cppreference.com

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
  • Recent changes
  • Offline version
  • What links here
  • Related changes
  • Upload file
  • Special pages
  • Printable version
  • Permanent link
  • Page information
  • In other languages
  • This page was last modified on 2 February 2024, at 15:13.
  • This page has been accessed 1,333,785 times.
  • Privacy policy
  • About cppreference.com
  • Disclaimers

Powered by MediaWiki

Copy assignment operators (C++ only)

The copy assignment operator lets you create a new object from an existing one by initialization. A copy assignment operator of a class A is a nonstatic non-template member function that has one of the following forms:

  • A::operator=(A)
  • A::operator=(A&)
  • A::operator=(const A&)
  • A::operator=(volatile A&)
  • A::operator=(const volatile A&)

If you do not declare a copy assignment operator for a class A , the compiler will implicitly declare one for you that is inline public.

The assignment x = y calls the implicitly defined copy assignment operator of B , which calls the user-defined copy assignment operator A::operator=(const A&) . The assignment w = z calls the user-defined operator A::operator=(A&) . The compiler will not allow the assignment i = j because an operator C::operator=(const C&) has not been defined.

The implicitly declared copy assignment operator of a class A will have the form A& A::operator=(const A&) if the following statements are true:

  • A direct or virtual base B of class A has a copy assignment operator whose parameter is of type const B& , const volatile B& , or B .
  • A non-static class type data member of type X that belongs to class A has a copy constructor whose parameter is of type const X& , const volatile X& , or X .

If the above are not true for a class A , the compiler will implicitly declare a copy assignment operator with the form A& A::operator=(A&) .

The implicitly declared copy assignment operator returns an lvalue reference to the operator's argument.

The copy assignment operator of a derived class hides the copy assignment operator of its base class.

  • Class A has a nonstatic data member of a const type or a reference type
  • Class A has a nonstatic data member of a type which has an inaccessible copy assignment operator
  • Class A is derived from a base class with an inaccessible copy assignment operator.

An implicitly defined copy assignment operator of a class A will first assign the direct base classes of A in the order that they appear in the definition of A . Next, the implicitly defined copy assignment operator will assign the nonstatic data members of A in the order of their declaration in the definition of A .

This browser is no longer supported.

Upgrade to Microsoft Edge to take advantage of the latest features, security updates, and technical support.

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

  • Graphics and multimedia
  • Language Features
  • Unix/Linux programming
  • Source Code
  • Standard Library
  • Tips and Tricks
  • Tools and Libraries
  • Windows API
  • Copy constructors, assignment operators,

Copy constructors, assignment operators, and exception safe assignment

*

Copy assignment operator

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 & . A type with a public copy assignment operator is CopyAssignable .

[ edit ] Syntax

[ edit ] 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
  • 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.

[ edit ] 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)

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 ] Deleted implicitly-declared copy assignment operator

The implicitly-declared or defaulted copy assignment operator for class T is defined as deleted in any of the following is true:

  • T has a non-static data member that is const
  • T has a non-static data member of a reference type.
  • T has a non-static data member that cannot be copy-assigned (has deleted, inaccessible, or ambiguous copy assignment operator)
  • T has direct or virtual base class that cannot be copy-assigned (has deleted, inaccessible, or ambiguous move assignment operator)
  • T has a user-declared move constructor
  • T has a user-declared move assignment operator

[ edit ] Trivial copy assignment operator

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

  • The operator is not user-provided (meaning, it is implicitly-defined or defaulted), and if it is defaulted, its signature is the same as implicitly-defined
  • 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) memeber 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 ] Implicitly-defined copy assignment operator

If the implicitly-declared copy assignment operator is not deleted or trivial, it is defined (that is, a function body is generated and compiled) by the compiler. 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, 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.

[ edit ] Notes

If both copy and move assignment operators are provided, overload resolution selects the move assignment if the argument is an rvalue (either prvalue such as a nameless temporary or xvalue such as the result of std::move ), and selects the copy assignment if the argument is 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.

[ edit ] Copy and swap

Copy assignment operator can be expressed in terms of copy constructor, destructor, and the swap() member function, if one is provided:

T & T :: operator = ( T arg ) { // copy/move constructor is called to construct arg     swap ( arg ) ;     // resources exchanged between *this and arg     return * this ; }   // destructor is called to release the resources formerly held by *this

For non-throwing swap(), this form provides strong exception guarantee . For rvalue arguments, this form automatically invokes the move constructor, and is sometimes referred to as "unifying assignment operator" (as in, both copy and move).

