Why/how does recursive IO work?

Question

Haskell IO is often explained in terms of the entire program being a pure function (main) that returns an IO value (often described as an imperative IO program)

Answer 1

The value main denotes is an infinite program:

main = do
  line <- getLine
  putStrLn line
  line <- getLine
  putStrLn line
  line <- getLine
  putStrLn line
  line <- getLine
  putStrLn line
  line <- getLine
  putStrLn line
  line <- getLine
  putStrLn line
  ...

But it's represented in memory as a recursive structure that references itself. That representation is finite, unless someone tries to unfold the entire thing to get a non-recursive representation of the entire program - that would never finish.

But just as you can probably figure out how to start executing the infinite program I wrote above without waiting for me to tell you "all" of it, so can Haskell's runtime system figure out how to execute main without unfolding the recursion up-front.

Haskell's lazy evaluation is actually interleaved with the runtime system's execution of the main IO program, so this works even for a function that returns an IO action which recursively invokes the function, like:

main = foo 1

foo :: Integer -> IO ()
foo x = do
  print x
  foo (x + 1)

Here foo 1 is not a recursive value (it contains foo 2, not foo 1), but it's still an infinite program. However this works just fine, because the program denoted by foo 1 is only generated lazily on-demand; it can be produced as the runtime system's execution of main goes along.

By default Haskell's laziness means that nothing is evaluated until it's needed, and then only "just enough" to get past the current block. Ultimately the source of all the "need" in "until it's needed" comes from the runtime system needing to know what the next step in the main program is so it can execute it. But it's only ever the next step; the rest of the program after that can remain unevaluated until after the next step has been fully executed. So infininte programs can be executed and do useful work so long as it's always only a finite amount of work to generate "one more step".