What functions can be called on a shared struct that's implicitly castable to immutable?

[deadalnix]

On Monday, 4 November 2013 at 09:34:52 UTC, Simen Kjærås wrote:
> On 04.11.2013 09:46, deadalnix wrote:
>> You are trying to solve a non problem here. s is not shared in 
>> the first
>> place, so you can start by fixing it here.
>>
>> Now, if S has some indirection, then it make sense to mark it 
>> as shared,
>> but then the trick you propose won't work.
>
> I believe we're talking past each other here. Perhaps a better 
> example:
>
> module foo;
>
> struct S {
>    immutable(int)[] arr;
>    void fuzz() const pure {
>    }
> }
>
> void bar(S s) {
>    s.fuzz();
> }
>
> void main() {
>    shared S* s = new shared S;
>    bar(*s);   // a
>    s.fuzz(); // b
> }
>
>
> This example demonstrates the exact same behavior as before, 
> and again I'm not doing a lot of anything special to get line a 
> to compile.
>
> Now, I'm still of the opinion that the 'trick' I propose works 
> - making a copy of s is cheap (as that's what the spec says 
> about copying structs), s is implicitly castable to immutable, 
> and the function to be called is guaranteed not to change s in 
> any way. From all that, it seems safe to call fuzz on a shared 
> S.

Any read of arr in fuzz won't be sequentially consistent, so it 
would be incorrect to let you call fuzz on a shared object.