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Copy Constructor in C++

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Competitive Programming in C++

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Pre-requisite: Constructor in C++ 

A copy constructor is a member function that initializes an object using another object of the same class. In simple terms, a constructor which creates an object by initializing it with an object of the same class, which has been created previously is known as a copy constructor .  

Copy constructor is used to initialize the members of a newly created object by copying the members of an already existing object.

Copy constructor takes a reference to an object of the same class as an argument.

The process of initializing members of an object through a copy constructor is known as copy initialization.

It is also called member-wise initialization because the copy constructor initializes one object with the existing object, both belonging to the same class on a member by member copy basis.

The copy constructor can be defined explicitly by the programmer. If the programmer does not define the copy constructor, the compiler does it for us.  

Syntax of Copy Constructor with Example

Syntax of Copy Constructor

Characteristics of Copy Constructor

1. The copy constructor is used to initialize the members of a newly created object by copying the members of an already existing object.

2. Copy constructor takes a reference to an object of the same class as an argument. If you pass the object by value in the copy constructor, it would result in a recursive call to the copy constructor itself. This happens because passing by value involves making a copy, and making a copy involves calling the copy constructor, leading to an infinite loop. Using a reference avoids this recursion. So we use reference of Objects to avoid infinite calls.

3. The process of initializing members of an object through a copy constructor is known as copy initialization.

4 . It is also called member-wise initialization because the copy constructor initializes one object with the existing object, both belonging to the same class on a member-by-member copy basis.

5. The copy constructor can be defined explicitly by the programmer. If the programmer does not define the copy constructor, the compiler does it for us.

Types of Copy Constructors

1. default copy constructor.

An implicitly defined copy constructor will copy the bases and members of an object in the same order that a constructor would initialize the bases and members of the object.

2. User Defined Copy Constructor 

A user-defined copy constructor is generally needed when an object owns pointers or non-shareable references, such as to a file, in which case a destructor and an assignment operator should also be written

When is the copy constructor called? 

In C++, a Copy Constructor may be called in the following cases: 

  • When an object of the class is returned by value. 
  • When an object of the class is passed (to a function) by value as an argument. 
  • When an object is constructed based on another object of the same class. 
  • When the compiler generates a temporary object.

It is, however, not guaranteed that a copy constructor will be called in all these cases, because the C++ Standard allows the compiler to optimize the copy away in certain cases, one example is the return value optimization (sometimes referred to as RVO).

Copy Elision

In copy elision , the compiler prevents the making of extra copies which results in saving space and better the program complexity(both time and space); Hence making the code more optimized.  

Now it is on the compiler to decide what it wants to print, it could either print the above output or it could print case 1 or case 2 below, and this is what Return Value Optimization is. In simple words, RVO is a technique that gives the compiler some additional power to terminate the temporary object created which results in changing the observable behavior/characteristics of the final program.

When is a user-defined copy constructor needed? 

If we don’t define our own copy constructor, the C++ compiler creates a default copy constructor for each class which does a member-wise copy between objects. The compiler-created copy constructor works fine in general. We need to define our own copy constructor only if an object has pointers or any runtime allocation of the resource like a file handle , a network connection, etc.  

The default constructor does only shallow copy.  

shallow copy in C++

Deep copy is possible only with a user-defined copy constructor. In a user-defined copy constructor, we make sure that pointers (or references) of copied objects point to new memory locations.  

Deep Copy in C++

Copy constructor vs Assignment Operator 

The main difference between Copy Constructor and Assignment Operator is that the Copy constructor makes a new memory storage every time it is called while the assignment operator does not make new memory storage.

Which of the following two statements calls the copy constructor and which one calls the assignment operator? 

A copy constructor is called when a new object is created from an existing object, as a copy of the existing object. The assignment operator is called when an already initialized object is assigned a new value from another existing object. In the above example (1) calls the copy constructor and (2) calls the assignment operator. See this for more details.

Example – Class Where a Copy Constructor is Required 

Following is a complete C++ program to demonstrate the use of the Copy constructor. In the following String class, we must write a copy constructor. 

What would be the problem if we remove the copy constructor from the above code? 

If we remove the copy constructor from the above program, we don’t get the expected output. The changes made to str2 reflect in str1 as well which is never expected.   

Can we make the copy constructor private?  

Yes, a copy constructor can be made private. When we make a copy constructor private in a class, objects of that class become non-copyable. This is particularly useful when our class has pointers or dynamically allocated resources. In such situations, we can either write our own copy constructor like the above String example or make a private copy constructor so that users get compiler errors rather than surprises at runtime.

