have been initialized, you can put them in a common initialization method and call it from your constructors and operator=.

Q Can you tell me how to call a Visual C++® class from C#, and what syntax I need to use for this?

Sunil Peddi

Q I have an application that is written in both C# (the GUI) and in classic C++ (some business logic). Now I need to call from a DLL written in C++ a function (or a method) in a DLL written in Visual C++ .NET. This one calls another DLL written in C#. The Visual C++ .NET DLL acts like a proxy. Is this possible? I was able to use LoadLibrary to call a function present in the Visual C++ .NET DLL, and I can receive a return value, but when I try to pass some parameters to the function in the Visual C++ .NET DLL, I get the following error:

How can I resolve this problem?

Giuseppe Dattilo

A I get a lot of questions about interoperability between the Microsoft® .NET Framework and native C++, so I don't mind revisiting this well-covered topic yet again. There are two directions you can go: calling the Framework from C++ or calling C++ from the Framework. I won't go into COM interop here as that's a separate issue best saved for another day.

Let's start with the easiest one first: calling the Framework from C++. The simplest and easiest way to call the Framework from your C++ program is to use the Managed Extensions. These Microsoft-specific C++ language extensions are designed to make calling the Framework as easy as including a couple of files and then using the classes as if they were written in C++. Here's a very simple C++ program that calls the Framework's Console class:

To use the Managed Extensions, all you need to do is import <mscorlib.dll> and whatever .NET assemblies contain the classes you plan to use. Don't forget to compile with /clr:

Your C++ code can use managed classes more or less as if they were ordinary C++ classes. For example, you can create Framework objects with operator new, and access them using C++ pointer syntax, as shown in the following:

Here, the String s is declared as pointer-to-String because String::Format returns a new String object.

The "Hello, world" and date/time programs seem childishly simple—and they are—but just remember that however complex your program is, however many .NET assemblies and classes you use, the basic idea is the same: use <mscorlib.dll> and whatever other assemblies you need, then create managed objects with new, and use pointer syntax to access them.

So much for calling the Framework from C++. What about going the other way, calling C++ from the Framework? Here the road forks into two options, depending on whether you want to call extern C functions or C++ class member functions. Again, I'll take the simpler case first: calling C functions from .NET. The easiest thing to do here is use P/Invoke. With P/Invoke, you declare the external functions as static methods of a class, using the DllImport attribute to specify that the function lives in an external DLL. In C# it looks like this:

This tells the compiler that MessageBox is a function in user32.dll that takes an IntPtr (HWND), two Strings, and an int. You can then call it from your C# program like so:

Of course, you don't need P/Invoke for MessageBox since the .NET Framework already has a MessageBox class, but there are plenty of API functions that aren't supported directly by the Framework, and then you need P/Invoke. And, of course, you can use P/Invoke to call C functions in your own DLLs. I've used C# in the example, but P/Invoke works with any .NET-based language like Visual Basic® .NET or JScript®.NET. The names are the same, only the syntax is different.

Note that I used IntPtr to declare the HWND. I could have got away with int, but you should always use IntPtr for any kind of handle such as HWND, HANDLE, or HDC. IntPtr will default to 32 or 64 bits depending on the platform, so you never have to worry about the size of the handle.

DllImport has various modifiers you can use to specify details about the imported function. In this example, CharSet=CharSet.Auto tells the Framework to pass Strings as Unicode or Ansi, depending on the target operating system. Another little-known modifier you can use is CallingConvention. Recall that in C, there are different calling conventions, which are the rules that specify how the compiler should pass arguments and return values from one function to another across the stack. The default CallingConvention for DllImport is CallingConvention.Winapi. This is actually a pseudo-convention that uses the default convention for the target platform; for example, StdCall (in which the callee cleans the stack) on Windows® platforms and CDecl (in which the caller cleans the stack) on Windows CE .NET. CDecl is also used with varargs functions like printf.

