Aloha from Honolulu,
Kiryl,
My original email had this:
>> First: in "Effective C++", item 15, p. 65:
>> "operator= default version C& C::operator=(const C&);"
>> The & follows the variable (?).
I assumed the "C" following "C & C::operator=(const "
was the variable--somehow it must have got lost on you. I understand
const is a type def.
My major concern was with line 8 (below)
>> 8: CAT (const CAT &); // copy constructor
But as Per W. pointed out in his reply on 8/24/04 8:49PM :
> int my_function(int &);
> _doesn't_ have the & last, since there is an "invisible" :) parameter
name
> following it. It is just a lazy short-form of declaration.
That the var is invisible was the info I needed.
Regarding "lhs" and "rhs", I was confused because "The Complete
Reference C++" never seems to use them. Perhaps he's keeping his code
as simple as possible for clarity sake. I continue to find his examples
most useful in gaining an understanding of the language. (My background
was Pascal through the '80s and early '90s--in MSDOS).
My thanks for all your help, Per especially.
Mahalo,
Roger
Kiryl V Mauryshchau wrote:
> > "operator= default version C& C::operator=(const &);"
> > The & follows the variable (?).
>
> Are you sure that it follows the variable. Maybe what you think is a
> variable in fact is a type definition. Check it.
>
> > 8: CAT (const CAT &);
>
> Here, "const CAT" is type, not variable. When you _declare_ some
> method, you can omit variable. Like in this case.
>
> > 27: CAT::CAT(const CAT & rhs)
>
> Here, "const CAT" is type, and "rhs" is variable/object. As I
> mentioned before, when you declare a function or method in the class
> (.h file), you may omit the variable/object in declaration. But when
> you write a function/method itself in implementation file, then you
> specify this variable/object.
>
> I am pretty sure, that in the first example above, what precedes the
> ampersand is type.
>
> Regards,
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