How to break const

[Timon Gehr]

On 06/20/2012 01:36 PM, Christophe Travert wrote:
> Timon Gehr , dans le message (digitalmars.D:170288), a écrit :
>> On 06/20/2012 09:16 AM, deadalnix wrote:
>>> Le 19/06/2012 17:49, Timon Gehr a écrit :
>>>>
>>>> The question is, what the meaning of 'const' references should be:
>>>>
>>>> 1. data cannot be changed transitively through the reference
>>>>
>>>> 2. the reference can reference both 'const' and 'immutable' data and
>>>> 'immutable' data can transitively not be changed through the
>>>> reference.
>>>>
>>>>
>>>> 1. requires transitive const for delegate context pointers, 2. does not.
>>>>
>>>
>>> No, 2. require 1., even if the initialization is broken.
>>>
>>> class Foo {
>>>       void delegate() dg;
>>>
>>>       this(immutable void delegate() dg) immutable {
>>>           thid.dg = dg;
>>>       }
>>> }
>>>
>>> Now, as delegate doesn't carry the constness of its context, an
>>> immutable instance of Foo can refers to something that isn't immutable.
>>
>> Clarification: 'const' means 'const'. No other qualifiers.
>>
>> There is no 'const' in that example code. 'immutable' obviously needs to
>> be transitive regardless of the particular interpretation of 'const'.
>>
> const means: maybe immutable.

Or maybe mutable. Therefore, interpretation '2.'

> Thus const needs to be transitive too.

Wrong. This is the (A==>B) ==> (B==>A) fallacy, where

A: 'const' is transitive
B: 'const' references cannot modify 'immutable' data

The conclusion regarding transitivity, given interpretation 2., is that
'const' needs to be transitive _if_ it is actually 'immutable'.

> If you apply different rules to const and to immutable, you are breaking
> the consistency of the langage.
>

Certainly not. This is like saying that applying different rules to
'const' and mutable is breaking the consistency of the language.
mutable is not transitive.