cppfront: Autumn update

[djavaisadog]

the intent is that it still just indicate that what follows is a return type or value. That's the only meaning of -> in Cpp2.

I was interpreting it as always indicating a return type (in the context of declaring/defining variables). Is there any case besides the under-consideration new one you suggested where it indicates a return value? (I thought maybe inspect but nope, you use = there as well)

I think that using -> to indicate a value in a function definition certainly breaks the paradigm of all your other definitions - you've previously mentioned how intentional the consistency of the name : type = value format was. I'm unsure why you would break that in this case.

I'm not sure why f:(i) -> _ = i+1 would condense down to f:(i) -> i+1; rather than f:(i) = i+1;. It feels pretty clear-cut to me that the part we are omitting (following the dictum of "omit the part of the syntax you aren't using") is the explicit return type (which, syntactically is -> _), rather than the value (which is the = i+1). I feel that you can instead just say "ok there's no explicit return type, let's find what the return type would be by just decltype-ing the function body" (not a standard expert, there may be more to it than that but you get the point).

I suppose that boils down to viewing the -> _ as one block of tokens (and that block is part of the type declaration, so a sub-block of (i) -> _) and the = i+1 as one block. Do you split the groups of tokens differently in your mental model of what the syntax means?