The introduction
The following code shows that when using runhaskell Haskell Garbage Collector releases the memory, when a is no longer used. It results in core dump while releasing variable a - for a purpose, to inspect the behaviour - a has got nullFunPtr as a finalizer.
module Main where
import Foreign.Ptr
import Foreign.ForeignPtr
main :: IO ()
main = do
a <- newForeignPtr nullFunPtr nullPtr
putStrLn "Hello World"
The problem
When running the same in ghci it does not release memory. How can I force ghci to release no longer used variables?
$ ghci
> import Foreign.Ptr
> import Foreign.ForeignPtr
> import System.Mem
> a <- newForeignPtr nullFunPtr nullPtr
> a <- return () -- rebinding variable a to show gc that I'm no longer using it
> performGC
> -- did not crash - GC didn't release memory
> ^D
Leaving GHCi.
[1] 4396 segmentation fault (core dumped) ghci
Memory was released on exit, but this is too late for me. I'm extending GHCi and using it for other purpose and I need to release the memory earlier - on demand or as fast as possible would be really great.
I know that I can call finalizeForeignPtr, but I'm using foreignPtr just for debug purposes. How can I release a in general in last example?
If there is no possibility to do it with ghci prompt, I can also modify ghci code. Maybe I can release this a by modyfing ghci Interactive Context or DynFlags? So far I've got no luck with my reaserch.
Tracing through the code we find that the value is stored in the field closure_env of the data type PersistentLinkerState, which is a ClosureEnv, i.e. a mapping from name to HValues. The relevant function in Linker.hs is
extendLinkEnv :: [(Name,HValue)] -> IO ()
-- Automatically discards shadowed bindings
extendLinkEnv new_bindings =
modifyPLS_ $ \pls ->
let new_closure_env = extendClosureEnv (closure_env pls) new_bindings
in return pls{ closure_env = new_closure_env }
and although the comment indicates that it should remove the shadowed binding, it does not, at least not the way you want it to.
The reason is, as AndrewC writes correctly: Although both variables have the same source code name, they are different to the compiler (they have a different Unique attached). We can observe this after adding some tracing to the function above:
*GHCiGC> a <- newForeignPtr nullFunPtr nullPtr
extendLinkEnv [a_azp]
*GHCiGC> a <- return ()
extendLinkEnv [a_aF0]
*GHCiGC> performGC
extendLinkEnv [it_aFL]
Removing bindings with the same source-name at this point should solve your GC problem, but I don’t know the compiler well enough to tell what else would break. I suggest you open a ticket, hopefully someone will know.
Confusion on binding vs. value
In the comments there seems to be some confusion about bindings and values. Consider this code:
> a <- return something
> b <- return somethingelse
> a <- return (b+b)
> b <- return anewthing
With the current implementation, the heap will consist of `
something
somethingelse
a thunk referencing the (+) operator and somethingelse
anewthing.
Furthermore the environment of the interpreter has references to all four heap values, so nothing can be GC’ed.
What remdezx rightly expected is that GHCi would drop the reference to something and somethingelse. This, in turn, would allow the run time system to garbage collect something (we assume no further references). GHCi still references the thunk, which in turn references somethingelse, so this would not be garbage collected.
Clearly the question was very implementation specific, and so is this answer :-)
Related
I came up with the following code for the raw-input problem, as discussed in SO Haskell read raw keyboard input. Unfortunately, when I go to ghci, run getAllInput, and hit the right arrow key, it never returns. Unless I kill it pretty quickly, it seems to eat all my memory so that other applications stop responding and I have to restart the OS. In the Activity Monitor I can see the memory for the ghc process go quickly into the gigabytes.
(1) I think the problem is in the recursive call of go, which is evaluated lazily by calling hReady before getChar; this means hReady keeps returning true and the stack grows forever. Is that plausible?
(2) I'm used to languages in which this would soon cause a stack overflow exception, so it wouldn't prevent me from working. Is there any general way to protect against this massive memory leak? Maybe starting ghci with a hard limit on memory use?
import System.IO
-- For example, should get "\ESC[C" from the user hitting the right arrow key.
getAllInput :: IO [Char]
getAllInput =
let
go :: IO [Char] -> IO [Char]
go chars = do
more <- hReady stdin
if more then go (added chars getChar) else chars
added :: IO [Char] -> IO Char -> IO [Char]
added chars char = do
chars1 <- chars
char1 <- char
return (chars1 ++ [char1])
in do
hSetBuffering stdin NoBuffering
firstChar <- getChar
go (return [firstChar])
I'm running ghci 7.10.3 in OS X 10.11.6. I cleaned up the code in some obvious ways, basically following the similar SO answer: putting the getChar call in its own line fixes the problem. But I'd like to understand this better in case it bites me again.
