First the Question - is the behavior below expected logically, or a bug to be reported for GHC?
The code below will leak memory (tested on ghc-8.8.4) because ghc seems to add join point and jumps to it at the end of the loop, building up the stack.
{-# OPTIONS_GHC -fno-full-laziness #-}
module Main where
import Control.Concurrent.Async (async,waitCatch)
import Data.IORef
import GHC.Conc
main :: IO ()
main = do
val <- newIORef 0 :: IO (IORef Int)
let loop1 = do
cval <- readIORef val
threadDelay 1
writeIORef val (cval + 1)
case cval > 100000000 of
True -> error "done"
False -> loop1
loop1 -- Deliberately, add this to cause space leak, but this statement is never executed because of case branching above
loop1Async <- async loop1
res <- waitCatch loop1Async
return ()
Compiling with -O2 -rtsopts -threaded and running with +RTS -s -hT -N will show space leak because of growing stack.
Looking at core output, it seems the leak is due to join (I guess it is a join point) and a jump to it at the end of the loop which grows the stack (if I have read the core correctly). Removing the last statement in loop1 fixes the leak.
ghc core output is here.
Update: Based on feedback in comments, it seems to be logical behavior, not a bug in ghc. So, an answer explaining stack increase would be good to have. This helps us understand what is going on here. ghc core output has been posted above.
I would certainly expect that line to cause the use of stack space. That extra call to loop1 is hardly irrelevant.
Change the test constant to 10 instead of a larger number, and replace the return () branch with print cval.
Compare the output you get with and without the "never executed" statement. You might find it's somewhat more executed than you think.
Related
I am trying to figure out how can i take the progress info from a Progress type (in Development.Shake.Progress) to output it before executing a command. The possible desired output would be:
[1/9] Compiling src/Window/Window.cpp
[2/9] Compiling src/Window/GlfwError.cpp
[3/9] Compiling src/Window/GlfwContext.cpp
[4/9] Compiling src/Util/MemTrack.cpp
...
For now i am simulating this using some IORef that keeps the total (initially set to the sum of the source files) and a count that i increase before executing each build command, but this seems like a hackish solution to me.
On top of that this solution seems to work correctly on clean builds, but misbehaves on partial builds as the sum that displayed is still the total of all the source files.
With access to a Progress data type i will be able to calculate this fraction correctly using its countSkipped, countBuild, and countTodo members (see Progress.hs:53), but i am still not sure how i can i achieve this.
Any help is appreciated.
Values of type Progress are currently only available as an argument to the function stored in shakeProgress. You can obtain the Progress whenever you want with:
{-# LANGUAGE RecordWildCards #-}
import Development.Shake
import Data.IORef
import Data.Monoid
import Control.Monad
main = do
ref <- newIORef $ return mempty
shakeArgs shakeOptions{shakeProgress = writeIORef ref} $ do
want ["test" ++ show i | i <- [1..5]]
"test*" %> \out -> do
Progress{..} <- liftIO $ join $ readIORef ref
putNormal $
"[" ++ show (countBuilt + countSkipped + 1) ++
"/" ++ show (countBuilt + countSkipped + countTodo) ++
"] " ++ out
writeFile' out ""
Here we create an IORef to squirrel away the argument passed to shakeProgress, then retrieve it later when running the rules. Running the above code I see:
[1/5] test5
[2/5] test4
[3/5] test3
[4/5] test2
[5/5] test1
Running at a higher level of parallelism gives less precise results - initially there are only 3 items in todo (Shake increments countTodo as it finds items todo, and spawns items as soon as it knows about any of them), and there are often two rules running at the same index (there is no information about how many are in progress). Given knowledge of your specific rules, you could refine the output, e.g. storing an IORef you increment to ensure the index was monotonic.
The reason this code is somewhat convoluted is that the Progress information was intended to be used for asynchronous progress messages, although your approach seems perfectly valid. It may be worth introducing a getProgress :: Action Progress function for synchronous progress messages.