[ edit ] Example

Learn C++

22.3 — Move constructors and move assignment

In lesson 22.1 -- Introduction to smart pointers and move semantics , we took a look at std::auto_ptr, discussed the desire for move semantics, and took a look at some of the downsides that occur when functions designed for copy semantics (copy constructors and copy assignment operators) are redefined to implement move semantics.

In this lesson, we’ll take a deeper look at how C++11 resolves these problems via move constructors and move assignment.

Recapping copy constructors and copy assignment

First, let’s take a moment to recap copy semantics.

Copy constructors are used to initialize a class by making a copy of an object of the same class. Copy assignment is used to copy one class object to another existing class object. By default, C++ will provide a copy constructor and copy assignment operator if one is not explicitly provided. These compiler-provided functions do shallow copies, which may cause problems for classes that allocate dynamic memory. So classes that deal with dynamic memory should override these functions to do deep copies.

Returning back to our Auto_ptr smart pointer class example from the first lesson in this chapter, let’s look at a version that implements a copy constructor and copy assignment operator that do deep copies, and a sample program that exercises them:

In this program, we’re using a function named generateResource() to create a smart pointer encapsulated resource, which is then passed back to function main(). Function main() then assigns that to an existing Auto_ptr3 object.

When this program is run, it prints:

(Note: You may only get 4 outputs if your compiler elides the return value from function generateResource())

That’s a lot of resource creation and destruction going on for such a simple program! What’s going on here?

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

  • Inside generateResource(), local variable res is created and initialized with a dynamically allocated Resource, which causes the first “Resource acquired”.
  • Res is returned back to main() by value. We return by value here because res is a local variable -- it can’t be returned by address or reference because res will be destroyed when generateResource() ends. So res is copy constructed into a temporary object. Since our copy constructor does a deep copy, a new Resource is allocated here, which causes the second “Resource acquired”.
  • Res goes out of scope, destroying the originally created Resource, which causes the first “Resource destroyed”.
  • The temporary object is assigned to mainres by copy assignment. Since our copy assignment also does a deep copy, a new Resource is allocated, causing yet another “Resource acquired”.
  • The assignment expression ends, and the temporary object goes out of expression scope and is destroyed, causing a “Resource destroyed”.
  • At the end of main(), mainres goes out of scope, and our final “Resource destroyed” is displayed.

So, in short, because we call the copy constructor once to copy construct res to a temporary, and copy assignment once to copy the temporary into mainres, we end up allocating and destroying 3 separate objects in total.

Inefficient, but at least it doesn’t crash!

However, with move semantics, we can do better.

Move constructors and move assignment

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).

Defining a move constructor and move assignment work analogously to their copy counterparts. However, whereas the copy flavors of these functions take a const l-value reference parameter (which will bind to just about anything), the move flavors of these functions use non-const rvalue reference parameters (which only bind to rvalues).

Here’s the same Auto_ptr3 class as above, with a move constructor and move assignment operator added. We’ve left in the deep-copying copy constructor and copy assignment operator for comparison purposes.

The move constructor and move assignment operator are simple. Instead of deep copying the source object (a) into the implicit object, we simply move (steal) the source object’s resources. This involves shallow copying the source pointer into the implicit object, then setting the source pointer to null.

When run, this program prints:

That’s much better!

The flow of the program is exactly the same as before. However, instead of calling the copy constructor and copy assignment operators, this program calls the move constructor and move assignment operators. Looking a little more deeply:

  • Res is returned back to main() by value. Res is move constructed into a temporary object, transferring the dynamically created object stored in res to the temporary object. We’ll talk about why this happens below.
  • Res goes out of scope. Because res no longer manages a pointer (it was moved to the temporary), nothing interesting happens here.
  • The temporary object is move assigned to mainres. This transfers the dynamically created object stored in the temporary to mainres.
  • The assignment expression ends, and the temporary object goes out of expression scope and is destroyed. However, because the temporary no longer manages a pointer (it was moved to mainres), nothing interesting happens here either.

So instead of copying our Resource twice (once for the copy constructor and once for the copy assignment), we transfer it twice. This is more efficient, as Resource is only constructed and destroyed once instead of three times.

