const/invariant

[Denton Cockburn]

On Wed, 05 Dec 2007 18:30:37 +1100, Derek Parnell wrote:

> On Wed, 5 Dec 2007 06:46:21 +0000 (UTC), Denton Cockburn wrote:
> 
>> given a class:
>> 
>> class C {}
>> 
>> what's the difference between:
>> 
>> const C c = new C;
>> and
>> invariant C c = new C;
>> 
>> From what I can see, neither are modifiable.  So what's different
>> (D2.008)?
> 
> Dunno, except that you have to say ...
> 
>  const C c = new C;
>  invariant C d = cast(invariant Foo)new C;
> 
> 
> And what I don't get is why would anyone bother with const/invariant for
> classes since you can't invoke any member functions when you do.
> 
> --------------
> class Foo
> {
>     int mbr;
> 
>        this() { mbr  = 6174; }
>     int get() {return mbr; }
> }
> 
> void main()
> {
>     const     Foo c = new Foo();
>     int a;
>     
>     a = c.get();
> }
> 
> 
>     
> --------------
> I get the error messages ...
> 
>   : function test.Foo.get () does not match parameter types () : Error:
>   c.get can only be called on a mutable object, not const(Foo)
> 
> Aside from the useless message text "function test.Foo.get () does not
> match parameter types ()" that doesn't actually tell me the problem is,
> I can't see why this is forbidden. I can see that get() doesn't change
> anything but the compiler still will not let me call it.
> 
> So if one can't call any member function in the const object, what is
> the point of it? Sure I can cast the const away but why would I want to
> do that for every invocation of a member function? I really don't get
> the point of const objects when implemented this way.
> 
> I thought it might be a way to tell the compiler that I can't call
> member functions that happen to change member values. That would make
> sense to me.

You can call const member functions.

Since the 2.008 changes, I'm missing the use of invariant for things that 
are passed by ref, like arrays and objects.