//body of the constructor
}
A “copy constructor” is a form of an overloaded constructor . A copy constructor is only called or invoked for initialization purpose. A copy constructor initializes the newly created object by another existing object.
When a copy constructor is used to initialize the newly created target object, then both the target object and the source object shares a different memory location. Changes done to the source object do not reflect in the target object. The general form of the copy constructor is
If the programmer does not create a copy constructor in a C++ program, then the compiler implicitly provides a copy constructor. An implicit copy constructor provided by the compiler does the member-wise copy of the source object. But, sometimes the member-wise copy is not sufficient, as the object may contain a pointer variable.
Copying a pointer variable means, we copy the address stored in the pointer variable, but we do not want to copy address stored in the pointer variable, instead, we want to copy what pointer points to. Hence, there is a need of explicit ‘copy constructor’ in the program to solve this kind of problems.
A copy constructor is invoked in three conditions as follow:
Let us understand copy constructor with an example.
In the code above, I had explicitly declared a constructor “copy( copy &c )”. This copy constructor is being called when object B is initialized using object A. Second time it is called when object C is being initialized using object A.
When object D is initialized using object A the copy constructor is not called because when D is being initialized it is already in the existence, not the newly created one. Hence, here the assignment operator is invoked.
The assignment operator is an assigning operator of C++. The “=” operator is used to invoke the assignment operator. It copies the data in one object identically to another object. The assignment operator copies one object to another member-wise. If you do not overload the assignment operator, it performs the bitwise copy. Therefore, you need to overload the assignment operator.
In above code when object A is assigned to object B the assignment operator is being invoked as both the objects are already in existence. Similarly, same is the case when object C is initialized with object A.
When the bitwise assignment is performed both the object shares the same memory location and changes in one object reflect in another object.
The Copy constructor is best for copying one object to another when the object contains raw pointers.
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The Copy constructor and the assignment operators are used to initializing one object to another object. The main difference between them is that the copy constructor creates a separate memory block for the new object. But the assignment operator does not make new memory space. It uses the reference variable to point to the previous memory block.
Assignment operator (syntax).
Let us see the detailed differences between Copy constructor and Assignment Operator.
Copy Constructor | Assignment Operator |
---|---|
The Copy constructor is basically an overloaded constructor | Assignment operator is basically an operator. |
This initializes the new object with an already existing object | This assigns the value of one object to another object both of which are already exists. |
Copy constructor is used when a new object is created with some existing object | This operator is used when we want to assign existing object to new object. |
Both the objects uses separate memory locations. | One memory location is used but different reference variables are pointing to the same location. |
If no copy constructor is defined in the class, the compiler provides one. | If the assignment operator is not overloaded then bitwise copy will be made |
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An implicit copy constructor
Prefer the implicit copy constructor
The copy constructor’s parameter must be a reference
When generateFraction returns a Fraction back to main , the copy constructor is implicitly called again. And when f2 is passed to printFraction , the copy constructor is called a third time.
The rule of three is a well known C++ principle that states that if a class requires a user-defined copy constructor, destructor, or copy assignment operator, then it probably requires all three. In C++11, this was expanded to the rule of five , which adds the move constructor and move assignment operator to the list.
When is a user-defined copy constructor needed, copy constructor vs assignment operator, how to make the copy constructor private, why does an argument to a copy constructor must be passed as a reference, why should the copy constructor argument be const in c++, understanding c++ copy constructor with example code.
