what is the difference between reference and pointer

[wenjiahe]

hi, buddy!I am a beginer here. thanks for your answer:
bow to everyone.
I am always confused by the reference and pointer. in the
code below I think it should return the Pointer 'this' but why
it return '*this', i think *this means the contents of the pointer.
who can help me? thanks a lot!
const Array &Array::operator =(const Array &right)
{
if(&right!=this)
{
delete [] ptr;
size = right.size;
ptr=new int[size];
assert(ptr!=0);
for(int i =0; i<size;i++)
ptr&#091;i&#093; = right.ptr&#091;i&#093;;
}
return *this;
}

this function should return a const reference of Array,right?

[kuphryn]

Correct.

The overloaded operator returns a constant reference to an object, thus allowing code:

Object A;
Object B = A;

The overloaded operator can in theory return a pointer to the object, but that does not work with the code above.

Change this line:

if(right!=*this)
{}

Kuphryn

[wenjiahe]

Originally posted by kuphryn
Correct.

The overloaded operator returns a constant reference to an object, thus allowing code:

Object A;
Object B = A;

The overloaded operator can in theory return a pointer to the object, but that does not work with the code above.

Change this line:

if(right!=*this)
{}

Kuphryn

thanks for you reply.
i think if(&right!=this) and if(right!=*this) will have the same effect. the code i paste above is a sample code of a book.i
have tested it. it's correct code. what i want to know is why
not return this but return *this.

[Andreas Masur]

Originally posted by wenjiahe
what i want to know is why
not return this but return *this.

Because 'this' is a pointer to the object and '*this' is the actual object. Since the operator will return a reference to the object and not the pointer you have to return the latter one...

[wenjiahe]

Originally posted by Andreas Masur
Because 'this' is a pointer to the object and '*this' is the actual object. Since the operator will return a reference to the object and not the pointer you have to return the latter one...

ok, you mean reference to an object equal the object itself,right?

[Andreas Masur]

Originally posted by wenjiahe
ok, you mean reference to an object equal the object itself,right?

Correct...a reference can be seen as a transparent way of looking to the object itself...