inout after function

[Dave Jones]

On Saturday, 25 November 2017 at 21:59:54 UTC, Ali Çehreli wrote:
> On 11/25/2017 01:51 PM, Dave Jones wrote:
> > What does the "inout" after front() do here...
> >
> >
> > @property ref inout(T) front() inout
> > {
> >      assert(_data.refCountedStore.isInitialized);
> >      return _data._payload[0];
> > }
> >
> > Cant seem to find an explanation in the docs or forums :(
>
> It's for member functions. Without it, and if you needed, you 
> would have to write separate functions for mutable, const, and 
> immutable objects of that type.
>
> For example, the following function works for all three 
> qualifications. It won't compile if you remove that inout:
>
> struct S {
>     int i;
>
>     @property ref inout(int) front() inout {
>         return i;
>     }
> }
>
> void main() {
>     auto m = S(1);
>     auto c = const(S)(2);
>     static assert(is(typeof(m.front) == int));
>     static assert(is(typeof(c.front) == const(int)));
> }
>
> Ali

So it makes it a const/immutable/mutable method depending on 
whether the instance it is called on is const/immutable/mutable?

So

>     @property ref inout(int) front() inout {
>         return i++;
>     }

Would fail if you called it on an immutable instance of S.