Why argument to a copy constructor must be passed as a reference?  

A copy constructor is called when an object is passed by value. Copy constructor itself is a function. So if we pass an argument by value in a copy constructor, a call to the copy constructor would be made to call the copy constructor which becomes a non-terminating chain of calls. Therefore compiler doesn’t allow parameters to be passed by value.

Why argument to a copy constructor should be const?

One reason for passing const reference is, that we should use const in C++ wherever possible so that objects are not accidentally modified. This is one good reason for passing reference as const , but there is more to it than ‘ Why argument to a copy constructor should be const?’

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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.

Copy Constructor
Assignment Operator
The Copy constructor is basically an overloaded constructor
Assignment operator is basically an operator.
This initializes the new object with an already existing object
This assigns the value of one object to another object both of which are already exists.
Copy constructor is used when a new object is created with some existing object
This operator is used when we want to assign existing object to new object.
Both the objects uses separate memory locations.
One memory location is used but different reference variables are pointing to the same location.
If no copy constructor is defined in the class, the compiler provides one.
If the assignment operator is not overloaded then bitwise copy will be made

Ankith Reddy

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Tech Differences

Know the Technical Differences

Difference Between Copy Constructor and Assignment Operator in C++

Copy-constructor-assignment-operator

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

Content: Copy Constructor Vs Assignment Operator

Comparison chart.

  • Key Differences
Basis for ComparisonCopy ConstructorAssignment Operator
BasicThe copy constructor is an overloaded constructor.The assignment operator is a bitwise operator.
MeaningThe copy constructor initializes the new object with an already existing object.The assignment operator assigns the value of one object to another object both of which are already in existence.
Syntaxclass_name(cont class_name &object_name) {
//body of the constructor
}
class_name Ob1, Ob2;
Ob2=Ob1;
Invokes(1)Copy constructor invokes when a new object is initialized with existing one.
(2)The object passed to a function as a non-reference parameter.
(3)The object is returned from the function.
The assignment operator is invoked only when assigning the existing object a new object.
Memory AllocationBoth the target object and the initializing object shares the different memory locations.Both the target object and the initializing object shares same allocated memory.
DefaultIf you do not define any copy constructor in the program, C++ compiler implicitly provides one.If you do not overload the "=" operator, then a bitwise copy will be made.

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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MyClass& other ); MyClass( MyClass& other ); MyClass( MyClass& other ); MyClass( MyClass& other );
MyClass* other );
MyClass { x; c; std::string s; };
MyClass& other ) : x( other.x ), c( other.c ), s( other.s ) {}
);
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, );
=( MyClass& other ) { x = other.x; c = other.c; s = other.s; * ; }
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<> MyArray<T>:: =( MyArray& rhs ) { ( != &rhs ) { [] pElements; pElements = T[ rhs.numElements ]; ( size_t i = 0; i < rhs.numElements; ++i ) pElements[ i ] = rhs.pElements[ i ]; numElements = rhs.numElements; } * ; }
<> MyArray<T>:: =( MyArray& rhs ) { MyArray tmp( rhs ); std::swap( numElements, tmp.numElements ); std::swap( pElements, tmp.pElements ); * ; }
< T > swap( T& one, T& two ) { T tmp( one ); one = two; two = tmp; }
<> MyArray<T>:: =( MyArray tmp ) { std::swap( numElements, tmp.numElements ); std::swap( pElements, tmp.pElements ); * ; }

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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 .

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Copy assignment operator

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A copy assignment operator of class T is a non-template non-static member function with the name operator = that takes exactly one parameter of type T , T & , const T & , volatile T & , or const volatile T & . For a type to be CopyAssignable , it must have a public copy assignment operator.

class_name class_name ( class_name ) (1)
class_name class_name ( const class_name ) (2)
class_name class_name ( const class_name ) = default; (3) (since C++11)
class_name class_name ( const class_name ) = delete; (4) (since C++11)

Explanation

  • Typical declaration of a copy assignment operator when copy-and-swap idiom can be used.
  • Typical declaration of a copy assignment operator when copy-and-swap idiom cannot be used (non-swappable type or degraded performance).
  • Forcing a copy assignment operator to be generated by the compiler.
  • Avoiding implicit copy assignment.

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 operator

If no user-defined copy assignment operators are provided for a class type ( struct , class , or union ), 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 & ) . (Note that 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 = ( const 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) exception 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.

Deleted implicitly-declared copy assignment operator

A implicitly-declared copy assignment operator for class T is defined as deleted if any of the following is true:

  • T has a user-declared move constructor;
  • T has a user-declared move assignment operator.