I was learning how to use the State monad and I noticed some odd behavior in terms of the order of execution. Removing the distracting bits that involve using the actual state, say I have the following code:
import Control.Monad
import Control.Monad.State
import Debug.Trace
mainAction :: State Int ()
mainAction = do
traceM "Starting the main action"
forM [0..2] (\i -> do
traceM $ "i is " ++ show i
forM [0..2] (\j -> do
traceM $ "j is " ++ show j
someSubaction i j
)
)
Running runState mainAction 1 in ghci produces the following output:
j is 2
j is 1
j is 0
i is 2
j is 2
j is 1
j is 0
i is 1
j is 2
j is 1
j is 0
i is 0
Outside for loop
which seems like the reverse order of execution of what might be expected. I thought that maybe this is a quirk of forM and tried it with sequence which specifically states that it runs its computation sequentially from left to right like so:
mainAction :: State Int ()
mainAction = do
traceM "Outside for loop"
sequence $ map handleI [0..2]
return ()
where
handleI i = do
traceM $ "i is " ++ show i
sequence $ map (handleJ i) [0..2]
handleJ i j = do
traceM $ "j is " ++ show j
someSubaction i j
However, the sequence version produces the same output. What is the actual logic in terms of the order of execution that is happening here?
Haskell is lazy, which means things are not executed immediately. Things are executed whenever their result is needed – but no sooner. Sometimes code isn't executed at all if its result isn't needed.
If you stick a bunch of trace calls in a pure function, you will see this laziness happening. The first thing that is needed will be executed first, so that's the trace call you see first.
When something says "the computation is run from left to right" what it means is that the result will be the same as if the computation was run from left to right. What actually happens under the hood might be very different.
This is in fact why it's a bad idea to do I/O inside pure functions. As you have discovered, you get "weird" results because the execution order can be pretty much anything that produces the correct result.
Why is this a good idea? When the language doesn't enforce a specific execution order (such as the traditional "top to bottom" order seen in imperative languages) the compiler is free to do a tonne of optimisations, such as for example not executing some code at all because its result isn't needed.
I would recommend you to not think too much about execution order in Haskell. There should be no reason to. Leave that up to the compiler. Think instead about which values you want. Does the function give the correct value? Then it works, regardless of which order it executes things in.
I thought that maybe this is a quirk of forM and tried it with sequence which specifically states that it runs its computation sequentially from left to right like so: [...]
You need to learn to make the following, tricky distinction:
The order of evaluation
The order of effects (a.k.a. "actions")
What forM, sequence and similar functions promise is that the effects will be ordered from left to right. So for example, the following is guaranteed to print characters in the same order that they occur in the string:
putStrLn :: String -> IO ()
putStrLn str = forM_ str putChar >> putChar '\n'
But that doesn't mean that expressions are evaluated in this left-to-right order. The program has to evaluate enough of the expressions to figure out what the next action is, but that often does not require evaluating everything in every expression involved in earlier actions.
Your example uses the State monad, which bottoms out to pure code, so that accentuates the order issues. The only thing that a traversal functions such as forM promises in this case is that gets inside the actions mapped to the list elements will see the effect of puts for elements to their left in the list.
I have created a timer in haskell. The problem is, it always returns 0. I think this is because of laziness, but I do not know how to fix it.
import System.CPUTime
timeit::IO ()->IO (Float)
timeit io=do
start <-getCPUTime
action <-seq start io
end <-seq action getCPUTime
return $! (fromIntegral $ end-start)/(10**12)
As you can see, I have throw in seq and $! galor, but to no avail. What do I do?
Here in an example run:
*Main> timeit test
What is your name?
Haskell
Your name is Haskell.
0.0
Here is some code I got to work
import Data.Time.Clock
timeit::IO ()->IO NominalDiffTime
timeit doit=do
start <- getCurrentTime
doit
end <- getCurrentTime
return (diffUTCTime end start)
Now for some discussion-
From the comments it seemed that you wanted real time, not cpu time, so that is what I wrote.