When a new object is generated from an existing object as a copy of the existing object, the copy function Object () { [native code] } is named. When a previously initialized object is given a new value from another object, the assignment operator is used. Function Object () { [native code] } is a member function that uses another object of the same class to initialize an object. The following is a general function prototype for a copy function Object () { [native code] }:
Classname(const classname &obj);
#include<iostream>
using namespace std;
class Point
int a, b;
Point(int x1, int y1) { a =x1; b = y1; }
// Copy constructor
Point(const Point &p1) {a = p1.a; b = p1.b; }
int getX() { return a; }
int getY() { return b; }
Point p1(10, 15); // Normal constructor is called here
Point p2 = p1; // Copy constructor is called here
// Let us access values assigned by constructors
cout << "p1.x = " << p1.getX() << ", p1.y = " << p1.getY();
cout << "\np2.x = " << p2.getX() << ", p2.y = " << p2.getY();
return 0;
According to C++ copy constructor the C++ compiler produces a default copy constructor Object() { [native code] } for each class if it doesn’t define own copy constructor Object() { [native code] }, which performs a member-wise copy between items. In general, the copy function Object() { [native code] } generated by the compiler works well. It just needs to define its own copy function Object() [native code] if an object has pointers or any runtime resource allocation, such as a file handle or a network connection.
When a new object is generated from an existing object as a copy of the existing object, the copy function Object() { [native code] } is named. When an already initialized object is given a new value from another object, the assignment operator is used.
Myclass c1,c2;
Myclass c3;
The above line illustrates the copy constructor
#include<cstring>
class String
char *s;
int size;
String(const char *str = NULL); // constructor
~String() { delete [] s; }// destructor
String(const String&); // copy constructor
void print() { cout << s << endl; } // Function to print string
void change(const char *); // Function to change
String::String(const char *str)
size = strlen(str);
s = new char[size+1];
strcpy(s, str);
void String::change(const char *str)
delete [] s;
String::String(const String& old_str)
size = old_str.size;
strcpy(s, old_str.s);
String str1("Welcome");
String str2 = str1;
str1.print(); // what is printed ?
str2.print();
str2.change("Welcome you all");
str1.print(); // what is printed now ?
In c++ copy constructor users don't get the predicted result if we delete the copy function Object() { [native code] } from the above program. Changes to str2 are reflected in str1 as well, which is unexpected.
char *s;
int size;
String(const char *str = NULL); // constructor
~String() { delete [] s; }// destructor
void print() { cout << s << endl; }
void change(const char *); // Function to change
String::String(const char *str)
size = strlen(str);
s = new char[size+1];
strcpy(s, str);
delete [] s;
String str1("Welcome");
String str2 = str1;
str1.print(); // what is printed ?
str2.print();
str2.change("Welcome you all");
str1.print(); // what is printed now ?
str2.print();
According to the C + + copy constructor it is possible to make a copy function Object() { [native code] }. When a copy function Object() { [native code] } in a class is made private, objects in that class become non-copyable. This is especially useful when the class contains pointers or resources that are dynamically allocated. In such cases, it can either write our own copy function Object() { [native code] }, as in the String example above, or create a private copy function Object() { [native code] }, in which case users will receive compiler errors rather than warnings.
According to C + + copy constructor, when an object is transferred by value, a copy function Object() { [native code] } is named. The copy function Object() { [native code] } is a function in and of itself. If an argument is passed by value in a copy function Object() { [native code] }, a call to copy function Object() { [native code] } is made to call copy function Object() { [native code] }, resulting in a non-terminating sequence of calls. As a result, the compiler won't let you move parameters by value.
According to C++ copy constructor, we pass an object by reference to the copy function Object() { [native code] }, and we usually pass it as a const reference. One justification for passing a const reference is that it can use const wherever possible in C++ to avoid unintentionally changing objects. This is one compelling excuse to transfer references as const, but there are others.
/* Class data members */
Test(Test &t) { /* Copy data members from t*/}
Test() { /* Initialize data members */ }
cout << "fun() Called\n";
Test t;
return t;
Test t1;
Test t2 = fun();
Need to change the copy constructor by the following:
Test(const Test &t) { cout << "Copy Constructor Called\n"; }
The fun() function returns a value. As a result, the compiler generates a temporary entity, which is then copied to t2 using the original program's copy function Object() { [native code] } (The temporary object is passed as an argument to the copy constructor). The compiler error occurs because compiler-created temporary objects cannot be bound to non-const references, despite the fact that the original program attempts to do so. It doesn't make sense to change something that isn't broken.