The calling convention is where Giuseppe ran into trouble. C++ uses yet a third calling convention: thiscall. With this convention, the compiler uses the hardware register ECX to pass the "this" pointer to class member functions that don't have variable arguments. Without knowing the exact details of Giuseppe's program, it sounds from the error message that he's trying to call a C++ member function that expects thiscall from a C# program that's using StdCall—oops!

Aside from calling conventions, another interoperability issue when calling C++ methods from the Framework is linkage: C and C++ use different forms of linkage because C++ requires name-mangling to support function overloading. That's why you have to use extern "C" when you declare C functions in C++ programs: so the compiler won't mangle the name. In Windows, the entire windows.h file (now winuser.h) is enclosed in extern "C" brackets.

While there may be a way to call C++ member functions in a DLL directly using P/Invoke and DllImport with the exact mangled names and CallingConvention=ThisCall, it's not something to attempt if you're in your right mind. The proper way to call C++ classes from managed code—option number two—is to wrap your C++ classes in managed wrappers. Wrapping can be tedious if you have lots of classes, but it's really the only way to go. Say you have a C++ class CWidget and you want to wrap it so .NET clients can use it. The basic formula looks something like this:

The pattern is the same for any class. You write a managed (__gc) class that holds a pointer to the native class, you write a constructor and destructor that allocate and destroy the instance, and you write wrapper methods that call the corresponding native C++ member functions. You don't have to wrap all the member functions, only the ones you want to expose to the managed world.

Figure 2 shows a simple but concrete example in full detail. CPerson is a class that holds the name of a person, with member functions GetName and SetName to change the name. Figure 3 shows the managed wrapper for CPerson. In the example, I converted Get/SetName to a property, so .NET-based programmers can use the property syntax. In C#, using it looks like this:

Figure 3 Managed Person Class

Figure 2 Native CPerson Class

Using properties is purely a matter of style; I could equally well have exposed two methods, GetName and SetName, as in the native class. But properties feel more like .NET. The wrapper class is an assembly like any other, but one that links with the native DLL. This is one of the cool benefits of the Managed Extensions: You can link directly with native C/C++ code. If you download and compile the source for my CPerson example, you'll see that the makefile generates two separate DLLs: person.dll implements a normal native DLL and mperson.dll is the managed assembly that wraps it. There are also two test programs: testcpp.exe, a native C++ program that calls the native person.dll and testcs.exe, which is written in C# and calls the managed wrapper mperson.dll (which in turn calls the native person.dll).

Figure 4** Interop Highway **

I've used a very simple example to highlight the fact that there are fundamentally only a few main highways across the border between the managed and native worlds (see Figure 4 ). If your C++ classes are at all complex, the biggest interop problem you'll encounter is converting parameters between native and managed types, a process called marshaling. The Managed Extensions do an admirable job of making this as painless as possible (for example, automatically converting primitive types and Strings), but there are times where you have to know something about what you're doing.

For example, you can't pass the address of a managed object or subobject to a native function without pinning it first. That's because managed objects live in the managed heap, which the garbage collector is free to rearrange. If the garbage collector moves an object, it can update all the managed references to that object—but it knows nothing of raw native pointers that live outside the managed world. That's what __pin is for; it tells the garbage collector: don't move this object. For strings, the Framework has a special function PtrToStringChars that returns a pinned pointer to the native characters. (Incidentally, for those curious-minded souls, PtrToStringChars is the only function as of this date defined in <vcclr.h>. Figure 5 shows the code.) I used PtrToStringChars in MPerson to set the Name (see Figure 3 ).

Figure 5 PtrToStringChars

Pinning isn't the only interop problem you'll encounter. Other problems arise if you have to deal with arrays, references, structs, and callbacks, or access a subobject within an object. This is where some of the more advanced techniques come in, such as StructLayout, boxing, __value types, and so on. You also need special code to handle exceptions (native or managed) and callbacks/delegates. But don't let these interop details obscure the big picture. First decide which way you're calling (from managed to native or the other way around), and if you're calling from managed to native, whether to use P/Invoke or a wrapper.