Otherwise, it is defined as defaulted.

A defaulted copy assignment operator for class T is defined as deleted if any of the following is true:

  • T has a non-static data member of non-class type (or array thereof) that is const ;
  • T has a non-static data member of a reference type;
  • T has a non-static data member or a direct or virtual base class that cannot be copy-assigned (overload resolution for the copy assignment fails, or selects a deleted or inaccessible function);
  • T is a union-like class , and has a variant member whose corresponding assignment operator is non-trivial.

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) , , and if it is defaulted, its signature is the same as implicitly-defined (until C++14) ;
  • 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;
has no non-static data members of -qualified type. (since C++14)

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.

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 . For union types, the implicitly-defined copy assignment copies the object representation (as by std::memmove ). For non-union class types ( class and struct ), the operator performs member-wise copy assignment of the object's bases and non-static members, in their initialization order, using built-in assignment for the scalars and copy assignment operator for class types.

The generation of the implicitly-defined copy assignment operator is deprecated (since C++11) if T has a user-declared destructor or user-declared copy constructor.

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.

Defect reports

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

DR Applied to Behavior as published Correct behavior
C++14 operator=(X&) = default was non-trivial made trivial

22.3 — Move constructors and move assignment

Related content

You now have enough context to understand the key insight behind move semantics.

Automatic l-values returned by value may be moved instead of copied

cppreference.com

Copy assignment operator.

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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.

Syntax Explanation Implicitly-declared copy assignment operator Implicitly-defined copy assignment operator Deleted copy assignment operator Trivial copy assignment operator Eligible copy assignment operator Notes Example Defect reports See also

[ 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.

return-type parameter-list  (1)
return-type parameter-list  function-body (2)
return-type parameter-list-no-default  (3) (since C++11)
return-type parameter-list  (4) (since C++11)
return-type class-name  parameter-list  function-body (5)
return-type class-name  parameter-list-no-default  (6) (since C++11)
class-name - the class whose copy assignment operator is being declared, the class type is given as in the descriptions below
parameter-list - a of only one parameter, which is of type , , const T&, volatile T& or const volatile T&
parameter-list-no-default - a of only one parameter, which is of type , , const T&, volatile T& or const volatile T& and does not have a default argument
function-body - the of the copy assignment operator
return-type - any type, but is favored in order to allow chaining asssignments

[ 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).

The implicitly-defined copy assignment operator for a class is if

is a , and that is of class type (or array thereof), the assignment operator selected to copy that member is a constexpr function.
(since C++14)
(until C++23)

The implicitly-defined copy assignment operator for a class is .

(since C++23)

The generation of the implicitly-defined copy assignment operator is deprecated if has a user-declared destructor or user-declared copy constructor.

(since C++11)

[ 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.

The implicitly-declared copy assignment operator for class is defined as deleted if declares a or .

(since C++11)

[ 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

A copy assignment operator is eligible if it is either user-declared or both implicitly-declared and definable.

(until C++11)

A copy assignment operator is eligible if it is not deleted.

(since C++11)
(until C++20)

A copy assignment operator is eligible if all following conditions are satisfied:

(if any) are satisfied. than any other copy assignment operator.
(since C++20)

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.

DR Applied to Behavior as published Correct behavior
C++98 the conditions where implicitly-declared copy assignment operators
are undefined did not consider multi-dimensional array types
consider these types
C++11 a volatile subobject made defaulted copy
assignment operators non-trivial ( )
triviality not affected
C++11 operator=(X&) = default was non-trivial made trivial
C++11 a defaulted copy assignment operator for class was not defined as deleted
if is abstract and has non-copy-assignable direct virtual base classes
the operator is defined
as deleted in this case
C++20 a copy assignment operator was not eligible if there
is another copy assignment operator which is more
constrained but does not satisfy its associated constraints
it can be eligible
in this case

[ 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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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

stackunderflow's user avatar

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

Community's user avatar

  • 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.

Vaughn Cato's user avatar

  • 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.

Tony Delroy's user avatar

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

witrus's user avatar

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copy constructor called instead of assignment operator

COMMENTS

  1. What's the difference between assignment operator and copy constructor?

    Copy constructor is called when a new object is created from an existing object, as a copy of the existing object. And assignment operator is called when an already initialized object is assigned a new value from another existing object. Example-. t2 = t1; // calls assignment operator, same as "t2.operator=(t1);"

  2. Copy Constructor vs Assignment Operator in C++

    But, there are some basic differences between them: Copy constructor. Assignment operator. It is called when a new object is created from an existing object, as a copy of the existing object. This operator is called when an already initialized object is assigned a new value from another existing object. It creates a separate memory block for ...

