Const function
Jesse Phillips
jessekphillips at gmail.com
Tue Dec 4 14:01:09 PST 2007
On Tue, 04 Dec 2007 13:48:10 -0500, Jason House wrote:
> Graham St Jack Wrote:
>
>> On Thu, 29 Nov 2007 02:16:40 -0500, Gilles G. wrote:
>>
>> > there are two ways to express function constness for now:
>> > const int foo();
>> > int foo() const;
>> > To my mind, both solutions are unintuitive. I would expect something
>> > like that:
>> > int const foo();
>> > Is there any big argument against this?
>>
>> I agree. A definition like:
>>
>> const T foo();
>>
>> looks to me like the returned T is const, and putting the const after
>> the function is way too non-D for me, so all that is left that makes
>> sense is:
>>
>> T const foo();
>
> I've seen Walter argue that he wants to be able to declare const
> functions in batch with const{
> T foo();
> T bar();
> }
>
> I guess you could say that he wants const{X;} and const X; to be the
> same, but doesn't want const(X) and const X to be the same. There's
> nothing like mixing two hotly debated threads together! I apologize in
> advance for it. I just couldn't resist since the parallels were so
> striking.
I would have to argue that const {W;} and const X; are the same while
const W* var; and const(W*) var; are different. As described in "Teach
Yourself C++ in 21 Days" a block/compound statement is used to act as one
statement. That is to say that only single statements are acceptable.
That is to say that {W;} is in fact only X.
Now look at the second case, const must be applied to something after it,
Janice assumes that it would stop at the *. However if we continue and
let W* var = X then we get const(X); with the latter being const(X) var;
My question is why does const X var; have to equal const(X) var? Wouldn't
it be legitimate to change ones thinking such that const X var is const(X
var) even thought it is not valid syntax.
I still have a lot to learn and to apply a lot of what I know, but after
reading posts and reading to understand const and its applications, I see
the current syntax to work well, even though I don't think it is
explained very well.
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