In Visual Studio® 2005 (which some of you may already have as beta bits), the Managed Extensions have been renamed and upgraded to something called C++/CLI. Think of the C++/CLI as Managed Extensions Version 2, or What the Managed Extensions Should Have Been. The changes are mostly a matter of syntax, though there are some important semantic changes, too. In general C++/CLI is designed to highlight rather than blur the distinction between managed and native objects. Using pointer syntax for managed objects was a clever and elegant idea, but in the end perhaps a little too clever because it obscures important differences between managed and native objects. C++/CLI introduces the key notion of handles for managed objects, so instead of using C pointer syntax for managed objects, the CLI uses ^ (hat):

As you no doubt noticed, there's also a gcnew operator to clarify when you're allocating objects on the managed heap as opposed to the native one. This has the added benefit that gcnew doesn't collide with C++ new, which can be overloaded or even redefined as a macro. C++/CLI has many other cool features designed to make interoperability as straightforward and intuitive as possible.

Send your questions and comments for Paul to   [email protected] .

Paul DiLascia is a freelance software consultant and Web/UI designer-at-large. He is the author of Windows ++: Writing Reusable Windows Code in C ++ (Addison-Wesley, 1992). In his spare time, Paul develops PixieLib, an MFC class library available from his Web site, www.dilascia.com .

Additional resources

• C++ Classes and Objects
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• C++ Abstraction
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Shallow Copy and Deep Copy in C++

In general, creating a copy of an object means to create an exact replica of the object having the same literal value, data type , and resources. There are two ways that are used by C++ compiler to create a copy of objects.

• Copy Constructor
• Assignment Operator

Depending upon the resources like dynamic memory held by the object, either we need to perform Shallow Copy or Deep Copy in order to create a replica of the object. In general, if the variables of an object have been dynamically allocated, then it is required to do a Deep Copy in order to create a copy of the object but one may wonder what is the shallow copy and deep copy? Don’t worry, GeeksforGeeks got you covered.

In this article, we will learn what is shallow copy, what is deep copy, how they are different from each other, where they happen by default and where we need one over another.

What is Shallow Copy?

In shallow copy, an object is created by simply copying the data of all variables of the original object. This works well if none of the variables of the object are defined in the heap section of memory but if some variables are dynamically allocated memory from heap section, then the copied object variable will also reference the same memory location.

This will create ambiguity and run-time errors, dangling pointer. Since both objects will reference to the same memory location, then change made by one will reflect those change in another object as well. Since we wanted to create a replica of the object, this purpose will not be filled by Shallow copy. 

Note: C++ compiler implicitly creates a copy constructor and overloads assignment operator in order to perform shallow copy at compile time.

Consider the below objects B1 and B2, having integer data members representing dimensions of a 3d box.

Now imagine one of the members is a pointer to the integer variable. Here, the shallow copy will only copy the address stored in the pointer leading both the object member pointing to the same object.

Below is the implementation of the above example:

What is Deep Copy?

In Deep copy, an object is created by copying data of all variables, and it also allocates similar memory resources with the same value to the object. In order to perform Deep copy, we need to explicitly define the copy constructor and assign dynamic memory as well, if required. Also, it is required to dynamically allocate memory to the variables in the other constructors, as well.

Consider the previous example again. If we deep copy the object, then each object will have their own copy of the data pointed by the breadth pointer.

Difference between the Shallow Copy and Deep Copy

The below table list the differences between the shallow copy and the deep copy in the tabular form:

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Tech Differences

Know the Technical Differences

Difference Between Copy Constructor and Assignment Operator in C++

Let us study the difference between the copy constructor and assignment operator.

Content: Copy Constructor Vs Assignment Operator

Comparison chart.

• Key Differences

Definition of Copy Constructor

A “copy constructor” is a form of an overloaded constructor . A copy constructor is only called or invoked for initialization purpose. A copy constructor initializes the newly created object by another existing object.

