Lazy evaluation and IO side effect confusion?

Read the " Tackling the awkward squad " paper by Simon Peyton Jones.

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This code (taken from Learn You A Haskell): main = do putStr " " putStr "I'm " putStrLn "Andy! " apparently desugars to main = putStr " " >>= (\_ -> putStr "I'm " >>= (\_ -> putStrLn "Andy! ")) Which, as I understand it can be interpretted as saying "In order to putStrLn "Andy!

" I first need to putStr "I'm ", and in order to do that I first need to putStr " "; I disagree with this interpretation, which is annoying because the compiler obviously doesn't and leaves me feeling confused. The problem I have with it is that the lambdas ignore their arguments, during lazy evaluation isn't this sort of thing supposed to be recognised and short-circuited? Also, sure, the binding returns an IO action, and when that IO action falls into main it gets executed.

But what's to stop it from printing " Andy! I'm "? I suspect it's whatever bind is doing.

Also, how does an IO action of type "IO ()" carry enough information to allow the runtime system to print " I'm Andy! "? How is that IO () different to an IO () than prints "" or writes to a file?

Consider another, from the wikipedia page for monad: Sugared version: do putStrLn "What is your name? " name >= (\_ -> getLine >>= (\name -> putStrLn ("Nice to meet you, " ++ name ++ "! "))) Similar story here.

I think I just need to see the definition of bind for IO and then it will be all clear. Something else that would help a lot is if someone could help me step through how the program actually gets evaluated and identify the exact moments when the side effects occur. Haskell functional-programming lazy-evaluation side-effects link|improve this question asked Nov 22 '11 at 11:18TheIronKnuckle1,324120 97% accept rate.

" You bet! The second argument to (>>=) is a particularly lazy function here, but the (>>=) function itself is not lazy. – Daniel Wagner Nov 22 '11 at 14:37.

Read the "Tackling the awkward squad" paper by Simon Peyton Jones. For related questions, see Pure Functional Language: Haskell , Is Haskell truly pure (is any language that deals with input and output outside the system)? In what sense is the IO Monad pure?

Haskell: actual IO monad implementation, in different language? Why does Haskell not have an I Monad (for input only, unlike the IO monad)? References for learning the theory behind pure functional languages such as Haskell?

Take any such explanation including mine with a grain of salt - no hand-waving can replace a rigorous peer-reviewed paper, and the explanations are necessarily over-simplifications. A very rough perspective is that >>= can be seen as a list constructor: data IO = Primitive and IO subsystem deconstructs the value of main and consumes that list. I.e.

`main is just a list. So you may want to take a look at the definition of Haskell entry point above main, bind is rather uninteresting. You can also read papers on history of haskell and look at earlier versions of IO subsystem for insight of what is going on.

Also look at C language is purely functional satiric post by Conal Elliott. The definition of functional purity is non-trivial and I remember a paper elaborating on the definition, but I don't remember the title.

It's interesting that everyone always explains >>= on IO values by analogy. Doesn't it have a definition somewhere in the prelude? Why does no one ever quote it?

Is bind over IO where the magic is happening? I'm reading the awkward squad paper right now and even SPJ shys away from actually defining bind – TheIronKnuckle Nov 22 '11 at 12:07 3 The IO type is abstract, so there's no one definition of >>= in the Haskell standard. Depending on how you implement IO you'll have different implementations of >>=.

If you delve into ghc you'll find that IO is a state monad and >>= is simply bind for a state monad. (There's more magic inside the compiler to make this efficient. ) – augustss Nov 22 '11 at 12:10 @ThelronKnuckle It is not an analogy on 'values'.

It is an analogy on another implementation of pure IO using the old main :: Request -> Response idea. Also the most interesting part is not of the bind, but of the runIO for IO monad. As IO monad is abstract, we must either provide an abstract explanation as SPJ did, or use some implementation strategy as I did.

– nponeccop Nov 22 '11 at 12:13.

Looking at IO in a real Haskell implementation will probably confuse more than it enlightens. But think of IO as being defined like this (this assumes you know GADTs): data IO a where Return a :: IO a Bind :: IO a -> (a -> IO b) -> IO be PutStr :: String -> IO () GetLine :: IO String instance Monad IO where return = Return (>>=) = Bind putStr :: String -> IO () putStr = PutStr getLine :: IO String getLine = GetLine So when you evaluate a program (of type IO ()) all it does is to build a data structure of type IO () that describes how the interaction with the world will happen once you execute it. You can then imagine the execution engine being written in, e.g. , C, and there is where all effects happen.

So main = do putStr " " putStr "I'm " putStrLn "Andy! " is the same as main = Bind (PutStr " ") (\ _ -> Bind (PutStr "I'm ") (\ _ -> PutStr "Andy! ")) And the sequencing of these comes from the way the execution engine works.

That said, I know of no Haskell implementation that actually does it this way. Real implementations tend to implement IO as a state monad with a token representing the real world being passed around (this is what guarantees sequencing), and primitives like putStr are just calls to C functions.

1 for the GADT approach and the 'Looking at IO in a real Haskell implementation will probably confuse more than it enlightens' thing. – nponeccop Nov 22 '11 at 14:15.

I think I just need to see the definition of bind for IO and then it will be all clear. Yes, you should do that. It's actually quite easy, and if I remeber correctly it goes like newtype IO = IO (RealWorld -> (a, RealWorld)) (IO f) >>= g = ioBind f g where ioBind :: (RealWorld -> (a, RealWorld)) -> (a -> IO b) -> RealWorld -> (b, RealWorld) ioBind f g rw = case f rw of (a, rw@RealWorld) -> case g a of IO be -> be rw The "trick" is that every IO value is actually basically a function, but to evaluate it you would need a token of type RealWorld.

There is only one instance that can supply such a value - the runtime system running main (and, of course, the function that must not be named).

I think this is more understandable if you think of the actions again as functions. Your binding example (do { foo Otherwise we fail in our action. This is different from ordinary functions because then Haskell really doesn't care if (arg) computes fully or at all, until the point it needs a bit of func(arg) to continue the program.

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