You have an infinite loop in go. If hReady returns true, then you call go again, which immediately calls hReady again, which will of course return true, and so on. You probably assume that added will be run because of go (added chars getChar), but it will not; it is just building an IO action and passing it to go as an argument, but that argument is only used when hReady returns False. The name chars is misleading -- chars is actually an I/O procedure which will return a list of characters when it is eventually run.
In general, when using monads, a "normal" function signature looks like:
:: Foo -> Bar -> Baz -> IO Quuz
That is, the monad (IO) is only present on the return value, not the arguments. A signature like
:: IO Foo -> IO Bar
usually indicates that something higher-order is going on, for example that this function will perhaps execute its argument multiple times or in a new context.
I recommend the signatures
go :: [Char] -> IO [Char]
added :: [Char] -> Char -> IO [Char]
and trying to get the program to compile from there.
You should also try changing added to be a pure function
added :: [Char] -> Char -> [Char]
because it doesn't actually have any side effects. The implementation and usage will need to be altered a little, though.
Summarizing the discussion...
(1) As explained by luqui's answer, the memory leak is caused by an infinite loop in which lazy evaluation prevents getChar from being called. Adding a line like c <- getChar forces the call to happen, so that go (added chars c) is now safe to call.
(2) Starting GHCi with a limited heap size, as in ghci getchars.hs +RTS -M100m, will interrupt the memory leak before it eats up all memory. See the GHC Users Guide for further details.
For example
main = do
let ls = [0..10000000]
print ls
This will create the array 'inplace', using O(1) memory.
The following edit causes the program to run out of memory while executing.
main = do
let ls = [0..10000000]
print ls
print ls
ls in this case must be kept in memory to be printed again. It would actually be heaps more memory efficient to recalculate the array again 'inplace' than to try to keep this in place. That's an aside though. My real question is "how and when does GHC communicate to the runtime system that ls can be destroyed while it's created in O(1) time?" I understand that liveness analysis can find this information, I'm just wondering where the information is used. Is it the garbage collector that is passed this info? Is it somehow compiled away? (If I look at the compiled core from GHC then both examples use eftInt, so if it's a compiler artifact then it must happen at a deeper level).
edit: My question was more about finding where this optimization took place. I thought maybe it was in the GC, which was fed some information from some liveness check in the compilation step. Due to the answers so far I'm probably wrong. This is most likely then happening at some lower level before core, so cmm perhaps?
edit2: Most of the answers here assume that the GC knows that ls is no longer referenced in the first example, and that it is referenced again in the second example. I know the basics of GC and I know that arrays are linked lists, etc. My question is exactly HOW the GC knows this. The answer could probably be only (a) it is getting extra information from the compiler, or (b) it doesn't need to know this, that this information is handled 100% by the compiler
ls here is a lazy list, not an array. In practice, it's closer to a stream or generator in another language.
The reason the first code works fine is that it never actually has the whole list in memory. ls is defined lazily and then consume element-by-element by print. As print is going along, there are no other references to the beginning of ls so list items can be garbage collected immediately.
In theory, GHC could realize that it's more efficient to not store the list in memory between the two prints but instead recompute it. However, this is not always desirable—a lot of code is actually faster if things are only evaluated once—and, more importantly, would make the execution model even more confusing for programmers.
This explanation is probably a lie, especially because I'm making it up as I go, but that shouldn't be a problem.
The essential mistake you're making is assuming that a value is live if a variable bound to it is in scope in a live expression. This is simply wrong. A value bound to a variable is only live as a result if it is actually mentioned in a live expression.
The job of the runtime is very simple
Execute the expression bound to main.
There is no 2.
We can think of this execution as involving a couple different steps that repeat over and over:
Figure out what to do now.
Figure out what to do next.
So we start with some main expression. From the start, the "root set" for GC consists of those names that are used in that main expression, not the things that are in scope in that expression. If I write
foo = "Hi!"
main = print "Bye!"
then since foo is not mentioned in main, it is not in the root set at the beginning, and since it is not even mentioned indirectly by anything mentioned by main, it is dead right from the start.
