How does copy constructor execute?

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I'm trying to understand copy constructor in detail. While doing so, i made the following example,

#include<iostream>

class Test
{
private:
        int a;
public:
        /*
        Test(const Test &t)                          // User defined copy constructor
        {
            a = t.a;
        } */
        Test()
        {
                a = 120;
        }
        int display()
        {
                return a ;
        }
        void set(int var)
        {
                a = var;
        }
};

int main()
{
        Test t1;
        std::cout << "t1.a " << t1.display() << std::endl;
        Test t2 = t1;                                           //Default copy constructor is called
        std::cout << "T2.a " << t2.display() << std::endl;
        t2.set(99);                                             //Changing the value
        std::cout << "t1.a " << t1.display() << std::endl;
        std::cout << "T2.a " << t2.display() << std::endl;
        return 0;
}

I read online that default copy constructor does "shallow copy" So it means if obj1 = obj2, what ever change i do in obj1 or obj2 even after the assignment needs to be reflected on both the object as they point to same location. But in this example when i change the value in one object it is not getting reflected in the other. Same result is obtained when using user defined copy constructor.

Can some one clarify this topic, whether shallow copy is happening or not!

Thanks!

Consider Shallow copy to mere assignment. So,

Test t2 = t1; 

means

t2.a = t1.a

Since a is an int, If you modify a, from t1, it will not reflect in t2. So for int shallow copy is indeed deep copy.

Consider the case a was of type int*. Now t2.a and t1.a both point to same memory locations. So if you modify the value at memory location t1.a, the same would be reflected via t2.a since they are infact pointing to same location.

How does copy constructor execute?, Consider Shallow copy to mere assignment. So, Test t2 = t1;. means t2.a = t1.a. Since a is an int , If you modify a , from t1 , it will not reflect in t2 . The copy constructor is a constructor which creates an object by initializing it with an object of the same class, which has been created previously. The copy constructor is used to − Initialize one object from another of the same type. Copy an object to pass it as an argument to a function. Copy an object to return it from a function.

A shallow copy is not something special that you need to remember as rule. Instead it is just something that happens as a consequence of using a reference or pointer. COnsider this example:

struct foo { 
     int* x;
};

int a = 4;
foo f{&a};

Here x points to a and if you copy f the new instances x will point to the same a. Thats a shallow copy. A deep copy would be to respect that not only x, but also what x points to, is an integral part of foo and needs to be copied as well. In general the compiler cannot decide what you want. Is x just a reference, or is what x refers to also part of foo? Hence what you get is the obvious: Only the members are copied not what they might refer to.

Now, if you copy the foo and then modify the value x points to, then this will modify the same a. A shallow copy was made. To my experience the terms deep and shallow copy are rather adding confusion than clarity. What you get for free is all members get copied (whether is is a shallow or deep copy). Only if you need more (copy also the pointee), you need to worry about shallow vs deep copies.

TL;DR: There is no deep/shallow copy in your example. For values this distinction does not make sense. Use a int* to see the effect.

C++ Copy Constructor, C++ Copy Constructor - The copy constructor is a constructor which creates an When the above code is compiled and executed, it produces the following� A copy constructor is called when an object is passed by value. Copy constructor itself is a function. So if we pass an argument by value in a copy constructor, a call to copy constructor would be made to call copy constructor which becomes a non-terminating chain of calls. Therefore compiler doesn’t allow parameters to be passed by value.

I read online that default copy constructor does "shallow copy"

This is not the right way of thinking of the copy constructor. The default copy constructor is just copying whatever members are in the type as if applying the copy constructor in turn on the members.

From reference:

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. For union types, the implicitly-defined copy constructor copies the object representation (as by std::memmove). For non-union class types (class and struct), the constructor performs full member-wise copy of the object's bases and non-static members, in their initialization order, using direct initialization.

So it is more like a deep-copy actually than a shallow-copy.

Copy constructor (C++), In the C++ programming language, a copy constructor is a special constructor for creating a uses the copy constructor to initialize as a copy of that temporary. // Temporary objects created while program execution are always of const type. Copy Constructor is a type of constructor which is used to create a copy of an already existing object of a class type. It is usually of the form X (X&), where X is the class name. The compiler provides a default Copy Constructor to all the classes.

The implicitly-defined copy constructor of a class simply copies each member [class.copy.ctor]/14. Copying the members basically means that every member of the new object (the one being copied into) is initialized from the corresponding member of the object being copied in the same way as if you had written

T member(original.member);

where T is the type of the member and original is the object being copied. If member is of class type, this effectively boils down to invoking the copy constructor of T. However, your member is a plain int, which is not a class type. There is no copy constructor to call (int does not have a copy constructor); the int in the new object is simply initialized from the int of the original object, which boils down to copying the value of the original int into the new int

Your compiler does not distinguish between deep copy versus shallow copy, it doesn't even know what "deep copy" or "shallow copy" would be supposed to mean. There is no such thing as a deep copy or a shallow copy in C++ at the language level. These are simply terms that are commonly used by programmers to talk about different approaches for copying objects that logically (but not physically) contain other objects…

C++ Copy Constructor, What is Copy Constructor in C++? Copy Constructor is a type of constructor which is used to create a copy of an already existing object of a class type. In C++, a Copy Constructor may be called in following cases: 1. When an object of the class is returned by value. 2.

C++ Notes: Copy Constructors, OOP: Copy Constructors. When copies of objects are made. A copy constructor is called whenever a new variable is created from an object. This happens in the� Compiler also creates a copy constructor if we don’t write our own copy constructor. Unlike default constructor, body of copy constructor created by compiler is not empty, it copies all data members of passed object to the object which is being created.

Copy constructor in C++, A copy constructor is a method of a class which initializes an object using another object of the same A copy constructor can be called in various scenarios. 29. } 30. 31. . 32. int get_y_cor(){. 33. return y_cor;. 34. } RUN� Prerequisite – Constructors in Java Like C++, Java also supports copy constructor. But, unlike C++, Java doesn’t create a default copy constructor if you don’t write your own. Following is an example Java program that shows a simple use of copy constructor.

Copy Constructor, A C++ compiler will create a copy constructor and assignment operator for your void Func::operator()() const: Task function executed by the function object. parameterized constructor :constructor with parameters is known as parameterized constructor.The parameterized constructor take its first argument as a reference to the instance being constructed known as self and the rest of the arguments are provided by the programmer.

Comments
  • shallow copy for int is deep copy. Try having int *a instead of int a
  • The default copy constructor does a bitwise copy.
  • You've been misinformed. A shallow copy does not mean that obj1 = obj2 causes obj1 and obj2 to refer to the same location. A shallow copy means that any members which are pointers are copied by value, rather than creating a separate copy of whatever the pointers point at. The distinction between deep and shallow copy only matters if the object contains pointers - which your example does not.
  • @user207421 - a "default" copy constructor does a member-wise copy. For basic types like int and for pointers that happens to be a bitwise copy. For each member (or base) that has an accessible copy constructor, it uses that copy constructor.
  • View it from a technical perspective. For the computer, an object is a set of numbers. Like the row of some table, with no name tag attached, no idea when each cell ends and without any knowledge about what the data is supposed to be. Could be integer data, could be char values, you don't know without knowing the class. That data is copied on a default copy constructor. Interesting here are especially pointers: those are addresses to some memory. Their raw data is that address. Which you copy, which means that after such a copy, both objects have the same address stored in that pointer.