When are the move constructor and move assignment called?

The move constructor and move assignment are called when those functions have been defined, and the argument for construction or assignment is an rvalue. Most typically, this rvalue will be a literal or temporary value.

The copy constructor and copy assignment are used otherwise (when the argument is an lvalue, or when the argument is an rvalue and the move constructor or move assignment functions aren’t defined).

Implicit move constructor and move assignment

The compiler will create an implicit move constructor and move assignment operator if all of the following are true:

  • There are no user-declared copy constructors or copy assignment operators.
  • There are no user-declared move constructors or move assignment operators.
  • There is no user-declared destructor.

The key insight behind move semantics

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

If we construct an object or do an assignment where the argument is an l-value, the only thing we can reasonably do is copy the l-value. We can’t assume it’s safe to alter the l-value, because it may be used again later in the program. If we have an expression “a = b” (where b is an lvalue), we wouldn’t reasonably expect b to be changed in any way.

However, if we construct an object or do an assignment where the argument is an r-value, then we know that r-value is just a temporary object of some kind. Instead of copying it (which can be expensive), we can simply transfer its resources (which is cheap) to the object we’re constructing or assigning. This is safe to do because the temporary will be destroyed at the end of the expression anyway, so we know it will never be used again!

C++11, through r-value references, gives us the ability to provide different behaviors when the argument is an r-value vs an l-value, enabling us to make smarter and more efficient decisions about how our objects should behave.

Key insight

Move semantics is an optimization opportunity.

Move functions should always leave both objects in a valid state

In the above examples, both the move constructor and move assignment functions set a.m_ptr to nullptr. This may seem extraneous -- after all, if a is a temporary r-value, why bother doing “cleanup” if parameter a is going to be destroyed anyway?

The answer is simple: When a goes out of scope, the destructor for a will be called, and a.m_ptr will be deleted. If at that point, a.m_ptr is still pointing to the same object as m_ptr , then m_ptr will be left as a dangling pointer. When the object containing m_ptr eventually gets used (or destroyed), we’ll get undefined behavior.

When implementing move semantics, it is important to ensure the moved-from object is left in a valid state, so that it will destruct properly (without creating undefined behavior).

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

In the generateResource() function of the Auto_ptr4 example above, when variable res is returned by value, it is moved instead of copied, even though res is an l-value. The C++ specification has a special rule that says automatic objects returned from a function by value can be moved even if they are l-values. This makes sense, since res was going to be destroyed at the end of the function anyway! We might as well steal its resources instead of making an expensive and unnecessary copy.

Although the compiler can move l-value return values, in some cases it may be able to do even better by simply eliding the copy altogether (which avoids the need to make a copy or do a move at all). In such a case, neither the copy constructor nor move constructor would be called.

Disabling copying

In the Auto_ptr4 class above, we left in the copy constructor and assignment operator for comparison purposes. But in move-enabled classes, it is sometimes desirable to delete the copy constructor and copy assignment functions to ensure copies aren’t made. In the case of our Auto_ptr class, we don’t want to copy our templated object T -- both because it’s expensive, and whatever class T is may not even support copying!

Here’s a version of Auto_ptr that supports move semantics but not copy semantics:

If you were to try to pass an Auto_ptr5 l-value to a function by value, the compiler would complain that the copy constructor required to initialize the function parameter has been deleted. This is good, because we should probably be passing Auto_ptr5 by const l-value reference anyway!

Auto_ptr5 is (finally) a good smart pointer class. And, in fact the standard library contains a class very much like this one (that you should use instead), named std::unique_ptr. We’ll talk more about std::unique_ptr later in this chapter.

Another example

Let’s take a look at another class that uses dynamic memory: a simple dynamic templated array. This class contains a deep-copying copy constructor and copy assignment operator.

Now let’s use this class in a program. To show you how this class performs when we allocate a million integers on the heap, we’re going to leverage the Timer class we developed in lesson 18.4 -- Timing your code . We’ll use the Timer class to time how fast our code runs, and show you the performance difference between copying and moving.

On one of the author’s machines, in release mode, this program executed in 0.00825559 seconds.

Now let’s run the same program again, replacing the copy constructor and copy assignment with a move constructor and move assignment.

On the same machine, this program executed in 0.0056 seconds.

Comparing the runtime of the two programs, (0.00825559 - 0.0056) / 0.00825559 * 100 = 32.1% faster!

