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Magnus Therning: State machine with conduit

Planet Haskell - Mon, 01/12/2015 - 6:00pm

Previously I wrote about a simple state machine in Haskell and then offered a simple example, an adder. As I wrote in the initial post I’ve been using this to implement communication protocols. That means IO, and for that I’ve been using conduit. In this post I thought I’d show how easy it is to wrap up the adder in conduit, read input from stdin and produce the result on stdout.

At the top level is a function composing three parts:

  1. input from stdin
  2. processing in the adder
  3. output on stdout

This can be written as this

process :: ResourceT IO () process = source $= calc $$ output where source = stdinC output = stdoutC calc = concatMapC unpack =$= wrapMachine calcMachine =$= filterC f =$= mapC (pack . show) where f (CalcResult _) = True f _ = False

The input to the adder is ordinary Chars so the ByteStrings generated by stdinC need to be converted before being handed off to the state machine, which is wrapped to make it a Conduit (see below). The state machine is generating more signals that I want printed, so I filter out everything but the CalcResults. Then it’s passed to show and turned into a ByteString before sent to stdoutC.

I think the implementation of wrapMachine shows off the simplicity of conduit rather well:

wrapMachine :: (Monad m) => Machine state event signal -> Conduit event m signal wrapMachine mach = do maybeEvent <- await case maybeEvent of Nothing -> return () Just event -> let (newMach, signals) = stepMachine mach event in do yield signals wrapMachine newMach

The main function is as short as this:

main :: IO () main = do hSetBuffering stdin NoBuffering hSetBuffering stdout NoBuffering runResourceT process

The result is about as exciting as expected

% runhaskell Calculator.hs 56 42 +CalcResult 98 17 56 +CalcResult 73

It never exits on its own so one has to use Ctrl-C to terminate. I thought I’d try to address that in the next post by adding a time-out to the adder machine.

Categories: Offsite Blogs - Mon, 01/12/2015 - 5:50pm
Categories: Offsite Blogs

goldfirere/units · GitHub - Mon, 01/12/2015 - 3:08pm
Categories: Offsite Blogs

goldfirere/units · GitHub - Mon, 01/12/2015 - 3:08pm
Categories: Offsite Blogs

JP Moresmau: EclipseFP 2.6.3 released

Planet Haskell - Mon, 01/12/2015 - 1:44pm
A new version of EclipseFP, the Eclipse plugins for Haskell development, has been released. The release notes can be found here.

As usual install or update from Eclipse by using the update site

Happy Haskell Hacking!
Categories: Offsite Blogs

Trying to compile a Haskell program statically, but ghc can't find ffi

Haskell on Reddit - Mon, 01/12/2015 - 11:39am

I'm running Fedora 20 trying to statically compile a very minimal Haskell program (I want to make an executable suitable for cgi-bin). The command I'm using is

ghc -O2 --make -static -optc-static -optl-static Test.hs

It exits with this error however:

Linking Test ... /usr/bin/ld: cannot find -lffi /usr/lib64/ghc-7.6.3/libHSrts.a(Linker.o): In function `addDLL': (.text+0x1a19): warning: Using 'dlopen' in statically linked applications requires at runtime the shared libraries from the glibc version used for linking collect2: error: ld returned exit status 1

/usr/bin/ld: cannot find -lffi seems to me to be the failure inducing error (the other message is a warning). So I think I need the to install the static version of the foreign function interface library, but I can't find it in the Fedora repo.

Has anybody else had this kind of problem before or know what is going on?

submitted by dissonantloos
[link] [5 comments]
Categories: Incoming News

why can it never go smoothly,language-c-0.4.7 edition

haskell-cafe - Mon, 01/12/2015 - 8:00am
Hi, setup-Simple-Cabal- The program 'happy' is required but it could not be found. so i installed happy. then it wanted alex. which i was also able to install. and then language-c-0.4.7 successfully installed. I have seen this sort of thing many times across all sorts of different packages. So my question is : are these dependency bugs that should be reported to the package maintainer ? or is that certain packages can't be built as a dependency target for some reason ? Thanks, Brian
Categories: Offsite Discussion

Asking for feedback on a small library - Constructing Transformer stacks and Interfaces at compile time

Haskell on Reddit - Mon, 01/12/2015 - 7:09am


in my quest on making programs extensible, but also hide implementation details between plug-ins or components I wrote a small library which uses Template Haskell to construct a Transformer stack at compile time. Transformers can be user defined newtypes which hide the underlying Reader, State or whatever implementation. For those newtyped Transformers, interface function can be defined. Using those functions, classes can be automatically generated, so that lifting is not necessary anymore.

