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.
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