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.
