Short-circuit AND within IO Monad

Question

I know the question has been asked there but I can't believe there is no straight answer.

I understand it's not good to hide side-effect inside a (&&) but in my case the side-effect are just checking something in the outstide world (existence of a file, check modification time etc, ask user a yes/no quistion).

So what is the haskell way to something like this, so that cond2 is not executed if cond1 is false.

cond1, cond2 :: IO bool


main = do
   cond <- liftM2 (&&) con1 con2
   if cond
   then   result1
   else   result2

I was expecting something like cond <- all [con1, con2] or equivalent, but I can't find anything.

Update

I can see lots of manual solution. I'm still puzzled that this function doesn't exists somewhere. One advantage of lazzy evaluation is it doesn't only short-circuit for hard-coded && like in C. It is really strange that when in imperative mode, Haskell can't even short-circuit &&. Although, all solution use somehow and if to short-circuit the evaluation. Is there not a way to make a generic lazzy liftM2 ?

Answer 1

This is what Pipes.Prelude.and does, going over a lazy stream of effectfully-generated conditionals and short-circuiting if any of them are False:

import Pipes (each)
import qualified Pipes.Prelude as Pipes

conds :: [IO Bool]
conds = ...

main = do
    cond <- Pipes.and (each conds >-> Pipes.sequence)
    print cond

Relevant links:

• and
• Pipes tutorial

Answer 2

This isn't different from what some others are saying, but isn't it simplest just to emulate the definition of and:

 andM = foldr (&&&) (return True)
  where ma &&& mb = ma >>= \p -> if p then mb else return p

then we get, say:

 > let test1 = putStrLn "This test succeeds" >> return True
 > let test2 = putStrLn "This test fails" >> return  False
 > andM [test1,test2,undefined,undefined]
 This test succeeds
 This test fails
 False

If andM didn't 'short-circuit', then the undefined cell would have been evaluated and returned an exception.

It is a little irritating that liftM2 (&&) doesn't work as one hoped.

Edit: I just noticed that, as one might have expected, this is defined in the monad-loops package http://hackage.haskell.org/package/monad-loops-0.4.2.1/docs/src/Control-Monad-Loops.html#andM