[Haskell-cafe] Looking for feedback on my beginner's Haskell resource
jo at durchholz.org
Tue Nov 6 06:42:15 UTC 2018
Am 05.11.18 um 23:27 schrieb Brandon Allbery:
> No state is modified, at least in ghc's implementation of IO.
That's what I'd expect.
> IO does carry "state" around, but never modifies it; it exists solely
> to establish a data dependency (passed to and returned from all IO
> actions; think s -> (a, s),
In Haskell, a data dependency can impose constraints on evaluation
order, but it isn't always linear: which subexpression is evaluated
first depends on what a pattern match requests (at least in Haskell:
Haskell's strict operation is the pattern match).
The ordering constraint becomes linear if each function calls just a
single other function. I'm not sure that that's what happens with IO;
input operations must allow choices and loops, making me wonder how
linearity is established. It also makes me wonder how an IO expression
would look like if fully evaluated; is it an infinite data structure,
made useful only through Haskell's laziness, or is it something that's
happening in the runtime?
The other thing that's confusing me is that I don't see anything that
starts the IO processing. There's no pattern match that triggers an
Not that this would explain much: If IO were constructed in a way that a
pattern match starts IO execution, there'd still be the question what
starts this first pattern match.
Then there's the open question what happens if a program has two IO
expressions. How does the runtime know which one to execute?
Forgive me for my basic questions; I have tried to understand Haskell,
but I never got the opportunity to really use it so I cannot easily test
More information about the Haskell-Cafe