A Copy constructor is an overloaded constructor used to declare and initialize an object from another object.
Syntax Of User-defined Copy Constructor:
Class_name(const class_name &old_object);
Consider the following situation:
class A { A(A &x) // copy constructor. { // copyconstructor. } }
In the above case, copy constructor can be called in the following ways
Let's see a simple example of the copy constructor.
// program of the copy constructor.
#include <iostream> using namespace std; class A { public: int x; A(int a) // parameterized constructor. { x=a; } A(A &i) // copy constructor { x = i.x; } }; int main() { A a1(20); // Calling the parameterized constructor. A a2(a1); // Calling the copy constructor. cout<<a2.x; return 0; }
Copy Constructor is called in the following scenarios:
Two types of copies are produced by the constructor
Example
Let's understand this through a simple example:
#include <iostream> using namespace std; class Demo { int a; int b; int *p; public: Demo() { p=new int; } void setdata(int x,int y,int z) { a=x; b=y; *p=z; } void showdata() { std::cout << "value of a is : " <<a<< std::endl; std::cout << "value of b is : " <<b<< std::endl; std::cout << "value of *p is : " <<*p<< std::endl; } }; int main() { Demo d1; d1.setdata(4,5,7); Demo d2 = d1; d2.showdata(); return 0; }
In the above case, a programmer has not defined any constructor, therefore, the statement Demo d2 = d1; calls the default constructor defined by the compiler. The default constructor creates the exact copy or shallow copy of the existing object. Thus, the pointer p of both the objects point to the same memory location. Therefore, when the memory of a field is freed, the memory of another field is also automatically freed as both the fields point to the same memory location. This problem is solved by the user-defined constructor that creates the Deep copy.
Deep copy dynamically allocates the memory for the copy and then copies the actual value, both the source and copy have distinct memory locations. In this way, both the source and copy are distinct and will not share the same memory location. Deep copy requires us to write the user-defined constructor.
Let's understand this through a simple example.
#include <iostream> using namespace std; class Demo { public: int a; int b; int *p; Demo() { p=new int; } Demo(Demo &d) { a = d.a; b = d.b; p = new int; *p = *(d.p); } void setdata(int x,int y,int z) { a=x; b=y; *p=z; } void showdata() { std::cout << "value of a is : " <<a<< std::endl; std::cout << "value of b is : " <<b<< std::endl; std::cout << "value of *p is : " <<*p<< std::endl; } }; int main() { Demo d1; d1.setdata(4,5,7); Demo d2 = d1; d2.showdata(); return 0; }
In the above case, a programmer has defined its own constructor, therefore the statement Demo d2 = d1; calls the copy constructor defined by the user. It creates the exact copy of the value types data and the object pointed by the pointer p. Deep copy does not create the copy of a reference type variable.
Copy Constructor | Assignment Operator |
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It is an overloaded constructor. | It is a bitwise operator. |
It initializes the new object with the existing object. | It assigns the value of one object to another object. |
Syntax of copy constructor: Class_name(const class_name &object_name) { // body of the constructor. } |
Syntax of Assignment operator: Class_name a,b; b = a; |
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The assignment operator is invoked when we assign the existing object to a new object. |
Both the existing object and new object shares the different memory locations. | Both the existing object and new object shares the same memory location. |
If a programmer does not define the copy constructor, the compiler will automatically generate the implicit default copy constructor. | If we do not overload the "=" operator, the bitwise copy will occur. |