Now suppose we take a more interesting example:
foo = "Hi!"
bar = "Bye!"
main = print foo >> print bar
Now foo is mentioned in main, so it starts out live. We evaluate main to weak head normal form to find out what to do, and we get, approximately,
(primitive operation that prints out "Hi!") >> print bar
Note that foo is no longer mentioned, so it is dead!
Now we execute that primitive operation, printing "Hi!", and our "to do list" is reduced to
print bar
Now we evaluate that to WHNF, and get, roughly,
(primitive operation to print "Bye!")
Now bar is dead. We print "Bye!" and exit.
Consider, now, the first program you described:
main = do
let ls = [0..10000000]
print ls
This desugars to
main =
let ls = [0..10000000]
in print ls
This is where we start. The "root set" at the beginning is everything mentioned in the in clause of the expression. So we conceptually have ls and print to start out. Now we can imagine that print, specialized to [Integer], looks something vaguely like the following (this is greatly simplified, and will print out the list differently, but that really doesn't matter).
print xs = case xs of
[] -> return ()
(y:ys) = printInteger y >> print ys
So when we start executing main (What do we do now? What will we do afterwards?), we are trying to calculate print ls. To do this, we pattern match on the first constructor of ls, which forces ls to be evaluated to WHNF. We find the second pattern, y:ys, matches, so we replace print ls with print Integer y >> print ys, where y points to the first element of ls and ys points to the thunk representing the second list constructor of ls. But note that ls itself is now dead! Nothing is pointing to it! So as print forces bits of the list, the bits it has already passed become dead.
In contrast, when you have
main =
let ls = ...
in print ls >> print ls
and you start executing, you start by calculating the thing to do first (print ls). You get
(printInteger y >> print ys) >> print ls
Everything is the same, except that the second part of the expression now points to ls. So even though the first part will be dropping pieces of the list as it goes, the second part will keep holding on to the beginning, keeping it all live.
Edit
Let me try explaining with something a little simpler than IO. Pretend that your program is an expression of type [Int], and the job of the runtime system is to print each element on its own line. So we can write
countup m n = if m == n then [] else m : countup (m+1)
main = countup 0 1000
The runtime system holds a value representing everything that it should print. Let's call the "current value" whatPrint. The RTS needs to follow a process:
Set whatPrint to main.
Is whatPrint empty? If so, I'm done, and can exit the program. If not, it is a cons, printNow : whatPrint'.
Calculate printNow and print it.
Set whatPrint to whatPrint'
Go to step 1.
In this model, the "root set" for garbage collection is just whatPrint.
In a real program, we don't produce a list; we produce an IO action. But such an action is also a lazy data structure (conceptually). You can think of >>=, return, and each primitive IO operation as a constructor for IO. Think of it as
data IO :: * -> * where
Return :: a -> IO a
Bind :: IO a -> (a -> IO b) -> IO b
PrintInt :: Int -> IO ()
ReadInt :: IO Int
...
The initial value of whatShouldIDo is main, but its value evolves over time. Only what it points to directly is in the root set. There is no magical analysis necessary.
Can IO actions in IO blocks call within unsafePerformIO be reordered?
I have effectively the IO function.
assembleInsts :: ... -> IO S.ByteString
assembleInsts ... = do
tmpInputFile <- generateUniqueTmpFile
writeFile tmpInputFile str
(ec,out,err) <- readProcessWithExitCode asm_exe [tmpInputFile] ""
-- asm generates binary output in tmpOutputFile
removeFile tmpInputFile
let tmpOutputFile = replaceExtension tmpIsaFile "bits" -- assembler creates this
bs <- S.readFile tmpOutputFile -- fails due to tmpOutputFile not existing
removeFile tmpOutputFile
return bs
where S.ByteString is a strict byte string.
Sadly, I need to call this in a tree of pure code far from the IO monad,
but since I the assembler behaves as a referentially transparent
(given unique files) tool, I figured for the time being I could make
an unsafe interface for the time being.
{-# NOINLINE assembleInstsUnsafe #-}
assembleInstsUnsafe :: ... -> S.ByteString
assembleInstsUnsafe args = unsafePerformIO (assembleInsts args)
In addition I added to the top of the module the following annotation
as per the documentation's (System.IO.Unsafe's) instructions.