The library can be found here. If the description above sounds too strange, please take a look at the example.

Ninja edit: Didn't submit to Hackage yet, because I wanted to hear first if people find it worthy. Thanks!

submitted by jrk-
[link] [3 comments]
Categories: Incoming News

Antti-Juhani Kaijanaho (ibid): Planet Haskell email is broken (UPDATE: fixed)

Planet Haskell - Mon, 01/12/2015 - 5:54am

I became aware about a week ago that Planet Haskell’s email address had not received any traffic for a while. It turns out the email servers are misconfigured. The issue remains unfixed as of this writing. Update: this issue has been fixed.

Please direct your Planet Haskell related mail directly to me ( for the duration of the problem.

Categories: Offsite Blogs


libraries list - Mon, 01/12/2015 - 5:42am
Discussion period: one month When handling sensitive information (like a user's password) it is desirable to only keep the data around for as short a time as possible. Specifically, relying on the garbage collector to clean it up is simply not good enough. I therefore propose that the following function to be added to the Data.ByteString.Unsafe module: -- | Overwrites the contents of a ByteString with \0 bytes. unsafeWipe :: ByteString -> IO () unsafeWipe bs = BS.unsafeUseAsCStringLen bs $ \(ptr, len) -> let go i | i < 0 = return () | otherwise = pokeElemOff ptr i 0 >> go (i - 1) in go (len - 1) It is added to the Unsafe module because it break referential transparency but since ByteStrings are always kept in pinned memory, it should not otherwise be considered unsafe. It could be used as follows: main = do passwd <- getPassword doSomethingWith passwd unsafeWipe passwd restOfProgram
Categories: Offsite Discussion

Equality Constraints (a ~ b)

glasgow-user - Mon, 01/12/2015 - 1:04am
Hi, I am trying to find more background on these. They don’t exist in the Haskell 2010 Language Report, they didn’t exist in ghc 6.8.2 but make an appearance in 7.0.1. The documentation in the manual is rather sparse and doesn’t contain a reference: section 7.11. Folk on #ghc referred me to I can find papers that refer to ~ in F_C (aka FC?) but as far as I can tell not in the Haskell language itself. Many thanks Dominic Steinitz dominic< at >
Categories: Offsite Discussion

Keegan McAllister: Continuations in C++ with fork

Planet Haskell - Mon, 01/12/2015 - 12:18am

[Update, Jan 2015: I've translated this code into Rust.]

While reading "Continuations in C" I came across an intriguing idea:

It is possible to simulate call/cc, or something like it, on Unix systems with system calls like fork() that literally duplicate the running process.

The author sets this idea aside, and instead discusses some code that uses setjmp/longjmp and stack copying. And there are several other continuation-like constructs available for C, such as POSIX getcontext. But the idea of implementing call/cc with fork stuck with me, if only for its amusement value. I'd seen fork used for computing with probability distributions, but I couldn't find an implementation of call/cc itself. So I decided to give it a shot, using my favorite esolang, C++.

Continuations are a famously mind-bending idea, and this article doesn't totally explain what they are or what they're good for. If you aren't familiar with continuations, you might catch on from the examples, or you might want to consult another source first (1, 2, 3, 4, 5, 6).

Small examples

I'll get to the implementation later, but right now let's see what these fork-based continuations can do. The interface looks like this.

template <typename T>
class cont {
void operator()(const T &x);

template <typename T>
T call_cc( std::function< T (cont<T>) > f );

std::function is a wrapper that can hold function-like values, such as function objects or C-style function pointers. So call_cc<T> will accept any function-like value that takes an argument of type cont<T> and returns a value of type T. This wrapper is the first of several C++11 features we'll use.

call_cc stands for "call with current continuation", and that's exactly what it does. call_cc(f) will call f, and return whatever f returns. The interesting part is that it passes to f an instance of our cont class, which represents all the stuff that's going to happen in the program after f returns. That cont object overloads operator() and so can be called like a function. If it's called with some argument x, the program behaves as though f had returned x.

