STM in Haskell(joeyh.name)

45 pointsby mzehrervor 13 Jahren11 comments

http://joeyh.name/blog/entry/the_newinwheezy_game:_STM/

11 comments

krigvor 13 Jahren

As a complete Haskell noob, I'm wondering about the push in this snippet:

  startDownload = do
    file <- atomically $ do
        f <- getQueuedFile
        push f currentDownLoads
        return f
    startDownloadThread file

How can that push be non-side-effecting and still do something sensible? If the snippet is replayed, won't multiple pushes happen? This may be a question only joeyh can answer, if this is something specific to his code base...

Naively, I would write something like (again, my Haskell not so good):

  startDownload = do
    file <- atomically $ getQueuedFile
    push f currentDownLoads
    startDownloadThread file

lelfvor 13 Jahren

You can think about your second version as without side-effects too. It's just in IO monad. Actually, all IO is kinda pure, just (IO smth) values fly behind the scenes. All dirty unpure things happen in runtime.

And in the first snippet it is in STM monad, and you can do, for example

  modifyTVar :: TVar a -> (a -> a) -> STM ()

You should understand monads more to grok it, it's quite beautiful.

svedinvor 13 Jahren

push doesn't have to be non-side-effecting. You are allowed to modify shared state, as long as it is a TVar (shared memory that supports transactions).

krigvor 13 Jahren

Right, of course. Thank you. :)

BasDirksvor 13 Jahren

Haskell isn't non-side-effecting. It's just a bit more sane and structured about when and how.

vemvvor 13 Jahren

Nobody said that anyway. I doubt there's such a thing as non-side-effecting languages :)

telvor 13 Jahren

Sure there are, they just are interpreted in side-effecting interpreters. Lambda calculus comes to mind.

FreeFullvor 13 Jahren

Anything that happens in the STM monad can be undone if something is wrong. If you for example write to a TVal but another thread was doing it at the same time, one of the threads won't have its changes go through until the transaction is retried. Of course this is all invisible to the person using STM.

tomevor 13 Jahren

Who said it was non-side-effecting?

krigvor 13 Jahren

I guess that's my misunderstanding. :)

jonkevor 13 Jahren

I do think this example http://computationalthoughts.blogspot.se/2008/03/some-exampl... is a lot better than the one in the article.