Haskell IORef array usage

Question

I am working on an program that uses a large list of IORef's to a data type. Which is the more memory/processor-efficient way to do this:

[IORef Foo]

or

IORef [Foo]

Ignore the fact that I am using lists instead of vectors or arrays.

On the other hand, [IORef] lets you change the number of items, while IORef [] doesn't. Therefore, a suggestion: use IORef [IORef]. — MigMit
– MigMit, Commented Jan 15, 2014 at 12:45

daniel gratzer · Accepted Answer · 2014-01-15 06:02:26Z

9

With [IORef Foo] you can update elements easily, with IORef [Foo], you can only update the whole list. Since you're likely wanting to efficiently update elements without copying, you want [IORef Foo]. Think, you want a list of mutable things, not a mutable list of immutable things.

As an example

import Data.IORef

makeTest :: IO [IORef Int]
makeTest = sequence [newIORef 0, newIORef 1, newIORef 2]

main = do
  test <- makeTest
  readIORef (test !! 1) >>= print
  modifyIORef (test !! 1) (+1) -- Doesn't copy list
  readIORef (test !! 1) >>= print

edited Jan 15, 2014 at 6:02

answered Jan 15, 2014 at 5:44

daniel gratzer

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