  3. The copy constructor and assignment operator

    The copy constructor is for creating a new object. It copies an existing object to a newly constructed object.The copy constructor is used to initialize a new instance from an old instance. It is not necessarily called when passing variables by value into functions or as return values out of functions. The assignment operator is to deal with an ...

  4. Copy constructors and copy assignment operators (C++)

    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.

  5. Copy Constructor in C++

    Copy constructor and Assignment operator are similar as they are both used to initialize one object using another object. But, there are some basic differences between them: Copy constructor Assignment operator It is called when a new object is created from an existing object, as a copy of the existing objectThis operator is called when an already

  6. Copy Constructor and operator=

    This is because the assignment operator is being called instead of the copy constructor. By default, the assignment operator does a member-wise copy of the object, which in this case gives a shallow copy. However, C++ gives us the ability to override the default assignment operator, and we'll learn that today too.

  7. Difference Between Copy Constructor and Assignment Operator in C++

    The difference between a copy constructor and an assignment operator is that a copy constructor helps to create a copy of an already existing object without altering the original value of the created object, whereas an assignment operator helps to assign a new value to a data member or an object in the program. Kiran Kumar Panigrahi.

  8. 14.14

    The rule of three is a well known C++ principle that states that if a class requires a user-defined copy constructor, destructor, or copy assignment operator, then it probably requires all three. In C++11, this was expanded to the rule of five, which adds the move constructor and move assignment operator to the list.

  9. 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.

  10. Difference Between Copy Constructor and Assignment Operator in C++

    Copy constructor and assignment operator, are the two ways to initialize one object using another object. ... A copy constructor is only called or invoked for initialization purpose. A copy constructor initializes the newly created object by another existing object. ... instead, we want to copy what pointer points to. Hence, there is a need of ...

  11. PDF Copy Constructors and Assignment Operators

    the assignment operator is called only when assigning an existing object a new value. Otherwise, you're using the copy constructor. What C++ Does For You Unless you specify otherwise, C++ will automatically provide objects a basic copy constructor and assignment operator that simply invoke the copy constructors and assignment operators of all ...

  12. Copy constructors

    A copy constructor is a constructor which can be called with an argument of the same class type and copies the content of the argument without mutating the argument. Contents. 1 Syntax; ... generation of the implicitly-defined copy constructor is deprecated if T has a user-defined destructor or user-defined copy assignment operator. (since C++11)

  13. Copy constructors, assignment operators,

    The first line runs the copy constructor of T, which can throw; the remaining lines are assignment operators which can also throw. HOWEVER, if you have a type T for which the default std::swap() may result in either T's copy constructor or assignment operator throwing, you are

  14. C++ at Work: Copy Constructors, Assignment Operators, and More

    CopyObj(rhs); return *this; } A At first glance this seems like a simple question with a simple answer: just write a copy constructor that calls operator=. CFoo::CFoo(const CFoo& obj) {. *this = obj; } Or, alternatively, write a common copy method and call it from both your copy constructor and operator=, like so:

  15. Copy constructors and Assignment Operators

    No assignment operator is used in the first test-case. It just uses the initialization form called "copy initialization". Copy initialization does not consider explicit constructors when initializing the object. struct A {. A(); // explicit copy constructor. explicit A(A const&); // explicit constructor. explicit A(int);

  16. 21.12

    The purpose of the copy constructor and the copy assignment operator are almost equivalent -- both copy one object to another. However, the copy constructor initializes new objects, whereas the assignment operator replaces the contents of existing objects. ... cout << "Copy constructor called\n"; // just to prove it works } // Overloaded ...

  17. 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 operator If no user-defined copy assignment operators are provided for a class type ( struct , class , or union ), the compiler will always declare one as ...

  18. 22.3

    C++11 defines two new functions in service of move semantics: a move constructor, and a move assignment operator. Whereas the goal of the copy constructor and copy assignment is to make a copy of one object to another, the goal of the move constructor and move assignment is to move ownership of the resources from one object to another (which is typically much less expensive than making a copy).

  19. Why Copy Constructor is called here instead of normal Constructor and

    As to the rationale of why no assignment operator is called, assignment only occurs when one assigns two completely formed objects to each other. In case of: C c4 = c1; c1 is a completely constructed object while c4 yet does not exist all, When such an scenario arises and = is present, it doesn't really mean Assignment but it means Initialization.

  20. Copy assignment operator

    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.

  21. call copy constructor from assignment operator function

    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. FeatureValue::FeatureValue(const FeatureValue& other) : m_value(nullptr), m_size(0)