Re: [Haskell-cafe] why cannot i get the value of a IORef variable ?

Well, I apologize for starting this whole thread which involves so many dead kittens :( I was just trying to help answer a question :) I guess I assumed too much.. that someone would think to be careful when using a function with the word 'unsafe' in it... So, be warned, all Haskellers! Be careful when using any function that starts with the word 'unsafe'! You may kill a kitten! And I guess this says something about using 'unsafe' functions... http://upload.wikimedia.org/wikipedia/en/1/11/God-kills-kitten.jpg (NSFW)... Cheers and sorry all, Tim On Thu, Oct 22, 2009 at 1:59 PM, Gregory Crosswhite < gcross@phys.washington.edu> wrote:

Yes, I was once taught that "Every time you use unsafePerformIO, God kills a kitten," so every time I consider using it I first ask myself: is this really worth an innocent kitten's life?

Cheers, Greg

On Oct 22, 2009, at 11:32 AM, David Menendez wrote:

On Thu, Oct 22, 2009 at 2:23 AM, Gregory Crosswhite

...

wrote:

...

For clarity, one trick that uses "unsafePerformIO" which you may have seen posted on this list earlier today is the following way of creating a globally visible IORef:

import Data.IORef import System.IO.Unsafe

*** counter = unsafePerformIO $ newIORef 0 ***

Danger! If the monomorphism restriction is disabled, this ends up creating a value of type forall a. Num a => IORef a, which can be used to break type safety.

More generally,

cell :: IORef a cell = unsafePerformIO $ newIORef undefined

unsafeCoerce :: a -> b unsafeCoerce x = unsafePerformIO $ do writeIORef cell x readIORef cell

This way lies segmentation faults. That "unsafe" is there for a reason.

-- Dave Menendez <http://www.eyrie.org/~zednenem/ http://www.eyrie.org/%7Ezednenem/>

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