How to break const

[Mehrdad]

On Monday, 18 June 2012 at 14:41:02 UTC, deadalnix wrote:
> Le 18/06/2012 16:28, Artur Skawina a écrit :
>> It's fine, if you view a delegate as opaque.
>>
>
> No it isn't. You cannot ensure transitivity anywhere. This have 
> obvious, and severe drawback for concurrent programing 
> (implicit sharing is back) and GC performances (GC can eb crazy 
> fast when it come to transitive immutable data, see OCaml's GC 
> performances for instance).
>
>> 'this' being const does not preclude accessing the object that 
>> 'this'
>> points to via another, mutable, reference.
>>
>> Consider the alternative - you'd have to forbid storing any 
>> delegate
>> with a non-const non-value argument inside any object.
>>
>> And "breaking" const would then _still_ be possible and 
>> trivial.
>>
>
> No, and your example don't demonstrate that in any way. 
> Transitivity is maintained in the example below, because g 
> isn't a member of s, and if it were, then the example would 
> break at compile time.
>
>>    import std.stdio;
>>
>>    S*[const(S)*] g;
>>
>>    struct S {
>>       int i;
>>       this(int i) { this.i = i; g[&this] =&this; }
>>       void f() const { g[&this].i=666; }
>>    }
>>
>>    void main() {
>>       const s = S(42);
>>       writeln(s);
>>       s.f();
>>       writeln(s);
>>    }
>>
>> Yes, D's "pure" could help here, only it's misnamed (even if in
>> this particular case that term would be fitting).
>>
>
> purity is another beast altogether.



Interesting, making the delegate `pure' doesn't change anything 
either.

So 'pure' doesn't let you "infer something just by looking at the 
code either", right?