difference between copy constructor and overloaded assignment operator

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

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

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

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

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

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

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

The object-oriented programming idea is supported by the general-purpose, middle-level, case-sensitive, platform agnostic computer language C++. Bjarne Stroustrup developed the C++ programming language in 1979 at Bell Laboratories. Since C++ is a platform-independent programming language, it may be used on a wide range of operating systems, including Windows, Mac OS, and different UNIX versions.

Assignment operators are used to assign values to variables from the aforementioned groups.

Let's first study a little bit about constructors before moving on to copy constructors. When an object is formed, a specific method called a constructor-which has the same name as the class name with the parentheses "()"-is automatically called. Initializing the variables of a freshly generated object is done using the constructor.

A copy constructor is a form of constructor that uses an already-created object from the same class to initialize a new object.

Let's now go over the specifics of the notion and contrast and compare the features of the assignment operator and copy constructor.

The assignment operator is used to give a variable a value. The assignment operator has a variable name as its left operand and a value to that variable as its right operand. A compilation error will be triggered if neither operand's datatype matches the other.

Assignment operators come in various forms.

: = operator Only the value is given to the variable. For instance, if "a=10," variable "a" will be given the value of 10.

The += operator first multiplies the variable's current value by the value on the right side before assigning the new value.

The operator "-=" first subtracts the variable's current value from the value on the right side before appending the new value.

In order to assign a new value to a variable, the *= operator first multiplies the variable's current value by the value on the right side.

The operator /= first divides the variable's current value by the value on the right side before assigning the new value to the variable.

Below is an illustration of an assignment operator. The assignment operator is being used in this case to give values to several variables.

The two variables "a" and "b" were used in the example above, and we first set the value of "a" to 5 using the assignment operator "=". And we've given variable b the value of the a variable. The output from the aforementioned code is shown below.

Programmers frequently need to do this in order to duplicate an object without affecting the original. The copy constructor is used in these circumstances. The copy constructor produces an object by initializing it with a different object of the same class that has already been constructed. The copy constructor comes in two varieties.

The default copy constructor is created by the C++ compiler when the copy constructor is not declared, and it copies all member variables exactly as they are.

User-Defined Copy Constructor: This term refers to a copy constructor that has been defined by the user.

The syntax for Copy Constructor is -

All of these C++ concepts' primary functions are to assign values, but the key distinction between them is that while the copy constructor produces a new object and assigns the value, the assignment operator assigns the value to the data member of the same object rather than to a new object.

The key distinctions between assignment operator and copy constructor are shown in the following table.

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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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Home » Technology » IT » Programming » What is the Difference Between Copy Constructor and Assignment Operator

What is the Difference Between Copy Constructor and Assignment Operator

The main difference between copy constructor and assignment operator is that copy constructor is a type of constructor that helps to create a copy of an already existing object without affecting the values of the original object while assignment operator is an operator that helps to assign a new value to a variable in the program.

A constructor is a special method that helps to initialize an object when creating it. It has the same name as the class name and has no return type. A programmer can write a constructor to give initial values to the instance variables in the class. If there is no constructor in the program, the default constructor will be called. Copy constructor is a type of constructor that helps to create a copy of an existing object. On the other hand, assignment operator helps to assign a new value to a variable.

Key Areas Covered

1. What is Copy Constructor      – Definition, Functionality 2. What is Assignment Operator      – Definition, Functionality 3. What is the Difference Between Copy Constructor and Assignment Operator      – Comparison of Key Differences

Constructor, Copy Constructor, Assignment Operator, Variable

What is Copy Constructor

In programming, sometimes it is necessary to create a separate copy of an object without affecting the original object. Copy constructor is useful in these situations. It allows creating a replication of an existing object of the same class. Refer the below example.

Figure 1: Program with copy constructor

The class Triangle has two instance variables called base and height. In line 8, there is a parameterized constructor. It takes two arguments. These values are assigned to the instance variables base and height. In line 13, there is a copy constructor. It takes an argument of type Triangle. New object’s base value is assigned to the instance variable base. Similarly, the new object’s height value is assigned to the instance variable height. Furthermore, there is a method called calArea to calculate and return area.

In the main method, t1 and t2 are Triangle objects. The object t1 is passed when creating the t2 object. The copy constructor is called to create t2 object. Therefore, the base and the height of the t2 object is the same as the base and height of t1 object. Finally, both objects have the same area.    

What is Assignment Operator

An assignment operator is useful to assign a new value to a variable. The assignment operator is “=”.  When there is a statement as c = a + b; the summation of ‘a’ and ‘b’ is assigned to the variable ‘c’.

Figure 2: Program with assignment operator

The class Number has an instance variable called num. There is a no parameter constructor in line 7. However, there is a parameterized constructor in line 9. It takes an argument and assigns it to the instance variable using the assignment operator. In line 12, there is a method called display to display the number. In the main method, num1 and num2 are two objects of type Number. Printing num1 and num2 gives the references to those objects. The num3 is of type Number. In line 24, num1 is assigned to num3 using the assignment operator. Therefore, num3 is referring to num1 object. So, printing num3 gives the num1 reference.  

The assignment operator and its variations are as follows.

Difference Between Copy Constructor and Assignment Operator

Copy constructor is a special constructor for creating a new object as a copy of an existing object. In contrast, assignment operator is an operator that is used to assign a new value to a variable. These definitions explain the basic difference between copy constructor and assignment operator.