When a copy constructor is used to initialize the newly created target object, then both the target object and the source object shares a different memory location. Changes done to the source object do not reflect in the target object. The general form of the copy constructor is

If the programmer does not create a copy constructor in a C++ program, then the compiler implicitly provides a copy constructor. An implicit copy constructor provided by the compiler does the member-wise copy of the source object. But, sometimes the member-wise copy is not sufficient, as the object may contain a pointer variable.

Copying a pointer variable means, we copy the address stored in the pointer variable, but we do not want to copy address stored in the pointer variable, instead, we want to copy what pointer points to. Hence, there is a need of explicit ‘copy constructor’ in the program to solve this kind of problems.

A copy constructor is invoked in three conditions as follow:

• Copy constructor invokes when a new object is initialized with an existing one.
• The object passed to a function as a non-reference parameter.
• The object is returned from the function.

Let us understand copy constructor with an example.

In the code above, I had explicitly declared a constructor “copy( copy &c )”. This copy constructor is being called when object B is initialized using object A. Second time it is called when object C is being initialized using object A.

When object D is initialized using object A the copy constructor is not called because when D is being initialized it is already in the existence, not the newly created one. Hence, here the assignment operator is invoked.

Definition of Assignment Operator

The assignment operator is an assigning operator of C++.  The “=” operator is used to invoke the assignment operator. It copies the data in one object identically to another object. The assignment operator copies one object to another member-wise. If you do not overload the assignment operator, it performs the bitwise copy. Therefore, you need to overload the assignment operator.

In above code when object A is assigned to object B the assignment operator is being invoked as both the objects are already in existence. Similarly, same is the case when object C is initialized with object A.

When the bitwise assignment is performed both the object shares the same memory location and changes in one object reflect in another object.

Key Differences Between Copy Constructor and Assignment Operator

• A copy constructor is an overloaded constructor whereas an assignment operator is a bitwise operator.
• Using copy constructor you can initialize a new object with an already existing object. On the other hand, an assignment operator copies one object to the other object, both of which are already in existence.
• A copy constructor is initialized whenever a new object is initialized with an already existing object, when an object is passed to a function as a non-reference parameter, or when an object is returned from a function. On the other hand, an assignment operator is invoked only when an object is being assigned to another object.
• When an object is being initialized using copy constructor, the initializing object and the initialized object shares the different memory location. On the other hand, when an object is being initialized using an assignment operator then the initialized and initializing objects share the same memory location.
• If you do not explicitly define a copy constructor then the compiler provides one. On the other hand, if you do not overload an assignment operator then a bitwise copy operation is performed.

The Copy constructor is best for copying one object to another when the object contains raw pointers.

Related Differences:

• Difference Between & and &&
• Difference Between Recursion and Iteration
• Difference Between new and malloc( )
• Difference Between Inheritance and Polymorphism
• Difference Between Constructor and Destructor

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Table of Contents

When is a user-defined copy constructor needed, copy constructor vs assignment operator, how to make the copy constructor private, why does an argument to a copy constructor must be passed as a reference, why should the copy constructor argument be const in c++, understanding c++ copy constructor with example code.

When a new object is generated from an existing object as a copy of the existing object, the copy function Object () { [native code] } is named. When a previously initialized object is given a new value from another object, the assignment operator is used. Function Object () { [native code] } is a member function that uses another object of the same class to initialize an object. The following is a general function prototype for a copy function Object () { [native code] }:

Classname(const classname &obj);

#include<iostream>

using namespace std; 

class Point

     int a, b;

     Point(int x1, int y1) { a =x1; b = y1; } 

     // Copy constructor

     Point(const Point &p1) {a = p1.a; b = p1.b; } 

     int getX()            { return a; }

     int getY()            { return b; }

     Point p1(10, 15); // Normal constructor is called here

     Point p2 = p1; // Copy constructor is called here

     // Let us access values assigned by constructors

     cout << "p1.x = " << p1.getX() << ", p1.y = " << p1.getY();

     cout << "\np2.x = " << p2.getX() << ", p2.y = " << p2.getY();

     return 0;

According to C++ copy constructor the C++ compiler produces a default copy constructor Object() { [native code] } for each class if it doesn’t define own copy constructor Object() { [native code] }, which performs a member-wise copy between items. In general, the copy function Object() { [native code] } generated by the compiler works well. It just needs to define its own copy function Object() [native code] if an object has pointers or any runtime resource allocation, such as a file handle or a network connection.

