Re: Operator Cast () Reference?

From:
Paul Bibbings <paul.bibbings@gmail.com>
Newsgroups:
comp.lang.c++
Date:
Thu, 10 Jun 2010 15:23:09 +0100
Message-ID:
<87r5kf7yr6.fsf@gmail.com>
Immortal Nephi <Immortal_Nephi@hotmail.com> writes:

     Someone posted his Byte class code on the previous thread. I have a
question about operator cast ().


I think that would be me.

Please explain the difference
between local object and reference object. Why do you need reference
object?


If you remember, the Byte class was introduced as a `proxy' for your
Array class, which stored a pointer to an array of unsigned char. A
Byte instance was returned by your Array::op[], so:

   Byte Array::operator[](int index) { return pData[index]; }

It is created from pData[index] using the non-explicit constructor:

   Byte::Byte(unsigned char&);

You will see that it stores the unsigned char at pData[index] *by
reference*. This is the whole point of using the proxy. Basically it
`represents' the stored data so that any operation upon the proxy is
*really* an operation upon the data it represents (stores).

The reason that you need:

   Byte::operator unsigned char&();

and not:

   Byte::operator unsigned char();

is that you need to allow assignment to the underlying unsigned char
through a Byte proxy. Consider:

   Array array(4);
   // ...
   array[0] = 42U;

The last call is equivalent to:

   Byte b(array[0]);
   b.operator unsigned char&() = 42U; // LHS returns reference to array[0]

Using a reference here 42U gets assigned to the actual unsigned char at
array[0]. If you consider the alternative, which would be:

   Byte b(array[0]);
   b.operator unsigned char() = 42U; // LHS returns *temporary*
                                       // array[0] *not* changed

then what you would be attempting to do is assigne 42U to a *temporary*
returned by b.operator unsigned char(), which is not what you want.

Regards

Paul Bibbings

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