{-# OPTIONS -fno-cse #-}
module Gen.IsaAsm where
(I tried to also add -fnofull-laziness as well, as per a reference that
I consulted, but this was rejected by the compiler. I don't think that
case applies here though.)
Running in ghci it reports the following error.
*** Exception: C:\Users\trbauer\AppData\Local\Temp\tempfile_13516_0.dat: openBinaryFile: does not exist (No such file or directory)
But if I remove removeFile tmpOutputFile, then it magically works.
Hence, it seems like the removeFile is executing ahead of the process termination.
Is this possible? The bytestring is strict, and I even tried to force the output at one point with a:
S.length bs `seq` return ()
before the removeFile.
Is there a way to dump intermediate code to find out what's going on?
(Maybe I can trace this with Process Monitor or something to find out.)
Unfortunately, I'd like to clean up within this operation (remove the file).
I think the exe version might work, but under ghci it fails (interpreted).
I am using GHC 7.6.3 from the last Haskell Platform.
I know unsafePerformIO is a really big hammer and has other risks associated with it, but it would really limit the complexity of my software change.
This may not be applicable, since it is based on assumptions unspecified in your question. In particular, this answer is based on the following two assumptions. S, which is unspecified, is Data.ByteString.Lazy and tmpDatFile, which is undefined, is tmpOutputFile.
import qualified Data.ByteString.Lazy as S
...
let tmpDatFile = tmpOutputFile
Possible Cause
If these assumptions are true, removeFile will run too early, even without the use of unsafePerformIO. The following code
import System.Directory
import qualified Data.ByteString.Lazy as S
assembleInsts = do
-- prepare a file, like asm might have generated
let tmpOutputFile = "dataFile.txt"
writeFile tmpOutputFile "a bit of text"
-- read the prepared file
let tmpDatFile = tmpOutputFile
bs <- S.readFile tmpOutputFile
removeFile tmpDatFile
return bs
main = do
bs <- assembleInsts
print bs
Results in the error
lazyIOfail.hs: DeleteFile "dataFile.txt": permission denied (The process cannot access the file because it is being used by another process.)
Removing the line removeFile tmpDatFile will make this code execute correctly, just like you describe, but leaving behind the temporary file isn't what is desired.
Possible Solution
Changing the import S to
import qualified Data.ByteString as S
instead results in the correct output,
"a bit of text".
Explanation
The documentation for Data.ByteSting.Lazy's readFile states that it will
Read an entire file lazily into a ByteString. The Handle will be held open until EOF is encountered.
Internally, readfile accomplishes this by calling unsafeInterleaveIO. unsafeInterleaveIO defers execution of the IO code until the term it returns is evaluated.
hGetContentsN :: Int -> Handle -> IO ByteString
hGetContentsN k h = lazyRead -- TODO close on exceptions
where
lazyRead = unsafeInterleaveIO loop
loop = do
c <- S.hGetSome h k -- only blocks if there is no data available
if S.null c
then do hClose h >> return Empty
else do cs <- lazyRead
return (Chunk c cs)
Because nothing tries to look at the constructor of the bs defined in the example above until it is printed, which doesn't happen until after removeFile has been executed, no chunks are read from the file (and the file is not closed) before removeFile is executed. Therefore, when removeFile is executed, the Handle opened by readFile is still open, and the file can't be removed.
Even if you are using unsafePerformIO, IO actions should not be reordered. If you want to be sure of that, you can use the -ddump-simpl flag to see the intermediate Core language which GHC produces, or even one of the other -dump-* flags showing all the compilation intermediate steps up to assembly.
I am aware that this answers what you asked, and not what you actually need, but you can rule out GHC bugs at least. It seems unlikely there's a bug affecting this in GHC, though.
Totally my fault.... sorry everyone. GHC does not reorder IO actions in an IO block under the above stated conditions as mentioned by those above. The assembler was just failing to assemble the output and create the assumed file. I simply forgot to check the exit code or the output stream of the assembler. I assumed the input to be syntactically correct since it is generated, the assembler rejected it and simply failed to create the file. It gave a valid error code and error diagnostic too, so that was really bad on my part. I may have been using readProcess the first time around, which raises an exception on a non-zero exit, but must have eventually changed this. I think the assembler had a bug where it didn't correctly indicate a failing exit code for some cases, and I had to change from readProcessWithExitCode.
