Why can not we use * for copy constructor, why reference is used?

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Hello Programmers,

I would like to know why only reference is used in copy constructor, why not pointer?

I used pointer, I get perfect result.

Please put some light on this. Below is the code that I wrote, I am able to see expected result. The same result I get when use reference.

Is there any specific reason not to use pointer?

C++

#include "stdafx.h"
#include<iostream>
#include<stdio.h>

using namespace std;

class MyString
{
  public:
  char *str1;
  int i;
  MyString(char * , int );
  MyString();
  MyString (MyString* );
};

MyString::MyString()
{


}

MyString::MyString( char * str2, int j)
{

   str1 = str2;
   i= j;

}

MyString::MyString (MyString * obj3)
{

   int len = strlen(obj3->str1);

   str1 =  (char*)malloc(len +1);
  strcpy( str1, obj3->str1); 
  i= obj3->i;

}

int _tmain(int argc, _TCHAR* argv[])
{
	 MyString obj1("HelloWorld", 10);

 MyString obj2(&obj1);

  cout<<obj2.str1;
  cout<<obj2.i;

  int j;
  cin >>j;

	return 0;
}}

Regards,
Joy

Posted 5-Jun-14 21:41pm

joyjhonson

Updated 5-Jun-14 21:55pm

CPallini

v2

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CPallini · Answer 1 · 2014-06-05T22:06:00

Solution 1

You question could be rephrased this way: "why should we use references instead of pointers?"
You may find a reason here: "[8.6] When should I use references, and when should I use pointers?"[^].

Please note this

C++

MyString (MyString* );

should be instead

C++

MyString (const MyString* );

Posted 5-Jun-14 22:06pm

CPallini

Comments

[no name] 6-Jun-14 4:21am

Agree, 5ed.

Sergey Chepurin · Answer 2 · 2014-06-05T22:32:00

Solution 2

Best answer, in my opinion -

"A null pointer wouldn't make sense. And using a pointer wouldn't allow copying temporaries, so you couldn't do things like:

MyClass
func()
{
    //  ...
    return MyClass(...);
}

Posted 5-Jun-14 22:32pm

Sergey Chepurin

Stefan_Lang · Answer 3 · 2014-06-05T23:52:00

That's how the copy constructor is defined. The constructor you defined is, per definition, not a copy constructor at all. The copy constructor is the one that will be implicitely called whenever you pass an argument by value to a function. Since it's implicit, you can't specify to take the address of the original object.

You're free to define a constructor that takes a pointer instead and make it behave the same. But that is not the one that will be called implicitely.