The System.Time libary is deprecated, so I switched you to the Data.Time library.
NominalDiffTime holds the time in seconds....
NominalDiffTime ignores leap seconds! In the unlikely event that a leap second is added during the running of the program, a 10 second delay will show up as 11 seconds. I googled around on how to fix this, and although Data.Time does have a DiffTime type to account for this, there doesn't seem to be a simple way to generate a DiffTime from UTCTime. I think you may be able to use the Posix time libraries and get seconds from Jan 1, 1970.... Then take the diff, but this seems like to much hastle for a pretty rare bug. If you are writing software to safely land airplanes however, please dig a bit deeper and fix this problem. :)
I've used this before and it works well for approximations. Don't rely on it for very precise times, but it should be accurate to within a millisecond. I've never tested its accuracy, so use at your own risk
import System.CPUTime
timeit :: IO a -> IO (a, Integer)
timeit action = do
start <- getCPUTime
value <- action
end <- getCPUTime
return (value, end - start)
No explicit strictness required.
question
I want a program that will write a sequence like,
1
...
10000000
to a file. What's the simplest code one can write, and get decent performance? My intuition is that there is some lack-of-buffering problem. My C code runs at 100 MB/s, whereas by reference the Linux command line utility dd runs at 9 GB/s 3 GB/s (sorry for the imprecision, see comments -- I'm more interested in the big picture orders-of-magnitude though).
One would think this would be a solved problem by now ... i.e. any modern compiler would make it immediate to write such programs that perform reasonably well ...
C code
#include <stdio.h>
int main(int argc, char **argv) {
int len = 10000000;
for (int a = 1; a <= len; a++) {
printf ("%d\n", a);
}
return 0;
}
I'm compiling with clang -O3. A performance skeleton which calls putchar('\n') 8 times gets comparable performance.
Haskell code
A naiive Haskell implementation runs at 13 MiB/sec, compiling with ghc -O2 -optc-O3 -optc-ffast-math -fllvm -fforce-recomp -funbox-strict-fields. (I haven't recompiled my libraries with -fllvm, perhaps I need to do that.) Code:
import Control.Monad
main = forM [1..10000000 :: Int] $ \j -> putStrLn (show j)
My best stab with Haskell runs even slower, at 17 MiB/sec. The problem is I can't find a good way to convert Vector's into ByteString's (perhaps there's a solution using iteratees?).
import qualified Data.Vector.Unboxed as V
import Data.Vector.Unboxed (Vector, Unbox, (!))
writeVector :: (Unbox a, Show a) => Vector a -> IO ()
writeVector v = V.mapM_ (System.IO.putStrLn . show) v
main = writeVector (V.generate 10000000 id)
It seems that writing ByteString's is fast, as demonstrated by this code, writing an equivalent number of characters,
import Data.ByteString.Char8 as B
main = B.putStrLn (B.replicate 76000000 '\n')
This gets 1.3 GB/s, which isn't as fast as dd, but obviously much better.
Some completely unscientific benchmarking first:
All programmes have been compiled with the default optimisation level (-O3 for gcc, -O2 for GHC) and run with
time ./prog > outfile
As a baseline, the C programme took 1.07s to produce a ~76MB (78888897 bytes) file, roughly 70MB/s throughput.
The "naive" Haskell programme (forM [1 .. 10000000] $ \j -> putStrLn (show j)) took 8.64s, about 8.8MB/s.
The same with forM_ instead of forM took 5.64s, about 13.5MB/s.
The ByteString version from dflemstr's answer took 9.13s, about 8.3MB/s.
The Text version from dflemstr's answer took 5.64s, about 13.5MB/s.
The Vector version from the question took 5.54s, about 13.7MB/s.
main = mapM_ (C.putStrLn . C.pack . show) $ [1 :: Int .. 10000000], where C is Data.ByteString.Char8, took 4.25s, about 17.9MB/s.
putStr . unlines . map show $ [1 :: Int .. 10000000] took 3.06s, about 24.8MB/s.