Do not implement move semantics using std::swap

Since the goal of move semantics is to move a resource from a source object to a destination object, you might think about implementing the move constructor and move assignment operator using std::swap() . However, this is a bad idea, as std::swap() calls both the move constructor and move assignment on move-capable objects, which would result in an infinite recursion. You can see this happen in the following example:

This prints:

And so on… until the stack overflows.

You can implement the move constructor and move assignment using your own swap function, as long as your swap member function does not call the move constructor or move assignment. Here’s an example of how that can be done:

This works as expected, and prints:

Deleting the move constructor and move assignment

You can delete the move constructor and move assignment using the = delete syntax in the exact same way you can delete the copy constructor and copy assignment.

If you delete the copy constructor, the compiler will not generate an implicit move constructor (making your objects neither copyable nor movable). Therefore, when deleting the copy constructor, it is useful to be explicit about what behavior you want from your move constructors. Either explicitly delete them (making it clear this is the desired behavior), or default them (making the class move-only).

The rule of five says that if the copy constructor, copy assignment, move constructor, move assignment, or destructor are defined or deleted, then each of those functions should be defined or deleted.

While deleting only the move constructor and move assignment may seem like a good idea if you want a copyable but not movable object, this has the unfortunate consequence of making the class not returnable by value in cases where mandatory copy elision does not apply. This happens because a deleted move constructor is still declared, and thus is eligible for overload resolution. And return by value will favor a deleted move constructor over a non-deleted copy constructor. This is illustrated by the following program:

guest

  • C++ Data Types
  • C++ Input/Output
  • C++ Pointers
  • C++ Interview Questions
  • C++ Programs
  • C++ Cheatsheet
  • C++ Projects
  • C++ Exception Handling
  • C++ Memory Management

Related Articles

  • Solve Coding Problems
  • RAPTOR Tool - A Flowchart Interpreter
  • Features of C++
  • is_void template in C++
  • Date and Time Parsing in C++
  • Features of C++ 17
  • Nested Classes in C++
  • Local Classes in C++
  • OpenCV C++ Windows Setup using Visual Studio 2019
  • Overflow of Values While Typecasting in C ++
  • Signal Handling in C++
  • Applications and Popularities of C++
  • History of C++
  • Templates and Static variables in C++
  • A Puzzle on C/C++ R-Value Expressions
  • C++ Program that will fill whole memory
  • Introduction to Parallel Programming with OpenMP in C++
  • Is assignment operator inherited?
  • Different Ways to Initialize a Variable in C++

Copy Constructor vs Assignment Operator 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:

Consider the following C++ program. 

Explanation: Here, t2 = t1;  calls the assignment operator , same as t2.operator=(t1); and   Test t3 = t1;  calls the copy constructor , same as Test t3(t1);

Must Read: When is a Copy Constructor Called in C++?

Please Login to comment...

  • anshikajain26
  • connorsfitzgerald
  • 10 Best ChatGPT Prompts for Lawyers 2024
  • What is Meta’s new V-JEPA model? [Explained]
  • What is Chaiverse & How it Works?
  • Top 10 Mailchimp Alternatives (Free) - 2024
  • Dev Scripter 2024 - Biggest Technical Writing Event By GeeksforGeeks

Improve your Coding Skills with Practice

 alt=

What kind of Experience do you want to share?

IMAGES

  1. C++ Tutorial: Operator Overloading Part 5

    copy assignment operator c implementation

  2. Assignment Operators in C

    copy assignment operator c implementation

  3. Assignment Operators in C » PREP INSTA

    copy assignment operator c implementation

  4. Assignment Operators in C

    copy assignment operator c implementation

  5. C programming +=

    copy assignment operator c implementation

  6. [100% Working Code]

    copy assignment operator c implementation

VIDEO

  1. C++ Practical

  2. Lecture 74||Size of Operator||C programming Tutorials||Complete Playlist||Placements||

  3. Assignment Operator in C Programming

  4. Assignment Operator│C programming│Part# 15│Learn CSE Malayalam

  5. Working with Assignment operators in C| C Tutorials for Beginners

  6. C++ example program

COMMENTS

  1. Copy assignment operator

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

  2. Implementing the copy constructor in terms of operator=

    8 Answers Sorted by: 35 If all members of MyClass have a default constructor, yes. Note that usually it is the other way around: class MyClass { public: MyClass (MyClass const&); // Implemented void swap (MyClass&) throw (); // Implemented MyClass& operator= (MyClass rhs) { rhs.swap (*this); return *this; } };