The types reflect this usage. The type parameter T in cont<T> is the return type of the function passed to call_cc. It's also the type of values accepted by cont<T>::operator().

Here's a small example.

int f(cont<int> k) {
std::cout << "f called" << std::endl;
std::cout << "k returns" << std::endl;
return 0;

int main() {
std::cout << "f returns " << call_cc<int>(f) << std::endl;

When we run this code we get:

f called
f returns 1

We don't see the "k returns" message. Instead, calling k(1) bails out of f early, and forces it to return 1. This would happen even if we passed k to some deeply nested function call, and invoked it there.

This nonlocal return is kind of like throwing an exception, and is not that surprising. More exciting things happen if a continuation outlives the function call it came from.

boost::optional< cont<int> > global_k;

int g(cont<int> k) {
std::cout << "g called" << std::endl;
global_k = k;
return 0;

int main() {
std::cout << "g returns " << call_cc<int>(g) << std::endl;

if (global_k)

When we run this, we get:

g called
g returns 0
g returns 1

g is called once, and returns twice! When called, g saves the current continuation in a global variable. After g returns, main calls that continuation, and g returns again with a different value.

What value should global_k have before g is called? There's no such thing as a "default" or "uninitialized" cont<T>. We solve this problem by wrapping it with boost::optional. We use the resulting object much like a pointer, checking for "null" and then dereferencing. The difference is that boost::optional manages storage for the underlying value, if any.

Why isn't this code an infinite loop? Because invoking a cont<T> also resets global state to the values it had when the continuation was captured. The second time g returns, global_k has been reset to the "null" optional value. This is unlike Scheme's call/cc and most other continuation systems. It turns out to be a serious limitation, though it's sometimes convenient. The reason for this behavior is that invoking a continuation is implemented as a transfer of control to another process. More on that later.


We can use continuations to implement backtracking, as found in logic programming languages. Here is a suitable interface.

bool guess();
void fail();

We will use guess as though it has a magical ability to predict the future. We assume it will only return true if doing so results in a program that never calls fail. Here is the implementation.

boost::optional< cont<bool> > checkpoint;

bool guess() {
return call_cc<bool>( [](cont<bool> k) {
checkpoint = k;
return true;
} );

void fail() {
if (checkpoint) {
} else {
std::cerr << "Nothing to be done." << std::endl;

guess invokes call_cc on a lambda expression, which saves the current continuation and returns true. A subsequent call to fail will invoke this continuation, retrying execution in a world where guess had returned false instead. In Scheme et al, we would store a whole stack of continuations. But invoking our cont<bool> resets global state, including the checkpoint variable itself, so we only need to explicitly track the most recent continuation.

Now we can implement the integer factoring example from "Continuations in C".

int integer(int m, int n) {
for (int i=m; i<=n; i++) {
if (guess())
return i;

void factor(int n) {
const int i = integer(2, 100);
const int j = integer(2, 100);

if (i*j != n)

std::cout << i << " * " << j << " = " << n << std::endl;

factor(n) will guess two integers, and fail if their product is not n. Calling factor(391) will produce the output

17 * 23 = 391

after a moment's delay. In fact, you might see this after your shell prompt has returned, because the output is produced by a thousand-generation descendant of the process your shell created.

Solving a maze

For a more substantial use of backtracking, let's solve a maze.

const int maze_size = 15;
char maze[] =
" | |\n"
"|--+ | | | |\n"
"| | | | --+ |\n"
"| | | |\n"
"|-+---+--+- | |\n"
"| | | |\n"
"| | | ---+-+- |\n"
"| | | |\n"
"| +-+-+--| |\n"
"| | | |--- |\n"
"| | |\n"
"|--- -+-------|\n"
"| \n"
"+------------- \n";

void solve_maze() {
int x=0, y=0;

while ((x != maze_size-1)
|| (y != maze_size-1)) {

if (guess()) x++;
else if (guess()) x--;
else if (guess()) y++;
else y--;

if ( (x < 0) || (x >= maze_size) ||
(y < 0) || (y >= maze_size) )

const int i = y*(maze_size+1) + x;
if (maze[i] != ' ')
maze[i] = 'X';

for (char c : maze) {
if (c == 'X')
std::cout << "\e[1;32mX\e[0m";
std::cout << c;

Whether code or prose, the algorithm is pretty simple. Start at the upper-left corner. As long as we haven't reached the lower-right corner, guess a direction to move. Fail if we go off the edge, run into a wall, or find ourselves on a square we already visited.

