c implement assignment operator in terms of copy constructor

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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.

[ edit ] Syntax

[ edit ] Explanation

The copy constructor is called whenever an object is initialized (by direct-initialization or copy-initialization ) from another object of the same type (unless overload resolution selects a better match or the call is elided ), which includes

• initialization: T a = b ; or T a ( b ) ; , where b is of type T ;
• function argument passing: f ( a ) ; , where a is of type T and f is void f ( T t ) ;
• function return: return a ; inside a function such as T f ( ) , where a is of type T , which has no move constructor .

[ edit ] Implicitly-declared copy constructor

If no user-defined copy constructors are provided for a class type, the compiler will always declare a copy constructor as a non- explicit inline public member of its class. This implicitly-declared copy constructor has the form T :: T ( const T & ) if all of the following are true:

• each direct and virtual base B of T has a copy constructor whose parameters are of type 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 constructor whose parameters are of type const M & or const volatile M & .

Otherwise, the implicitly-declared copy constructor is T :: T ( T & ) .

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

A class can have multiple copy constructors, e.g. both T :: T ( const T & ) and T :: T ( T & ) .

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

[ edit ] Implicitly-defined copy constructor

If the implicitly-declared copy constructor is not deleted, 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++11) . For union types, the implicitly-defined copy constructor copies the object representation (as by std::memmove ). For non-union class types, the constructor performs full member-wise copy of the object's direct base subobjects and member subobjects, in their initialization order, using direct initialization. For each non-static data member of a reference type, the copy constructor binds the reference to the same object or function to which the source reference is bound.

[ edit ] Deleted copy constructor

The implicitly-declared or explicitly-defaulted (since C++11) copy constructor 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 potentially constructed subobject of class type M (or possibly multi-dimensional array thereof) such that
• M has a destructor that is deleted or (since C++11) inaccessible from the copy constructor, or
• the overload resolution as applied to find M 's copy constructor
• 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 constructor

The copy constructor for class T is trivial if all of the following are true:

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

A trivial copy constructor for a non-union class effectively copies every scalar subobject (including, recursively, subobject of subobjects and so forth) of the argument and performs no other action. However, padding bytes need not be copied, and even the object representations of the copied subobjects need not be the same as long as their values are identical.

TriviallyCopyable objects can be copied by copying their object representations manually, e.g. with std::memmove . All data types compatible with the C language (POD types) are trivially copyable.

[ edit ] Eligible copy constructor

Triviality of eligible copy constructors determines whether the class is an implicit-lifetime type , and whether the class is a trivially copyable type .

[ edit ] Notes

In many situations, copy constructors are optimized out even if they would produce observable side-effects, see copy elision .

[ 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 assignment
• copy elision
• default constructor
• aggregate initialization
• constant initialization
• copy initialization
• default initialization
• direct initialization
• initializer list
• list initialization
• reference initialization
• value initialization
• zero initialization
• move assignment
• move constructor
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Copy 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 & . For a type to be CopyAssignable , it must have a public copy assignment operator.

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;

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.

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C++ At Work

Copy Constructors, Assignment Operators, and More

Paul DiLascia

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

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

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

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

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

Figure 1 Copy Constructors and Assignment Operators

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

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

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

As opposed to:

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

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

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

Sunil Peddi

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

How can I resolve this problem?

Giuseppe Dattilo

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

Figure 3 Managed Person Class

Figure 2 Native CPerson Class

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

Figure 4** Interop Highway **

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

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

Figure 5 PtrToStringChars

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

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

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

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

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

Additional resources

14.14 — Introduction to the copy constructor

The copy constructor

This is illustrated in the following example:

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Copy Constructor in C++

A copy constructor is a type of constructor 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 .  

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.

Syntax of Copy Constructor in C++

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

Here, the const qualifier is optional but is added so that we do not modify the obj by mistake.

Syntax of Copy Constructor

Examples of Copy Constructor in C++

Example 1: user defined copy constructor.

If the programmer does not define the copy constructor, the compiler does it for us.

Example 2: 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.

Need of User Defined Copy Constructor

If we don’t define our own copy constructor, the C++ compiler creates a default copy constructor for each class which works fine in general. However, 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 because the default constructor does only shallow copy.

Shallow Copy means that only the pointers will be copied not the actual resources that the pointers are pointing to. This can lead to dangling pointers if the original object is deleted.

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 copy of the dynamic resource allocated manually in the copy constructor using new operators.

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. 

Note: Such classes also need the overloaded assignment operator. See this article for more info – C++ Assignment Operator Overloading

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 c hanges made to str2 reflect in str1 as well which is never expected. Also, if the str1 is destroyed, the str2’s data member s will be pointing to the deallocated memory.

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).

Refer to this article for more details – When is a Copy Constructor Called in C++?

Copy Elision

In copy elision, the compiler prevents the making of extra copies by making the use to techniques such as NRVO and RVO which results in saving space and better the program complexity (both time and space); Hence making the code more optimized.

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.

Frequently Asked Questions in C++ Copy Constructors

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?  

If you pass the object by value in the copy constructor, it will 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 recursion. Using a reference avoids this recursion. So, we use reference of objects to avoid infinite calls.

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?’

Related Articles:

• Constructors in C++

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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. How can I pretty-print it? Please point to errors(if any) and suggestions.

