[Haskell-cafe] IORef memory leak

[Daniel van den Eijkel]

Yes I guessed that.
Thanks,
Daniel

Claus Reinke schrieb:
>>> It is not possible to write a modifyIORef that *doesn't* leak memory!
>>>   
>> Why? Or can one read about it somewhere?
>
> Possibly, Don meant that 'modifyIORef' is defined in a way that does 
> not allow to enforce evaluation of the result of the modification
> function (a typical problem with fmap-style library functions):
>
>    modifyIORef ref f = readIORef ref >>= writeIORef ref . f
>
> No matter whether 'f' is strict or not, the 'writeIORef r' doesn't
> evaluate its result, just stores the unevaluated application:
>
>    > r<-newIORef 0
>    > modifyIORef r (\x->trace "done" $ x+1)
>    > modifyIORef r (\x->trace "done" $ x+1)
>    > readIORef r
>    done
>    done
>    2
>
> If it had been defined like this instead
>
>    mRef r ($) f = readIORef r >>= (writeIORef r $) . f
>
> it would be possible to transform the strictness of 'writeIORef r'
> to match that of 'f':
>
>    > r<-newIORef 0
>    > mRef r ($) (\x->trace "done" $ x+1)
>    > mRef r ($) (\x->trace "done" $ x+1)
>    > readIORef r
>    done
>    done
>    2
>    > r<-newIORef 0
>    > mRef r ($!) (\x->trace "done" $ x+1)
>    done
>    > mRef r ($!) (\x->trace "done" $ x+1)
>    done
>    > readIORef r
>    2
>
> Claus
>
>
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