[Haskell-cafe] Re: Syntax of 'do'

[Maurí­cio]

 >> It sounds like you tried to redefine (>>) and
 >> (>>=) and make 'do' use the new definitions.
 >> This is not possible, regardless of what types
 >> you give (>>) and (>>=).
 >
 > Watch out for rebindable syntax: (...)
 >
 > At first reading, I thought that
 > -XNoImplicitPrelude was required to turn this
 > on. But now I'm not sure: (...)

I wrote this test to check your sugestion. It does
build with -XNoImplicitPrelude, but not without
it:

----------
module Test where {
import Prelude hiding ( ( >> ) , ( >>= ) ) ;

  data PseudoMonad a = PseudoMonad a ;
  ( >> ) = \(PseudoMonad x) (PseudoMonad _) -> PseudoMonad x ;
  ( >>= ) = (\(PseudoMonad a) f -> f a)
      :: PseudoMonad Integer -> (Integer -> PseudoMonad Integer)
      -> PseudoMonad Integer;
  plusOne n = (PseudoMonad (n + 1))
      :: PseudoMonad Integer;
  c = (PseudoMonad 1) >> ((PseudoMonad 2) >>= (\n -> plusOne n));
  d = do {(PseudoMonad 1) ; a <- (PseudoMonad 2) ; plusOne a }
}
----------

It's interesting that the types involved in >>=
etc. should still be like "t t1", that's why I had
to create PseudoMonad. Using just Integer (i.e., 2
 >> 3 would be valid) doesn't work, even if all
operators are defined accordingly.

Best,
Maurício