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

[Simen Kjærås]

On 04.11.2013 06:53, deadalnix wrote:
> On Sunday, 3 November 2013 at 21:55:22 UTC, Simen Kjærås wrote:
>> Consider:
>>
>> module foo;
>>
>> struct S {
>>     immutable(int)[] arr;
>>     void fuzz() const pure {
>>     }
>> }
>>
>> void bar(S s) {
>>     s.fuzz();
>> }
>>
>> void main() {
>>     shared S s;
>>     bar(s);   // a
>>     s.fuzz(); // b
>> }
>>
>>
>> In this case, the line marked 'a' works perfectly - it compiles and 
>> does what I'd expect it to.
>>
>> However,the line marked 'b' does not compile - " non-shared const 
>> method foo.S.fuzz is not callable using a shared mutable object ".
>>
>
> It is because a imply a pass b value, when b a pass by reference.
Indeed. That's why I wrote that one line further down. That's not the 
point. The point is I have to jump through no hoops to get line a to 
compile - no 'assumeUnshared', no cast, no explicit copying.

On that basis, I argue that line b could be made to work by silently 
creating a copy of s, because the call to fuzz is guaranteed not to 
change s. There may be problems with this that I have not considered. If 
you know of any, please do tell.

I guess it might be conceivable that fuzz waits for a synchronization 
message (but is that really possible in a pure const function?), which 
will never happen for the copy, but is this even a case we want to support?

--
   Simen