Re: Copy Constructor

From:
 Roy <royashish@gmail.com>
Newsgroups:
comp.lang.c++
Date:
Mon, 30 Jul 2007 02:52:54 -0700
Message-ID:
<1185789174.128111.92750@57g2000hsv.googlegroups.com>
On Jul 30, 1:18 pm, Dima Stopel <bravo...@gmail.com> wrote:

On Jul 30, 7:24 am, Ian Collins <ian-n...@hotmail.com> wrote:

Roy wrote:

On Jul 30, 9:00 am, Ian Collins <ian-n...@hotmail.com> wrote:

royash...@gmail.com wrote:

   // Destructor
   ~cpchk(){
   cout<<"Destructor called"<<endl;
   if(name!=0)
           free(name);
   }
   a=b;
Heres the code i am talking about , Kindly comment .

Two serious errors:

You attempt to free name twice (why the check for 0?), whether or not it
was allocated by malloc.

--
Ian Collins.


Hi Ian ,

I have checked for 0 , as in the constructor i equate the pointer to
0 , that way my program knows how to check
against freeing a pointer not allocated memory .


Freeing NULL is a no-op, so you don't have to check.

In your case, you have no way of telling whether the name points to
dynamic memory.

I didn't get your point of freeing name twice . There are two
different Objects in the program . Both need to
have their memory freed ?


Once you assign one to the other, both have a name pointer pointing to
the same address. Try printing the pointer value in your destructor.

--
Ian Collins.


I that one of the problems is in the following function:

        // Methods for name
        char* getname()const{
                cout<<"Address:Name>>"<<&name<<endl;
                return name;
        }

You print the pointer to pointer to name and not the actual value of
the pointer.
If you change this line: cout<<"Address:Name>>"<<&name<<endl;
to this line: cout<<"Address:Name>>"<<(void*)name<<endl;
you will get the following output:

Constructor called
Constructor called

Before Assignent
Object A
0012FF50
Address:Name>>00356328
Hitesh
Address:Num>>0012FF54
10
Object B
0012FF40
Address:Name>>00356370
Rajesh
Address:Num>>0012FF44
30

After Assignent
Object A
0012FF50
Address:Name>>00356370
Rajesh
Address:Num>>0012FF54
30
Object B
0012FF40
Address:Name>>00356370
Rajesh
Address:Num>>0012FF44
30
Press any key to continue . . .
Destructor called
Destructor called

After the copy operation the name addresses are the same. This is what
you have expected, not ?


Yes , your observation with the code is correct , Its a fault and
actually printing the pointer to a pointer .
and yes after the copy operation the name addresses are the
same :-) .

Thanks
Dima

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