954,504 Members — Technology Publication meets Social Media
Username:
Password:
Lost login information?
Have something to say? Contribute New Article Reply to this Article

copy constructors quick question

So when two objects are declared say class shoe shoe1 and class shoe shoe_copy and then shoe1's variables are set, if the line of code: shoe_copy=shoe1; what would that be doing? After all what is the real data structure of the objects of class shoe?

CPPRULZ
Junior Poster in Training
91 posts since Aug 2008
Reputation Points: 10
Solved Threads: 0
 

shoe_copy=shoe1; will call the assignment operator, not the copy constructor. The copy constructor is called for example when you pass an object to a function by value.

// copy constructor called (notice I pass by value, not by reference)
void f(Shoe s)
{
}

minas1
Junior Poster in Training
82 posts since Nov 2008
Reputation Points: 13
Solved Threads: 8
 

yes but what willl shoe_copy=shoe1 do?

CPPRULZ
Junior Poster in Training
91 posts since Aug 2008
Reputation Points: 10
Solved Threads: 0
 
yes but what willl shoe_copy=shoe1 do?

Call the copy constructor.

Let me explain:

Shoe shoe1; // calls the constructor
Shoe shoe_copy(shoe1); // copy constructor
Shoe shoe_copy = shoe1; // copy constructor, even if it seems it's the assignment operator;

shoe_copy = shoe1; // assignment operator

minas1
Junior Poster in Training
82 posts since Nov 2008
Reputation Points: 13
Solved Threads: 8
 

but the last example shoe_copy=shoe1; - does that set the adresses equal or what does it physically do to shoe_copy.

CPPRULZ
Junior Poster in Training
91 posts since Aug 2008
Reputation Points: 10
Solved Threads: 0
 

It will not modify and addresses but it will set all of the values of shoe_copy to the values of shoe1. It wont call a copy constructor. I have seen this called a "shallow copy" ( here )

death_oclock
Posting Whiz
393 posts since Apr 2006
Reputation Points: 129
Solved Threads: 45
 

Ohhhh... okay that is probably the most informative clear article I have read on the subject and I thank you for that excellent link.

So when p=q calls the copy constructor is this accessing a compiler-defined ovverloaded version of the = operator?

CPPRULZ
Junior Poster in Training
91 posts since Aug 2008
Reputation Points: 10
Solved Threads: 0
 

sorry that last post was badly stated as a question, what I meant was: When p=q calls an overloaded assignment operator and preforms a shallow copy is this done using the automatic copy constructor that is called when an object is initialized to another?
So if i explicitly defined a complex copy constructor and said p=q that used the assignment operator would it preform a complex copy?

CPPRULZ
Junior Poster in Training
91 posts since Aug 2008
Reputation Points: 10
Solved Threads: 0
 

No, the copy constructor is used only when called explicitly or when an object is being initialized. You could, however, have the assignment operator call your copy constructor.

death_oclock
Posting Whiz
393 posts since Apr 2006
Reputation Points: 129
Solved Threads: 45
 
No, the copy constructor is used only when called explicitly or when an object is being initialized. You could, however, have the assignment operator call your copy constructor.


Yet another context where copy constructor works:

class C {...};
void f(C); // by value
...
C c;
f(c); // <= parameter/argument binding.
ArkM
Postaholic
2,001 posts since Jul 2008
Reputation Points: 1,234
Solved Threads: 348
 

Oh okay thank you, also how would you create a complex copy constructor? I succeeded in making an explicit shallow copy constructor however in the line "value=c.value;" I have been trying to put *value=*c.value to make a complex copy but it has failed breaking at 0xcccccccccc saying: "Unhandled exception at 0x011518af in copy_constructor_test-complex.exe: 0xC0000005: Access violation writing location 0xcccccccc."
here is the working program without the small edit:
box.h

class box{
public:
	int *value;
	box()
	{
		value=new int;
		*value=0;
	};
	box(box &c)
	{*value=*c.value;};
};

main.cpp

#include<iostream>
#include"box.h"
using namespace std;
int main ()
{
	box a;
	*a.value=5;
	box b(a);   //calls the shallow copy constructor that makes value point to the same dynamic memory
	cout<<"*b.value after copied from a where *a.value is 5: "<<*b.value<<endl;
	system("pause");
	*a.value=10;
	cout<<endl<<"*b.value after *a.value=10; : "<<*b.value<<endl;// this will output 10 because &value of both object versions are the same
	system("pause");
	cout<<endl<<endl<<"The change of *a.value has effected *b.value since both object versions of value point to the same dynamic memory. Here are the 2 object versions of value: "<<endl<<" a.value: "<<a.value<<endl<<" b.value: "<<b.value<<endl;
	system("pause");
	return 0;
}

Thank you.

CPPRULZ
Junior Poster in Training
91 posts since Aug 2008
Reputation Points: 10
Solved Threads: 0
 

The copy constructor is used when the class contain a data member which is a pointer....
when you say object a=b you did not want to make the pointers in the two objects have the same address....by doing this if ,for example object a has been destroyed there will be an run time error .....

and using the equal sign will not enforce the program to use the copy constructor...so you have either to call it explicitly or overload the equal sign

MMahmoud
Newbie Poster
2 posts since Oct 2008
Reputation Points: 10
Solved Threads: 1
 

try this in the copy constructor
box(box &c)
{
value= new int;
*value=*c.value;
}

MMahmoud
Newbie Poster
2 posts since Oct 2008
Reputation Points: 10
Solved Threads: 1
 

This question has already been solved

Post: Markdown Syntax: Formatting Help
You