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According to C++ copy constructor, when an object is made, a function Object() { [native code] } is a special type of member function that is called automatically. A function Object() { [native code] } in C++ has the same name as the class it belongs to, and it does not have a return form.
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Copy constructors.
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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.
Syntax Explanation Implicitly-declared copy constructor Implicitly-defined copy constructor Deleted copy constructor Trivial copy constructor Eligible copy constructor Notes Example Defect reports See also |
class-name parameter-list | (1) | ||||||||
class-name parameter-list function-body | (2) | ||||||||
class-name single-parameter-list | (3) | (since C++11) | |||||||
class-name parameter-list | (4) | (since C++11) | |||||||
class-name class-name parameter-list function-body | (5) | ||||||||
class-name class-name single-parameter-list | (6) | (since C++11) | |||||||
class-name | - | the class whose copy constructor is being declared |
parameter-list | - | a non-empty satisfying all following conditions: , the first parameter is of type T&, const T&, volatile T& or const volatile T&, and . |
single-parameter-list | - | a of only one parameter, which is of type T&, const T&, volatile T& or const volatile T& and does not have a default argument |
function-body | - | the of the copy constructor |
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
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:
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 & ) .
Even if some user-defined copy constructors are present, the user may still force the implicit copy constructor declaration with the keyword default. | (since C++11) |
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) .
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.
If this satisfies the requirements of a (until C++23) (since C++23), the generated copy constructor is constexpr. The generation of the implicitly-defined copy constructor is deprecated if has a user-defined destructor or user-defined copy assignment operator. | (since C++11) |
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:
has a non-static data member of rvalue reference type. | (since C++11) |
The implicitly-declared copy constructor for class is defined as deleted if declares a or . | (since C++11) |
The copy constructor for class T is trivial if all of the following are true:
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.
A copy constructor is eligible if it is either user-declared or both implicitly-declared and definable. | (until C++11) |
A copy constructor is eligible if it is not deleted. | (since C++11) (until C++20) |
A copy constructor is eligible if all following conditions are satisfied: (if any) are satisfied. than any other copy constructor. | (since C++20) |
Triviality of eligible copy constructors determines whether the class is an implicit-lifetime type , and whether the class is a trivially copyable type .
In many situations, copy constructors are optimized out even if they would produce observable side-effects, see copy elision .
[ edit ] defect reports.
The following behavior-changing defect reports were applied retroactively to previously published C++ standards.
DR | Applied to | Behavior as published | Correct behavior |
---|---|---|---|
C++98 | the conditions where implicitly-declared copy constructors are undefined did not consider multi-dimensional array types | consider these types | |
C++11 | volatile members make copy non-trivial ( ) | triviality not affected | |
C++11 | X(X&) = default was non-trivial | made trivial | |
C++20 | a copy constructor was not eligible if there is another copy constructor which is more constrained but does not satisfy its associated constraints | it can be eligible in this case |
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This topic describes how to write a move constructor and a move assignment operator for a C++ class. A move constructor enables the resources owned by an rvalue object to be moved into an lvalue without copying. For more information about move semantics, see Rvalue Reference Declarator: && .
This topic builds upon the following C++ class, MemoryBlock , which manages a memory buffer.
The following procedures describe how to write a move constructor and a move assignment operator for the example C++ class.
Define an empty constructor method that takes an rvalue reference to the class type as its parameter, as demonstrated in the following example:
In the move constructor, assign the class data members from the source object to the object that is being constructed:
Assign the data members of the source object to default values. This prevents the destructor from freeing resources (such as memory) multiple times:
Define an empty assignment operator that takes an rvalue reference to the class type as its parameter and returns a reference to the class type, as demonstrated in the following example:
In the move assignment operator, add a conditional statement that performs no operation if you try to assign the object to itself.
In the conditional statement, free any resources (such as memory) from the object that is being assigned to.