I am still not sure why the error went away when I elided the removeFile.
I thought about deleting the question, but I a hoping the suggestions above help others debug similar (more valid) problems as well. I've been burned by the lazy IO thing Cirdec mentioned, and the -ddump-simpl flag mentioned by chi is good to know as well.
I'm creating a small program to use with an irc bot that should take a string and then evaluate the string. For this I'm using the hint package, which work very well for my needs. The problem that I now have is that I want to be able to prevent evaluation of expressions that take a vary long to calculate e.g. 2^1000000000.
I tried using the System.Timeout package like this:
import Data.Maybe
import Language.Haskell.Interpreter
import System.Timeout
import System.Environment (getArgs)
main :: IO()
main = do
r <- timeout 500000 $ runInterpreter $ hEval arg
case r of
Nothing -> putStrLn "Timed out!"
Just x ->
case x of
Left err -> putStrLn (show err)
Right a -> putStrLn a
hEval e = do
setImportsQ [("Prelude", Nothing),("Data.List",Nothing)]
a <- eval e
return $ take 200 a
But it's not working, the timeout does not fire unless I put in such a short time that nothing can be evaluated. I read on the page for the Timeout package that it could have problems with some modules and have to let theme finish but my understanding is not good enough to know if Hint is such a module.
So any help on this would be appreciated, even if it's just to tell me that this isn't going to work.
GHC threads are cooperative. They can only yield or be terminated by asynchronous exceptions when they perform a memory allocation. This normally works fine, but someone malicious can write a tight loop that runs for a significant time without allocating.
The mueval package was created to deal with things like this. It's implemented in terms of hint, but with a lot of extra safety added in various ways.
Consider the following code snippet
import qualified Foreign.Concurrent
import Foreign.Ptr (nullPtr)
main :: IO ()
main = do
putStrLn "start"
a <- Foreign.Concurrent.newForeignPtr nullPtr $
putStrLn "a was deleted"
putStrLn "end"
It produces the following output:
start
end
I would had expected to see "a was deleted" somewhere after start..
I don't know what's going on. I have a few guesses:
The garbage collector doesn't collect remaining objects when the program finishes
putStrLn stops working after main finishes. (btw I tried same thing with foreignly imported puts and got the same results)
My understanding of ForeignPtr is lacking
GHC bug? (env: GHC 6.10.3, Intel Mac)
When using Foreign.ForeignPtr.newForeignPtr instead of Foreign.Concurrent.newForeignPtr it seems to work:
{-# LANGUAGE ForeignFunctionInterface #-}
import Foreign.C.String (CString, newCString)
import Foreign.ForeignPtr (newForeignPtr)
import Foreign.Ptr (FunPtr)
foreign import ccall "&puts" puts :: FunPtr (CString -> IO ())
main :: IO ()
main = do
putStrLn "start"
message <- newCString "a was \"deleted\""
a <- newForeignPtr puts message
putStrLn "end"
outputs:
start
end
a was "deleted"
From the documentation of Foreign.Foreign.newForeignPtr:
Note that there is no guarantee on how soon the finaliser is executed after the last reference was dropped; this depends on the details of the Haskell storage manager. Indeed, there is no guarantee that the finalizer is executed at all; a program may exit with finalizers outstanding.
So you're running into undefined behaviour: i.e., anything can happen, and it may change from platform to platform (as we saw under Windows) or release to release.
The cause of the difference in behaviour you're seeing between the two functions may be hinted at by the documentation for Foreign.Concurrent.newForeignPtr:
These finalizers necessarily run in a separate thread...
If the finalizers for the Foreign.Foreign version of the function use the main thread, but the Foreign.Concurrent ones use a separate thread, it could well be that the main thread shuts down without waiting for other threads to complete their work, so the other threads never get to run the finalization.
Of course, the docs for the Foreign.Concurrent version do claim,
The only guarantee is that the finalizer runs before the program terminates.
I'm not sure that they actually ought to be claiming this, since if the finalizers are running in other threads, they can take an arbitrary amount of time to do their work (even block forever), and thus the main thread would never be able to force the program to exit. That would conflict with this from Control.Concurrent:
In a standalone GHC program, only the main thread is required to terminate in order for the process to terminate. Thus all other forked threads will simply terminate at the same time as the main thread (the terminology for this kind of behaviour is "daemonic threads").