Haskell IORef - an answer vs. a function to get an answer

Question

I'm trying to understand how IORefs are really used, and I'm having trouble following the sample code I found on https://www.seas.upenn.edu/~cis194/spring15/lectures/12-unsafe.html

newCounter :: IO (IO Int)
newCounter = do
  r <- newIORef 0
  return $ do
    v <- readIORef r
    writeIORef r (v + 1)
    return v

printCounts :: IO ()
printCounts = do
  c <- newCounter
  print =<< c
  print =<< c
  print =<< c

When printCounts executes "c <- newCounter", why doesn't c get the result of doing the work in the newCounter "return $ do" block, which seems like it should get assigned to the constant "IO 0" the first time it is called and then never change? Instead, c seems to get assigned the function defined in that "return $ do" block, which is then executed anew every time printCounts gets to another "print =<< c." It seems that the answer somehow lies in newCounter having the double nested "IO (IO Int)" type, but I can't follow why that makes c a function to be re-executed when called instead of a constant evaluated just once.

Answer 1

You can think of IO as a type of programs. newCounter :: IO (IO Int) is a program that outputs a program. More precisely, newCounter allocates a new counter, and returns a program that, when run, increments the counter and returns its old value. newCounter doesn't execute the program it returns. It would if you wrote instead:

newCounter :: IO (IO Int)
newCounter = do 
  r <- newIORef 0
  let p = do              -- name the counter program p
        v <- readIORef r
        writeIORef r (v + 1)
        return v
  p          -- run the counter program once
  return p   -- you can still return it to run again later

---

You can also use equational reasoning to unfold printCounts into a sequence of primitives. All versions of printCounts below are equivalent programs:

-- original definition
printCounts :: IO ()
printCounts = do
  c <- newCounter
  print =<< c
  print =<< c
  print =<< c

-- by definition of newCounter...

printCounts = do
  c <- do
    r <- newIORef 0
    return $ do
      v <- readIORef r
      writeIORef r (v + 1)
      return v
  print =<< c
  print =<< c
  print =<< c

-- by the monad laws (quite hand-wavy for brevity)
-- do
--   c <- do
--     X
--     Y
--   .....
-- =
-- do
--   X
--   c <- 
--     Y
--   .....
--
-- (more formally,
--  ((m >>= \x -> k x) >>= h) = (m >>= (\x -> k x >>= h)))

printCounts = do
  r <- newIORef 0
  c <-
    return $ do
      v <- readIORef r
      writeIORef r (v + 1)
      return v
  print =<< c
  print =<< c
  print =<< c

-- c <- return X
-- =
-- let c = X
--
-- (more formally, ((return X) >>= (\c -> k c)) = (k X)

printCounts = do
  r <- newIORef 0
  let c = do
        v <- readIORef r
        writeIORef r (v + 1)
        return v
  print =<< c
  print =<< c
  print =<< c

-- let-substitution

printCounts = do
  r <- newIORef 0
  print =<< do
        v <- readIORef r
        writeIORef r (v + 1)
        return v
  print =<< do
        v <- readIORef r
        writeIORef r (v + 1)
        return v
  print =<< do
        v <- readIORef r
        writeIORef r (v + 1)
        return v

-- after many more applications of monad laws and a bit of renaming to avoid shadowing
-- (in particular, one important step is ((return v >>= print) = (print v)))

printCounts = do
  r <- newIORef 0
  v1 <- readIORef r
  writeIORef r (v1 + 1)
  print v1
  v2 <- readIORef r
  writeIORef r (v2 + 1)
  print v2
  v3 <- readIORef r
  writeIORef r (v3 + 1)
  print v3

In the final version, you can see that printCounts quite literally allocates a counter and increments it three times, printing each intermediate value.

One key step is the let-substitution one, where the counter program gets duplicated, which is why it gets to run three times. let x = p; ... is different from x <- p; ..., which runs p, and binds x to the result rather than the program p itself.