The following example frees the _data member from the object that is being assigned to:
Follow steps 2 and 3 in the first procedure to transfer the data members from the source object to the object that is being constructed:
Return a reference to the current object, as shown in the following example:
The following example shows the complete move constructor and move assignment operator for the MemoryBlock class:
The following example shows how move semantics can improve the performance of your applications. The example adds two elements to a vector object and then inserts a new element between the two existing elements. The vector class uses move semantics to perform the insertion operation efficiently by moving the elements of the vector instead of copying them.
This example produces the following output:
Before Visual Studio 2010, this example produced the following output:
The version of this example that uses move semantics is more efficient than the version that does not use move semantics because it performs fewer copy, memory allocation, and memory deallocation operations.
To prevent resource leaks, always free resources (such as memory, file handles, and sockets) in the move assignment operator.
To prevent the unrecoverable destruction of resources, properly handle self-assignment in the move assignment operator.
If you provide both a move constructor and a move assignment operator for your class, you can eliminate redundant code by writing the move constructor to call the move assignment operator. The following example shows a revised version of the move constructor that calls the move assignment operator:
The std::move function converts the lvalue other to an rvalue.
Rvalue Reference Declarator: && std::move
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I have the following code to test out my understanding of basic pointers in C++:
This code works exactly as expected as is, it prints out:
However if I remove the copy constructor it prints out something like:
I don't understand why this is the case. My understanding is that the copy constructor is used when the assignment is to an object that doesn't exist. Here intVal1 does exist, so why isn't the assignment operator called?
The copy constructor is not used during assignment. Copy constructor in your case is used when passing arguments to displayInteger function. The second parameter is passed by value, meaning that it is initailized by copy contructor.
Your version of copy constructor performs deep copying of data owned by the class (just like your assignment operator does). So, everything works correctly with your version of copy constructor.
If you remove your own copy constructor, the compiler will generate one for you implicitly. The compiler-generated copy constructor will perform shallow copying of the object. This will violate the "Rule of Three" and destroy the functionality of your class, which is exactly what you observe in your experiment. Basically, the first call to displayInteger damages your intVal1 object and the second call to displayInteger damages your intVal2 object. After that both of your objects are broken, which is why the third displayInteger call displays garbage.
If you change the declaration of displayInteger to
your code will "work" even without an explicit copy constructor. But it is not a good idea to ignore the "Rule of Three" in any case. A class implemented in this way either has to obey the "Rule of Three" or has to be made non-copyable.
The problem you're experiencing is caused by the default copy constructor, which copies the pointer but doesn't associate it with newly allocated memory (like your implementation of copy constructor does). When you pass object by value, a copy is created and when the execution goes out of scope, this copy is destructed. delete from the destructor invalidates the value pointer of intVal1 object, making it dangling pointer , dereferencing of which causes undefined behavior .
Debug outputs might be used to understand the behavior of your code:
Now output of this code:
ctor ctor(int) copy ctor 0x9ed4028 0 destructor copy ctor 0x9ed4038 10 destructor assignment copy ctor 0x9ed4048 10 destructor destructor destructor
which shows that copy constructor is used when passing objects by value. However, important to notice here is the destructor called upon the return from your function. And if you remove your implementation of copy constructor, then the output is:
ctor ctor(int) 0x8134008 0 destructor 0x8134018 10 destructor assignment 0x8134008 135479296 destructor destructor destructor
showing that the first copy called delete on the same pointer (pointing to 0x8134008 ) as was used by third copy later, where the memory pointed by this dangling pointer has been used.
Think about this call:
You are creating a copy of intVal1 into the intObj parameter of displayInteger :
That copy will be pointing to the same int that intVal1 is. When displayInteger returns, intObj is destroyed, which will cause the int to be destroyed, and the pointer in intVal1 to be pointing to an invalid object. At that point all bets are off (A.K.A. undefined behavior) if you try to access the value. A similar thing happens for intVal2 .
At a more general level, by removing the copy constructor, you are violating the Rule of Three, which typically leads to these kinds of problems.
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But, there are some basic differences between them: Copy constructor. Assignment operator. It is called when a new object is created from an existing object, as a copy of the existing object. This operator is called when an already initialized object is assigned a new value from another existing object. It creates a separate memory block for ...
