Yep, that did the trick.

IIRC, the const in front means const(this)

Looks like a compiler bug. The trailing const should work.

Walter Bright wrote:
> Looks like a compiler bug. The trailing const should work.

No, not a bug.  That seems to work fine.
What doesn't work is doing it *without* const either in front or at the end.  And that's correct behavior.

--bb

On Tue, 04 Mar 2008 08:33:00 +0900, Bill Baxter wrote:

> and apparently const at the end means that too, now.
> 
>    	Foo opMul(const Foo b) const  { // also ok
>                ...
>          }
> 
> 
> --bb

I think that was said to be something that's going to be fixed (at some
point).

On Mon, 03 Mar 2008 14:47:56 -0800, Walter Bright wrote:

> Looks like a compiler bug. The trailing const should work.

So the trailing const will be kept?

Why have it both ways?

(even though there are other things that can be specified in multiple ways)

*joking* Not very pythonic... *joking*

On 03/03/2008, Bill Baxter <dnewsgroup@billbaxter.com> wrote:
> I can't keep up here.  Is a trailing 'const' legal D syntax now?

For member functions, yes.

On Tue, 04 Mar 2008 00:31:38 +0100, Bill Baxter <dnewsgroup@billbaxter.com> wrote:

> Walter Bright wrote:
>> Looks like a compiler bug. The trailing const should work.
>
> No, not a bug.  That seems to work fine.
> What doesn't work is doing it *without* const either in front or at the end.  And that's correct behavior.
>
> --bb


Bill, take a look at my code. The opMul with trailing const clearly changes this.bar, thus it's a bug.

Simen Kjaeraas wrote:
> On Tue, 04 Mar 2008 00:31:38 +0100, Bill Baxter <dnewsgroup@billbaxter.com> wrote:
> 
>> Walter Bright wrote:
>>> Looks like a compiler bug. The trailing const should work.
>>
>> No, not a bug.  That seems to work fine.
>> What doesn't work is doing it *without* const either in front or at the end.  And that's correct behavior.
>>
>> --bb
> 
> 
> Bill, take a look at my code. The opMul with trailing const clearly changes this.bar, thus it's a bug.

Ah, ok.  My mistake.  I don't usually look at the groups with threading on because the threads just get too darn deep. And Walter's reply contained no context, so I thought he was responding to Denton's part of the thread about things compiling.

--bb

Denton Cockburn wrote:
> Yep, that did the trick.
> 
> IIRC, the const in front means const(this)
> 

and apparently const at the end means that too, now.

        Foo opMul(const Foo [b]) const  { // also ok
               ...
         }


--bb

On Thu, 06 Mar 2008 03:04:39 +0300, Bill Baxter <dnewsgroup@billbaxter.com> wrote:

> Denton Cockburn wrote:
>> Yep, that did the trick.
>>
>> IIRC, the const in front means const(this)
>>
>
> and apparently const at the end means that too, now.
>
>         Foo opMul(const Foo [b]) const  { // also ok
>                ...
>          }
>
>
> --bb


This syntax looks less confusing to me. Why keep both?