[Haskell-cafe] Should "do 1" compile

[Claus Reinke]

>> foo = do (1 :: Int)
> 
> While intuitively this should be disallowed, it seems a pity that
> desugaring couldn't be totally separated from typechecking. Hmm.

or perhaps not. while a type-free desugaring, followed by
type-checking, seems more modular, i'd rather see any type
errors in terms of the sugared syntax. which kind of rules
out total separation, doesn't it?

'do {e}' --> 'e' could be interpreted in a type-free sense, but
'do {e1;e2}' --> 'e1 >> e2' already exposes the type constraints.

i prefer to think of it as a kind of eta-expansion: i don't know
the type of 'e', but i do know something about the type of 
'\x->e x' and i would like to know something about the type
of 'do {e}'.

monad laws have the same issue, btw:

'return () >> e' --> 'e' only makes sense for monadic 'e'

claus