  3. 21.12

    The copy assignment operator (operator=) is used to copy values from one object to another already existing object. Related content As of C++11, C++ also supports "Move assignment". We discuss move assignment in lesson 22.3 -- Move constructors and move assignment . Copy assignment vs Copy constructor

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

    8 contributors Feedback Note Starting in C++11, two kinds of assignment are supported in the language: copy assignment and move assignment. In this article "assignment" means copy assignment unless explicitly stated otherwise. For information about move assignment, see Move Constructors and Move Assignment Operators (C++).

  5. Copy assignment operators (C++ only)

    An implicitly defined copy assignment operator of a class A will first assign the direct base classes of A in the order that they appear in the definition of A. ... Note: This document describes the syntax, semantics, and IBM z/OS® XL C/C++ implementation of the C and C++ programming languages. For a general-purpose C or C++ standard reference ...

  6. 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

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

    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=.

  8. Copy constructors, assignment operators,

    1 2 3 4 MyClass ( const MyClass& other ); MyClass ( MyClass& other ); MyClass ( volatile const MyClass& other ); MyClass ( volatile MyClass& other ); Note that none of the following constructors, despite the fact that they could do the same thing as a copy constructor, are copy constructors: 1 2

  9. c++

    Copy constructor and assignment operator implementation choices - Asked 11 years, 7 months ago Modified 11 years, 7 months ago Viewed 2k times 6 I recently revisited the copy constructor, assignment operator, copy swap idom seen here: What is the copy-and-swap idiom? and many other places -

  10. PDF Copy Constructors and Assignment Operators

    assignment operators. While C++ will automatically provide these functions if you don't explicitly define them, in many cases you'll need to manually control how your objects are duplicated. This handout discusses copy constructors and assignment operators, including both high-level concepts and practical implementation techniques. Assignment ...

  11. Copy assignment operator

    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&. A type with a public copy assignment operator is CopyAssignable . Syntax Explanation

  12. 22.3

    By default, C++ will provide a copy constructor and copy assignment operator if one is not explicitly provided. These compiler-provided functions do shallow copies, which may cause problems for classes that allocate dynamic memory. So classes that deal with dynamic memory should override these functions to do deep copies.

  13. c++

    I'm just pushing (biased) style habits on you. With primitive types, it makes no difference, but with objects as members, assignment via initialization list constructs into the object rather than using the assignment operator. That means initialization list usage saves a wasted default construction and doesn't require a operator=.

  14. PDF CS11 Intro C++

    •This is called the copy-assignment operator •The default copy-assignment operator generated by C++ simply copies the values of all members from the RHS into the LHS •To write our own version of the copy -assignment operator, implement this member operator-overload function: Complex & Complex::operator=(constComplex &c);

  15. assignment operator overloading in c++

    { if(this == &rhs) return *this; itsRadius = rhs.getRadius(); return *this; } My Copy Constructor is this: SimpleCircle::SimpleCircle(const SimpleCircle & rhs) { itsRadius = rhs.getRadius(); } In the above operator overloading code, copy constructor is called as there is a new object is being created; hence I used the below code:

  16. c++

    A binary tree , its copy constructor and assignment operator. I implemented a binary tree in the following code. Its node's copy constructor and assignment operator should copy itself and all its descendents. Similarity a node's destructor should delete itself and all nodes descended from it.The print function just prints each node in a new line.

  17. c++

    2 I made the following operator overloading test: #include <iostream> #include <string> using namespace std; class TestClass { string ClassName; public: TestClass(string Name) { ClassName = Name; cout << ClassName << " constructed." << endl; } ~TestClass() { cout << ClassName << " destructed." << endl; }

  18. Copy Constructor vs Assignment Operator 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: Consider the following C++ program. CPP #include <iostream> #include <stdio.h> using namespace std; class Test { public: Test () {} Test (const Test& t) {

  19. c++

    Assignment operator and deep copy Ask Question Asked 10 years, 8 months ago Modified 10 years, 8 months ago Viewed 10k times 4 I'm learning C++ and there is something I don't get about assignment operators. As far as I understand, they are supposed to provide deep copies of object. Here is an example