• constructor

• 1 \$\begingroup\$ 1) I would make three functions for printing: {pre, in, post}-order. 2) Use std::shared_ptr<T> instead of raw pointers - you will not need to implement your own destructor in that case. \$\endgroup\$ –  Ryan Dougherty Commented May 19, 2015 at 5:28
• \$\begingroup\$ An interesting addition would be to try and implement the move assignment operator and move constructor as well. \$\endgroup\$ –  Miklas Commented May 19, 2015 at 9:38

2 Answers 2

Assignment operator..

The assignment operator is not 100% exception safe. You should not modify the current object while you have not yet finished making the copy.

It should look more like this:

Now that looks like a lot of hard work. There is an easier way to achieve exactly the same effect. You can use the copy and swap idiom.

Calling delete on a null pointer is valid and does nothing.

So we can simplify this too.

Why do you need a make_copy ?

It is quite normal to call the copy constructor directly.

Print statement

You can make the print function a method. Also to make it more versatile you should pass a stream to which you want to print (so it can also print to file). If you want a no-argument version just make the stream parameter default to std::cout .

This makes it easy to define the output operator for your class.

• \$\begingroup\$ Why are you deleting left and right in assignment operator's catch block? \$\endgroup\$ –  bisarch Commented May 19, 2015 at 23:32
• \$\begingroup\$ Also, why didn't you use std::swap in the copy and swap idiom based assignment operator? \$\endgroup\$ –  bisarch Commented May 19, 2015 at 23:45
• \$\begingroup\$ @sank: Opps. That's wrong. Fixed. \$\endgroup\$ –  Loki Astari Commented May 20, 2015 at 20:16
• \$\begingroup\$ @sank: Normally in the copy and swap idium you would use the class's own swap method (which is also called by a custom swap function). The swap method would use swap to swap the state of the objects internal members. That way you only have one place where the state of the object is swapped (and thus one place to update when you update the state of the object). \$\endgroup\$ –  Loki Astari Commented May 20, 2015 at 20:19

I don't think that this is the right way to create a copy, because this will create a shallow copy, deleting one tree might delete the other tree also.

• 2 \$\begingroup\$ Are you sure? According to the description, “its node's copy constructor and assignment operator should copy itself and all its descendents.” \$\endgroup\$ –  Martin R Commented Sep 6, 2022 at 8:26

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Implement C++ assignment operator in terms of constructor

Suppose you want to implement a resource-managing class in C++. You cannot use the Rule of Zero or Rule of Five Defaults , so you actually need to implement copy and move constructor, copy and move assignment operator, and destructor. In this question, Iʼll use Box as an example, but this can apply to lots of different types.

(Note that a better Box implementation would have more features like noexcept and constexpr functions, explicit forwarding constructors based on value_type ʼs constructors, allocator support, etc; here Iʼm implementing just whatʼs necessary for the question and the tests. It might also save some code with std::unique_ptr , but that would make it a less-clear example)

Note that the assignment operators share lots of code with each other, with their respective constructors, and with the destructor. This would be somewhat lessened if I didnʼt want to allow assignment to moved-from Box en, but would be more apparent in a more complicated class.

Digression: Copy-and-Swap

One standard way of handling this is with the Copy-And-Swap Idiom (also in this context called the Rule of Four and a Half) , which also gets you a strong exception guarantee if swap is nothrow :

This allows you to write only one assignment operator (taking other by value lets the compiler move the other to the parameter if possible, and does the copying for you if necessary), and that assignment operator is simple (assuming you already have swap ). However, as the linked article says, that has issues such as doing an extra allocation and keeping extra copies of the contents around during the operation.

What I havenʼt seen before is something Iʼll call a Destroy-and-Initialize assignment operator. Since weʼre already doing all the work in the constructor, and an assigned-to version of the object should be identical to a copy-constructed one, why not use the constructor, as follows:

This still does an extra allocation like Copy-and-Swap does, but only in the copy assignment case instead of the move assignment case, and it does it after destroying one of the copies of T , so it doesnʼt fail under resource constraints.

• Has this been proposed before, and if so where can I read more about it?
• Is this UB in some cases, such as if the Box is a subobject of something else, or is it allowed to destroy and re-construct subobjects?
• Does this have any downsides I havenʼt mentioned, like not being constexpr compatible?
• Are there other options to avoid assignment operator code reuse, like this and the Rule of Four and a Half, when you canʼt just = default them?
• copy-and-swap
• rule-of-zero
• rule-of-five

• 2 This approach will have UB if any of the non-static member of the class are const qualified or of a reference type: stackoverflow.com/a/58415092/4342498 –  NathanOliver Commented Nov 30, 2021 at 21:55
• @NathanOliver Yeah, but nothing else would work either , unless you want to special-case the handling of that member and it makes sense to have that not change on assignment. –  Daniel H Commented Nov 30, 2021 at 22:34
• That approach will cause invalid object if copy constructor throws. –  Phil1970 Commented Dec 1, 2021 at 0:18
• @Phil1970 Ah, youʼre right, and at least in my example Box thereʼs no way to avoid it without just specifying that it requires T be nothrow copy constructible, which isnʼt otherwise required. And even then since the copy constructor is allocating memory, that might fail. It could work for move assignment in this case, which avoids some of the code reuse but not as much. –  Daniel H Commented Dec 1, 2021 at 0:47
• I think that Copy and swap is adequate 99% of the time. The extra copy (when required) make the assignment transactional which is generally a good thing. If you really need more control, you can always write the operator yourself. –  Phil1970 Commented Dec 2, 2021 at 1:54

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