On Sunday, 26 November 2017 at 04:51:08 UTC, Adam D. Ruppe wrote:
On Sunday, 26 November 2017 at 01:35:01 UTC, Dave Jones wrote:
So it makes it a const/immutable/mutable method depending on
whether the instance it is called on is
const/immutable/mutable?
On the outside, yes.
So
@property ref inout(int) front() inout {
return i++;
}
Would fail if you called it on an immutable instance of S.
That wouldn't compile in any case: on the inside of the
function, inout == const (this is the only way the one function
can be used for all three). The inout propagation is just seen
at the call site.
Ahh ok, makes sense now.
