casting away const and then mutating

[Jonathan M Davis via Digitalmars-d]

On Friday, 24 July 2015 at 20:08:11 UTC, Steven Schveighoffer 
wrote:
> On 7/24/15 3:35 PM, Jonathan M Davis wrote:
>> Your case works, because the const reference is gone after the 
>> cast, and
>> there are no others that were created from the point that it 
>> temporarily
>> became const. So, it's a very special case. And maybe the rule 
>> can be
>> worded in a way that incorporates that nicely, whereas simply 
>> saying
>> that it's undefined behavior to cast away const and mutate 
>> would not
>> allow it. But we cannot say that it's defined behavior to cast 
>> away
>> const and mutate simply because you know that the data is 
>> mutable, or we
>> do not have physical const, and const provides no real 
>> guarantees.
>
> I agree, we can't just make the general case that you can cast 
> away const if you know the data is mutable, given some 
> configuration of function calls. There has to be complete 
> visibility to the compiler within the same function to allow 
> the possibility that some mutable data changed.
>
> We can start with "casting away const and mutating, even if you 
> know the underlying data is mutable, is UB, except for these 
> situations:..."

The only except that makes any sense to me is when you're casting 
away const from the last const reference, so there are no const 
references left for the compiler to make any assumptions - so the 
case where you're trying to mimic inout. Something like

----
int x;
const int *y = &x;
*(cast(int *)y) = 5;
----

should be completely invalid IMHO. I don't see any reason to make 
it valid to cast away const and mutate just because the compiler 
can see that that's what you're doing, especially when it doesn't 
buy you anything, since you have access to the mutable reference 
anyway. Allowing it would just complicate things.

It might be possible to word the spec in a way to essentially 
allow you to do your own inout when inout doesn't cut it, since 
you're not really violating what const is supposed to guarantee, 
but for the rest, I say leave it undefined, because in that case 
you are violating it.

- Jonathan M Davis