A manual loop,
main = putStr $ go 1
where
go :: Int -> String
go i
| i > 10000000 = ""
| otherwise = shows i . showChar '\n' $ go (i+1)
took 2.32s, about 32.75MB/s.
main = putStrLn $ replicate 78888896 'a' took 1.15s, about 66MB/s.
main = C.putStrLn $ C.replicate 78888896 'a' where C is Data.ByteString.Char8, took 0.143s, about 530MB/s, roughly the same figures for lazy ByteStrings.
What can we learn from that?
First, don't use forM or mapM unless you really want to collect the results. Performancewise, that sucks.
Then, ByteString output can be very fast (10.), but if the construction of the ByteString to output is slow (3.), you end up with slower code than the naive String output.
What's so terrible about 3.? Well, all the involved Strings are very short. So you get a list of
Chunk "1234567" Empty
and between any two such, a Chunk "\n" Empty is put, then the resulting list is concatenated, which means all these Emptys are tossed away when a ... (Chunk "1234567" (Chunk "\n" (Chunk "1234568" (...)))) is built. That's a lot of wasteful construct-deconstruct-reconstruct going on. Speed comparable to that of the Text and the fixed "naive" String version can be achieved by packing to strict ByteStrings and using fromChunks (and Data.List.intersperse for the newlines). Better performance, slightly better than 6., can be obtained by eliminating the costly singletons. If you glue the newlines to the Strings, using \k -> shows k "\n" instead of show, the concatenation has to deal with half as many slightly longer ByteStrings, which pays off.
I'm not familiar enough with the internals of either text or vector to offer more than a semi-educated guess concerning the reasons for the observed performance, so I'll leave them out. Suffice it to say that the performance gain is marginal at best compared to the fixed naive String version.
Now, 6. shows that ByteString output is faster than String output, enough that in this case the additional work of packing is more than compensated. However, don't be fooled by that to believe that is always so. If the Strings to pack are long, the packing can take more time than the String output.
But ten million invocations of putStrLn, be it the String or the ByteString version, take a lot of time. It's faster to grab the stdout Handle just once and construct the output String in non-IO code. unlines already does well, but we still suffer from the construction of the list map show [1 .. 10^7]. Unfortunately, the compiler didn't manage to eliminate that (but it eliminated [1 .. 10^7], that's already pretty good). So let's do it ourselves, leading to 8. That's not too terrible, but still takes more than twice as long as the C programme.
One can make a faster Haskell programme by going low-level and directly filling ByteStrings without going through String via show, but I don't know if the C speed is reachable. Anyway, that low-level code isn't very pretty, so I'll spare you what I have, but sometimes one has to get one's hands dirty if speed matters.
Using lazy byte strings gives you some buffering, because the string will be written instantly and more numbers will only be produced as they are needed. This code shows the basic idea (there might be some optimizations that could be made):
import qualified Data.ByteString.Lazy.Char8 as ByteString
main =
ByteString.putStrLn .
ByteString.intercalate (ByteString.singleton '\n') .
map (ByteString.pack . show) $
([1..10000000] :: [Int])
I still use Strings for the numbers here, which leads to horrible slowdowns. If we switch to the text library instead of the bytestring library, we get access to "native" show functions for ints, and can do this:
import Data.Monoid
import Data.List
import Data.Text.Lazy.IO as Text
import Data.Text.Lazy.Builder as Text
import Data.Text.Lazy.Builder.Int as Text
main :: IO ()
main =
Text.putStrLn .
Text.toLazyText .
mconcat .
intersperse (Text.singleton '\n') .
map Text.decimal $
([1..10000000] :: [Int])
I don't know how you are measuring the "speed" of these programs (with the pv tool?) but I imagine that one of these procedures will be the fastest trivial program you can get.
If you are going for maximum performance, then it helps to take a holistic view; i.e., you want to write a function that maps from [Int] to series of system calls that write chunks of memory to a file.
