Re: Reference Counting

From:
"Protoman" <Protoman2050@gmail.com>
Newsgroups:
comp.lang.c++
Date:
13 Jul 2006 12:34:59 -0700
Message-ID:
<1152819299.911505.158290@p79g2000cwp.googlegroups.com>
Thomas J. Gritzan wrote:

Protoman schrieb:

OK, here's the new SmrtPtr class:

// COPYRIGHT CMDR DOUGLAS I. PEREIRA 07/10/06
// ALL UNAUTHORIZED THIRD PARTY USE IS PROHIBITED
// I HAVE WORKED VERY HARD AND HAVE SPENT HOURS OF
// TIME DEVELOPING THIS. DO NOT COPY IT OR I WILL SUE YOU
// ************************************************************
#pragma once
#include <iostream>
#include "SmrtPtrDB.hpp"
using namespace std;
class NullPtr{};
template<class T>
class SmrtPtr
{
public:
explicit SmrtPtr<T>(T* obj=0):ptr(obj), DataBase(new SmrtPtrDB){}
SmrtPtr<T>(const SmrtPtr<T>& rhs):ptr(rhs.ptr), DataBase(new
SmrtPtrDB(rhs.DataBase->status())){DataBase->add();}


The smartpointers should _share_ a common SmrtPtrDB, if they share a
common object pointer. Here you create another copy of SmrtPtrDB in the
copy constructor.

~SmrtPtr<T>()
{
DataBase->sub();
if(DataBase->status()==0)
{delete ptr;cout << "Deleted." << endl;}
else cout << "Out of scope. " << endl;
}


You don't delete the SmrtPtrDB.

void operator=(T val){if(ptr==0)throw NullPtr();else *ptr=val;}


Remove this. The user should dereference the smart pointer to access its
value.

void operator=(T* val){ptr=val;}


Here you should release the old ptr/SmrtPtrDB and create a new pair.
Something like this:

void operator=(T* val)
{
   SmrtPtr<T> temp(val);
   swap(temp);
}

SmrtPtr<T>& operator=(const SmrtPtr<T>& rhs)
{
if(this!=&rhs)
{
SmrtPtr<T> temp(rhs);
swap(temp);
}
else return *this;
}


You return *this in the "else" but not in the "if". Why?

bool operator==(const SmrtPtr<T>& rhs)const{if(ptr==rhs.ptr)return
true;else return false;}
bool operator!=(const SmrtPtr<T>& rhs)const{if(ptr!=rhs.ptr)return
true;else return false;}
bool operator<=(const SmrtPtr<T>& rhs)const{if(ptr<=rhs.ptr)return
true;else return false;}
bool operator>=(const SmrtPtr<T>& rhs)const{if(ptr>=rhs.ptr)return
true;else return false;}
int status(){return DataBase->status();}
T& operator*()const{if(ptr==0)throw NullPtr();else return *ptr;}
T* operator->()const{if(ptr==0)throw NullPtr();else return ptr;}
operator T*()const{if(ptr==0)throw NullPtr();else return ptr;}
private:
void swap(SmrtPtr<T>& rhs){this=&rhs;}


Better this way:

void swap(SmrtPtr<T>& rhs)
{
std::swap(ptr,rhs.ptr);
std::swap(DataBase,rhs.DataBase);
}

mutable SmrtPtrDB* DataBase;
T* ptr;
};


Well, your operator= should return *this by reference, as every
operator= does.

Anyway, if you won't get your indentation right, I won't read your code
anymore. It is horrible...

--
Thomas


For swap(), how about *this=rhs? Will that work?

Here's the stuff I fixed:

~SmrtPtr<T>()
{
DataBase->sub();
if(DataBase->status()==0)
{
delete ptr;
delete DataBase;
cout << "Deleted." << endl;
}
else
{
delete DataBase;
cout << "Out of scope. " << endl;
}
}

void operator=(T* val)
{
SmrtPtr<T> temp(val);
swap(temp);
}

SmrtPtr<T>& operator=(const SmrtPtr<T>& rhs)
{
SmrtPtr<T> temp(rhs);
swap(temp);
return *this;
}

void swap(const SmrtPtr<T>& rhs){*this=rhs;}

Sorry about my indenting, it looks normal on my screen.

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