Re: Operator Cast () Reference?

From:
"Francesco S. Carta" <entuland@gmail.com>
Newsgroups:
comp.lang.c++
Date:
Thu, 10 Jun 2010 09:06:04 -0700 (PDT)
Message-ID:
<1c3da021-3c56-47b0-b653-ae8624adb496@z10g2000yqb.googlegroups.com>
Paul Bibbings <paul.bibbi...@gmail.com> wrote:

Immortal Nephi <Immortal_Ne...@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 t=

o 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.


Furthermore, the last assignment above will be rejected by the
compiler, thus ensuring that one will not get what one doesn't
want ;-)

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