Lazy bytestrings are good representation for a sequence of chunks of memory. Mapping a lazy bytestring to a series of systems calls that write chunks of memory is what L.hPut is doing (assuming an import qualified Data.ByteString.Lazy as L). Hence, we just need a means to efficiently construct the corresponding lazy bytestring. This is what lazy bytestring builders are good at. With the new bytestring builder (here is the API documentation), the following code does the job.
import qualified Data.ByteString.Lazy as L
import Data.ByteString.Lazy.Builder (toLazyByteString, charUtf8)
import Data.ByteString.Lazy.Builder.ASCII (intDec)
import Data.Foldable (foldMap)
import Data.Monoid (mappend)
import System.IO (openFile, IOMode(..))
main :: IO ()
main = do
h <- openFile "/dev/null" WriteMode
L.hPut h $ toLazyByteString $
foldMap ((charUtf8 '\n' `mappend`) . intDec) [1..10000000]
Note that I output to /dev/null to avoid interference by the disk driver. The effort of moving the data to the OS remains the same. On my machine, the above code runs in 0.45 seconds, which is 12 times faster than the 5.4 seconds of your original code. This implies a throughput of 168 MB/s. We can squeeze out an additional 30% speed (220 MB/s) using bounded encodings].
import qualified Data.ByteString.Lazy.Builder.BasicEncoding as E
L.hPut h $ toLazyByteString $
E.encodeListWithB
((\x -> (x, '\n')) E.>$< E.intDec `E.pairB` E.charUtf8)
[1..10000000]
Their syntax looks a bit quirky because a BoundedEncoding a specifies the conversion of a Haskell value of type a to a bounded-length sequence of bytes such that the bound can be computed at compile-time. This allows functions such as E.encodeListWithB to perform some additional optimizations for implementing the actual filling of the buffer. See the the documentation of Data.ByteString.Lazy.Builder.BasicEncoding in the above link to the API documentation (phew, stupid hyperlink limit for new users) for more information.
Here is the source of all my benchmarks.
The conclusion is that we can get very good performance from a declarative solution provided that we understand the cost model of our implementation and use the right datastructures. Whenever constructing a packed sequence of values (e.g., a sequence of bytes represented as a bytestring), then the right datastructure to use is a bytestring Builder.
I have used the libraries criterion and cmdargs.
When I compile the program completely without cmdargs and run it e.g. ./prog --help then I get some unwanted response from criterion about the possible options and the number of runs etc..
When I compile and run it as below the command line options are first picked up by my code then then read by criterion. Criterion then subsequently reports and error telling me that the option --byte is unknown. I have not seen anything in the criterion documentation how this could be switched off or worked around. Is there a way to clear out the command line options ofter I have read them? Otherwise I would need to use e.g. CPUTime instead of criterion, that is OK to me since I do to really require the loads of extra functionality and data that criterion delivers.
{-# LANGUAGE OverloadedStrings #-}
{-# LANGUAGE DeriveDataTypeable #-}
import System.Console.CmdArgs
data Strlen = Strlen {byte :: Int} deriving (Data, Typeable, Show)
strlen = cmdArgsMode $ Strlen {byte = def} &= summary "MessagePack benchmark v0.04"
main = do
n <- cmdArgsRun strlen
let datastring = take (byte n) $ randomRs ('a','z') (mkStdGen 3)
putStrLn "Starting..."
conn <- connect "192.168.35.62" 8081
defaultMain [bench "sendReceive" $ whnfIO (mywl conn datastring)]
Use System.Environment.withArgs. Parse the command line arguments first with cmdArgs, then pass what you haven't used to criterion:
main = do
(flags, remaining) <- parseArgsHowever
act according to flags
withArgs remaining $
defaultMain [ ... ]
Take a look at the criterion source. You should be able to write your own defaultMainWith function that handles args however you want, including ignoring them, or ignoring unknown args, or etc...