Flip default for return_ref

[MichaReiser]

This has come up in #718.

It's very easy to forget a return_ref on a non-copy type. On the other hand, returning a ref for a copy type may be undesired, but the performance consequences are often negligible.

Should we change the default for return_ref to be true and instead introduce a return_cloned.

Related: Should we introduce a return_deref? There have been a few occasions where I had an Option<T> and I ended up writing a manual accessor to get the deref behavior. That might also suggest that the syntax should be return(cloned), return(deref) return(ref)?

This is a rather subtle breaking change. We should change/decide on this before the first "stable" release.

[Veykril]

For the Copy case I'd also expect the compiler to change the ref-passing to copy passing in some occasions on its own so I think the "risk" there of that being less efficient is negligible. The bigger annoyance is that you in fact now have a reference to your type which means you need to deref it manually, but that can be worked around by well putting the marker attribute on it.

Generalizing it via return(<action>) seems to generally be a good move for future extensions imo. Either way 👍 from me for flipping the default, it sounds like what one would expect in terms of default behavior.

In an ideal world with specialization we could work around this via a trait with an assoc type that marks the return type and specializing on Copy to reduce the annotation burden entirely (I think).

[Veykril]

#582 related

[davidbarsky]

I'm glad you opened this issue: I strongly think Salsa should default to #[return_ref] and I'm onboard with generalizing via return(<action>).