Functionality with Objects

Functionality with objects is also a major difference between copy constructor and assignment operator. Copy constructor initializes the new object with an already existing object while assignment operator assigns the value of one object to another object which is already in existence.

Copy constructor helps to create a copy of an existing object while assignment operator helps to assign a new value to a variable. This is another difference between copy constructor and assignment operator.

The difference between copy constructor and assignment operator is that copy constructor is a type of constructor that helps to create a copy of an already existing object without affecting the values of the original object while assignment operator is an operator that helps to assign a new value to a variable in the program.

1. Thakur, Dinesh. “Copy Constructor in Java Example.” Computer Notes, Available here .

About the Author: Lithmee

Lithmee holds a Bachelor of Science degree in Computer Systems Engineering and is reading for her Master’s degree in Computer Science. She is passionate about sharing her knowldge in the areas of programming, data science, and computer systems.

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What's the difference between assignment operator and copy constructor in C++?

The Copy constructor and the assignment operators are used to initialize 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 reference variable to point to the previous memory block.

Copy Constructor (Syntax)

Assignment operator (syntax).

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

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Thursday, February 23, 2012

Difference between copy constructor and overloading assignment operator in c++.

• Copy constructor is used to create the new object and assigning values of other object to the newly created object
• Assignment operator is used to assign the values from one object to the already existing object. Here already existing is the main one.

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Why Copy Constructor is called here instead of normal Constructor and overloaded assignment operator? [duplicate]

Possible Duplicate: Is there a difference in C++ between copy initialization and direct initialization? Copy constructors and Assignment Operators

I have a class C in which I have overloaded Normal, copy constructor and assignment operator to print a trace of what is being called..

I wrote following pieces of code to test what is being called when?

This seems to baffle me since in this code though I am initializing at the time of declaration, it is through assignment operator and not copy constructor .. Am I understanding it wrong ??

• copy-constructor

• That syntax is merely a confusing syntactical sugar that makes the = sign not be an assignment. –  Seth Carnegie Commented May 15, 2012 at 6:48

3 Answers 3

Because C c4 = c1; is syntactically semantically equivalent to:

Both would invoke copy constructor in this particular case. However there is a subtle difference between "copy initialization" (1st syntax) and "direct initialization" (2nd syntax). Look at this answer.

Note : In "layman's terms", a variable (here c4 ) is constructed till the first ; (or , ' for multiple objects) is encountered; Till then everything is one or the other type of constructor.

In case of C c4 = c1; , compiler doesn't have to check for most vexing parse . However one can disable C c4 = c1; kind of syntax by declaring copy constructor explicit . For that matter any constructor can be made explicit and you can prevent = sign in the construction.

• Yeah.. that was my actual doubt .. Why C c4 = c1; is syntactically equivalent to C c4(C1) and not C C4; c4 = c1; Why was it made that way..? –  Anerudhan Gopal Commented May 15, 2012 at 6:49
• @AnerudhanGopal because that's just the way it is (and maybe for the reason that the former is more efficient). –  Seth Carnegie Commented May 15, 2012 at 6:50
• @AnerudhanGopal, in c++ for many constructs there are multiple syntax. this is one of them. I have edited my answer for more details. –  iammilind Commented May 15, 2012 at 6:59
• 1 The first part of the answer is still misleading which states, C c4 = c1; is syntactically equivalent to: C c4(c1); , It is not! They might behave similarly sometimes but they are definitely not the same. –  Alok Save Commented May 15, 2012 at 7:11
• 3 I have several problems with this answer. The first is purely linguistic: C c4 = c1; is clearly not syntactically equivalent to C c4( c1 ); . You probably meant to say semantically equivalent, which is only true if the initializing expression ( c1 in this case) has the same type as the new object. And finally, c4 is fully constructed at the end of the declarator, which may be before the ; , if you define two variables in the same statement: C c4 = c1, c5 = c4; . (Bad practice, IMHO, but perfectly legal.) –  James Kanze Commented May 15, 2012 at 7:53

is Copy Initialization .

It tries to convert c1 to an type C if it is already of not that type by finding a suitable conversion function and then uses the the created C instance for copy constructing a new C instance.

Note that though,

is Direct Initialization

And it is important to note that there is a difference between Copy Initialization and Direct Initialization they are not the same!

Why C c4 = c1; is syntactically equivalent to C c4(C1) and not C C4; c4 = c1; Why was it made that way? First, Copy Initialization & Direct Initialization are not the same. 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:

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. So Basically, what happens here is Initialization and not Assignment and the C++ Standard defines specific rules as to how this initialization should be done.

Because there's no assignment operator in:

Like the comma, the equal sign can be either an operator or punctuation. In an expression, it is always an operator, but in the above, the initialization expression is simply c1 ; the = sign is punctuation, indicating that copy initialization, rather than direct initialization, is to be used. (Direct initialization would be

and is, IMHO, generally preferable. In the case where the initialization expression has the same type as the new object, however, there is no difference.)

Not the answer you're looking for? Browse other questions tagged c++ copy-constructor or ask your own question .

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• What sort of impact did the discovery that water could be broken down (via electrolysis) into gas have?
• Does the overall mAh of the battery add up when batteries are parallel?
• Why do combinatorists care about Kazhdan–Lusztig polynomials?
• grep command fails with out-of-memory error
• TeXbook Exercise 21.10 Answer
• Can you successfully substitute pickled onions for baby onions in Coq Au Vin?
• The hat-check problem
• Sticker on caption phone says that using the captions can be illegal. Why?
• Why if gravity would be higher, designing a fully reusable rocket would be impossible?