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When a new object is generated from an existing object as a copy of the existing object, the copy function Object() { [native code] } is named. When an already initialized object is given a new value from another object, the assignment operator is used.

Myclass c1,c2;

Myclass c3;

The above line illustrates the copy constructor

Where Copy Constructor is Needed:

#include<cstring>

class String

     char *s;

     int size;

     String(const char *str = NULL); // constructor

     ~String() { delete [] s; }// destructor

     String(const String&); // copy constructor

     void print() { cout << s << endl; } // Function to print string

     void change(const char *); // Function to change

 String::String(const char *str)

     size = strlen(str);

     s = new char[size+1];

     strcpy(s, str);

 void String::change(const char *str)

     delete [] s;

String::String(const String& old_str)

     size = old_str.size;

     strcpy(s, old_str.s);

     String str1("Welcome");

     String str2 = str1;

     str1.print(); // what is printed ?

     str2.print();

     str2.change("Welcome you all"); 

     str1.print(); // what is printed now ?

What Happens if We Remove the Copy Constructor from the Above Code?

In  c++ copy constructor users don't get the predicted result if we delete the copy function Object() { [native code] } from the above program. Changes to str2 are reflected in str1 as well, which is unexpected.

         char *s;

         int size;

String(const char *str = NULL); // constructor

         ~String() { delete [] s;  }// destructor

         void print() { cout << s << endl; }

         void change(const char *);  // Function to change

String::String(const char *str)

         size = strlen(str);

         s = new char[size+1];

         strcpy(s, str);

         delete [] s;

         String str1("Welcome");

         String str2 = str1; 

str1.print(); // what is printed ?

str2.print(); 

str2.change("Welcome you all"); 

str1.print(); // what is printed now ?

str2.print();

According to the C + + copy constructor it is possible to make a copy function Object() { [native code] }. When a copy function Object() { [native code] } in a class is made private, objects in that class become non-copyable. This is especially useful when the class contains pointers or resources that are dynamically allocated. In such cases, it can either write our own copy function Object() { [native code] }, as in the String example above, or create a private copy function Object() { [native code] }, in which case users will receive compiler errors rather than warnings.

According to C + + copy constructor, when an object is transferred by value, a copy function Object() { [native code] } is named. The copy function Object() { [native code] } is a function in and of itself. If an argument is passed by value in a copy function Object() { [native code] }, a call to copy function Object() { [native code] } is made to call copy function Object() { [native code] }, resulting in a non-terminating sequence of calls. As a result, the compiler won't let you move parameters by value.

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According to C++ copy constructor, we pass an object by reference to the copy function Object() { [native code] }, and we usually pass it as a const reference. One justification for passing a const reference is that it can use const wherever possible in C++ to avoid unintentionally changing objects. This is one compelling excuse to transfer references as const, but there are others.

/* Class data members */

Test(Test &t) { /* Copy data members from t*/}

Test()  { /* Initialize data members */ }

     cout << "fun() Called\n";

     Test t;

     return t;

     Test t1;

     Test t2 = fun();

Need to change the copy constructor by the following:

Test(const Test &t) { cout << "Copy Constructor Called\n"; }

The fun() function returns a value. As a result, the compiler generates a temporary entity, which is then copied to t2 using the original program's copy function Object() { [native code] } (The temporary object is passed as an argument to the copy constructor). The compiler error occurs because compiler-created temporary objects cannot be bound to non-const references, despite the fact that the original program attempts to do so. It doesn't make sense to change something that isn't broken.

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According to C++ copy constructor, when an object is made, a function Object() { [native code] } is a special type of member function that is called automatically. A function Object() { [native code] } in C++ has the same name as the class it belongs to, and it does not have a return form.

