On Fri Aug 3 06:06:31 EDT Neil Mitchell wrote:
What are the semantics of do b >> f (<- a) where does the evaluation of a get lifted to?
I suggest the execution of (a) should be done immediately before the action obtained by applying the monadic function whose argument it is part of: do b >> ( a >>= \ra -> f ra) Similarly, I'd desugar if (<- a) then f (<- b) else g (<- c) to a >>= \ra -> if ra then (b >>= \rb -> f rb) else (c >>= \rc -> g rc) I think it would just be confusing to have to think about lines in the "do" block since "do" blocks are just syntactic sugar. The only possible thing that might be needed by the "do" block is to define what monad the action should be evaluated in. Perhaps the rule could be that if (<- a) occurs in some expression the compiler should search for the nearest enclosing "do" block to identify which monad (a) should be evaluated in, then should again search outwards from the expression (<- a) to find the nearest enclosing expression (mexp) which yields a result in that same monad, then desugar to (a >>= \ra -> mexp') where mexp' is obtained from mexp by replacing that occurrence of (<- a) by (ra). (Evaluations would be done in the same order as depth first traversal of their occurrences in mexp) Regarding objections by others about (<- a) being confused with a section, (<-) is not a valid operator therefore it can't be part of a section so no confusion should arise imho... Best regards, Brian.