Copy constructor is called when a new object is created from an existing object, as a copy of the existing object. And assignment operator is called when an already initialized object is assigned a new value from another existing object. Example-. t2 = t1; // calls assignment operator, same as "t2.operator=(t1);"
Use an assignment operator operator= that returns a reference to the class type and takes one parameter that's passed by const reference—for example ClassName& operator=(const ClassName& x);. Use the copy constructor. If you don't declare a copy constructor, the compiler generates a member-wise copy constructor for you.
Copy assignment vs Copy constructor. The purpose of the copy constructor and the copy assignment operator are almost equivalent -- both copy one object to another. However, the copy constructor initializes new objects, whereas the assignment operator replaces the contents of existing objects.
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: Copy constructor Assignment operator It is called when a new object is created from an existing object, as a copy of the ...
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.
The first line runs the copy constructor of T, which can throw; the remaining lines are assignment operators which can also throw. HOWEVER, if you have a type T for which the default std::swap() may result in either T's copy constructor or assignment operator throwing, you are
The difference between a copy constructor and an assignment operator is that a copy constructor helps to create a copy of an already existing object without altering the original value of the created object, whereas an assignment operator helps to assign a new value to a data member or an object in the program. Kiran Kumar Panigrahi.
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. Reference: 1.
Copy constructor and assignment operator, are the two ways to initialize one object using another object. The fundamental difference between the copy constructor and assignment operator is that the copy constructor allocates separate memory to both the objects, i.e. the newly created target object and the source object. The assignment operator allocates the same memory location to the newly ...
It's always the constructor taking an int that's called in this case. This is called an implicit conversion and is semantically equivalent to the following code: CTest b(5); The assignment operator is never invoked in an initialisation. Consider the following case: CTest b = CTest(5); Here, we call the constructor (taking an int) explicitly ...
Copy constructor vs assignment operator in C - The Copy constructor and the assignment operators are used to initializing one object to another object. The main difference between them is that the copy constructor creates a separate memory block for the new object. But the assignment operator does not make new memory space. It uses the reference
The rule of three is a well known C++ principle that states that if a class requires a user-defined copy constructor, destructor, or copy assignment operator, then it probably requires all three. In C++11, this was expanded to the rule of five , which adds the move constructor and move assignment operator to the list.
Copy Constructor vs Assignment Operator. When a new object is generated from an existing object as a copy of the existing object, the copy function Object() { [native code] } is named. When an already initialized object is given a new value from another object, the assignment operator is used. Myclass c1,c2; Myclass c3; c2=c1;
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 ...
The one thing to be careful of is to make sure that the swap method is a true swap, and not the default std::swap which uses the copy constructor and assignment operator itself. Typically a memberwise swap is used. std::swap works and is 'no-throw' guaranteed with all basic types and pointer types. Most smart pointers can also be swapped with a ...
Assignment Operators While the copy constructor is used to set up a new version of an object that's a duplicate of another object, the assignment operator is used to overwrite the value of an already-created object with the contents of another class instance. For example, the following code will invoke the assignment operator, not the copy ...
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 ;
1. What is the difference between the functionality of a copy constructor and an Assignment operator. Difference is that copy ctor constructs new object with a copy of existing one, assignment operator overrides fully constructed object with a copy. For example if you have a raw pointer to dynamically allocated memory in your class - copy ctor ...
This topic describes how to write a move constructor and a move assignment operator for a C++ class. A move constructor enables the resources owned by an rvalue object to be moved into an lvalue without copying. For more information about move semantics, see Rvalue Reference Declarator: &&. This topic builds upon the following C++ class ...
Sep 23, 2013 at 21:41. Rule of Five: When C++11 is used (as is becoming more and more prevalent nowadays), the rule of three is replaced by the "Rule of Five". That means that in addition to a copy constructor, destructor, (copy) assignment operator, now also a move constructor and move assignment operator should be defined. - Piotr99.