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call copy constructor from assignment operator function

I have a class with a point to dynamically allocated array, so I created copy constructor and assignment operator function. Since copy constructor and assignment operator function do the same work, I call copy constructor from the assignment operator function but get "error C2082: redefinition of formal parameter" . I am using Visual Studio 2012.

• visual-studio

4 Answers 4

The offending line isn't what you think it is. It actually declares a variable other of type FeatureValue . This is because constructors to not have names and cannot be called directly.

You can safely invoke the copy assignment operator from the constructor as long as the operator is not declared virtual.

This will works just dandy or you can use the typical guidelines for the copy & swap idiom suggested in Vaughn Cato's answer

• 3 why don't just m_value[i] = other.m_value[i] instead of using intermediate value, e.g. value[i] = other.m_value[i] ? –  stackunderflow Commented Jul 5, 2013 at 4:35

You can't directly call a constructor like you would any other method. What you are doing is actually declaring a variable called other of type FeatureValue .

Take a look at the copy-and-swap idiom for a good way to avoid duplication between the assignment operator and the copy constructor: What is the copy-and-swap idiom?

Even better, use a std::vector instead of new and delete . Then you don't need to write your own copy constructor or assignment operator.

• It's Bjoink all over again! –  Captain Obvlious Commented Jul 5, 2013 at 2:51
• Its uncanny how you guys both answered the same thing at the same time. –  Borgleader Commented Jul 5, 2013 at 2:51
• @Borgleader what you may find even more odd is I've know about Vaughn for well over about 25 years. He just didn't know about me ;) –  Captain Obvlious Commented Jul 5, 2013 at 2:52
• Ok. Is it possible to call copy constructor from the assignment operator function then? –  stackunderflow Commented Jul 5, 2013 at 2:54
• @zwx: No -- you could put the common code into a third function that both the constructor and the assignment operator call though. –  Vaughn Cato Commented Jul 5, 2013 at 2:59

Short answer - don't do it.

• When the copy constructor is called, it's constructing the new object with reference to the object being copied, but the default constructor does not run before the copy constructor. This means m_value has an indeterminate value when the copy constructor starts running - you can assign to it, but to read from it is undefined behaviour, and to delete[] it considerably worse (if anything can be worse than UD! ;-)). So, just leave out that delete[] line.

Next, if operator= tries to leverage the functionality from the copy constructor, it has to first release any existing data m_value is pointing at or it will be leaked. Most people try to do that as follows ( which is broken ) - I think this is what you were trying for:

The problem with this is that if the creation of FeatureValue fails (e.g. because new can't get the memory it wants), then the FeatureValue object is left with an invalid state (e.g. m_value might be pointing off into space). Later when the destructor runs and does a delete[] m_value , you have undefined behaviour (your program will probably crash).

You really should approach this more systematically... either writing it out step by step, or perhaps implementing a guaranteed non-throwing swap() method (easy to do... just std::swap() m_size and m_value , and using it ala:

That's easy and clean, but it has a couple minor performance/efficiency issues:

keeping any existing m_value array around longer than necessary, increasing peak memory usage... you could call clear() . In practice, most non-trivial programs wouldn't care about this unless the data structure in question was holding huge amounts of data (e.g. hundreds of megabytes or gigabytes for a PC app).

not even trying to reuse the existing m_value memory - instead always doing another new for other (that can lead to reduced memory usage but isn't always worthwhile).

Ultimately, the reasons there can be distinct copy constructor and operator= - rather than having the compiler automatically create one from the other - is that optimally efficient implementations can't - in general - leverage each other in the way you'd hoped.

The statement FeatureValue(other); actually invokes the copy constructor to create a new Featurevalue object,which has nothing to do with *this .

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22.3 — Move constructors and move assignment

Let’s take a closer look. There are 6 key steps that happen in this program (one for each printed message):

These functions do a memberwise move, which behaves as follows:

Move semantics is an optimization opportunity.