Hello Brian,
Sunday, August 20, 2006, 5:15:41 PM, you wrote:
> data R = R {_xRef :: !(IORef Int)}
> foo :: Int -> R -> IO ()
> foo i R{_xRef = xRef} = writeIORef xRef i
> is the above more efficient than:
> foo i r = writeIORef (_xRef r) i
i think that first _may_ be more efficient because of strictness
analysis. btw, if you want beter efficiency, you may use unboxed
references (http://haskell.org/haskellwiki/Library/ArrayRef)
--
Best regards,
Bulat mailto:[EMAIL PROTECTED]
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