Once we've reached the goal, we iterate over the char array and print it out with some rad ANSI color codes.

Once again, we're making good use of the fact that our continuations reset global state. That's why we see 'X' marks not on the failed detours, but only on a successful path through the maze. Here's what it looks like.

|--+ |X| | |
| | |X| --+ |
| |XXXXX| |
|-+---+--+-X| |
| |XXX | XXX|
| |X|X---+-+-X|
|X+-+-+--|XXX |
|X| | |--- |
|XXXX | |

Excess backtracking

We can run both examples in a single program.

int main() {

If we change the maze to be unsolvable, we'll get:

17 * 23 = 391
23 * 17 = 391
Nothing to be done.

Factoring 391 a different way won't change the maze layout, but the program doesn't know that. We can add a cut primitive to eliminate unwanted backtracking.

void cut() {
checkpoint = boost::none;

int main() {

The implementation

For such a crazy idea, the code to implement call_cc with fork is actually pretty reasonable. Here's the core of it.

template <typename T>
// static
T cont<T>::call_cc(call_cc_arg f) {
int fd[2];
int read_fd = fd[0];
int write_fd = fd[1];

if (fork()) {
// parent
return f( cont<T>(write_fd) );
} else {
// child
char buf[sizeof(T)];
if (read(read_fd, buf, sizeof(T)) < ssize_t(sizeof(T)))
return *reinterpret_cast<T*>(buf);

template <typename T>
void cont<T>::impl::invoke(const T &x) {
write(m_pipe, &x, sizeof(T));

To capture a continuation, we fork the process. The resulting processes share a pipe which was created before the fork. The parent process will call f immediately, passing a cont<T> object that holds onto the write end of this pipe. If that continuation is invoked with some argument x, the parent process will send x down the pipe and then exit. The child process wakes up from its read call, and returns x from call_cc.

There are a few more implementation details.

  • If the parent process exits, it will close the write end of the pipe, and the child's read will return 0, i.e. end-of-file. This prevents a buildup of unused continuation processes. But what if the parent deletes the last copy of some cont<T>, yet keeps running? We'd like to kill the corresponding child process immediately.

    This sounds like a use for a reference-counted smart pointer, but we want to hide this detail from the user. So we split off a private implementation class, cont<T>::impl, with a destructor that calls close. The user-facing class cont<T> holds a std::shared_ptr to a cont<T>::impl. And cont<T>::operator() simply calls cont<T>::impl::invoke through this pointer.

  • It would be nice to tell the compiler that cont<T>::operator() won't return, to avoid warnings like "control reaches end of non-void function". GCC provides the noreturn attribute for this purpose.

  • We want the cont<T> constructor to be private, so we had to make call_cc a static member function of that class. But the examples above use a free function call_cc<T>. It's easiest to implement the latter as a 1-line function that calls the former. The alternative is to make it a friend function of cont<T>, which requires some forward declarations and other noise.

There are a number of limitations too.

  • As noted, the forked child process doesn't see changes to the parent's global state. This precludes some interesting uses of continuations, like implementing coroutines. In fact, I had trouble coming up with any application other than backtracking. You could work around this limitation with shared memory, but it seemed like too much hassle.

  • Each captured continuation can only be invoked once. This is easiest to observe if the code using continuations also invokes fork directly. It could possibly be fixed with additional forking inside call_cc.

  • Calling a continuation sends the argument through a pipe using a naive byte-for-byte copy. So the argument needs to be Plain Old Data, and had better not contain pointers to anything not shared by the two processes. This means we can't send continuations through other continuations, sad to say.

  • I left out the error handling you would expect in serious code, because this is anything but.

  • Likewise, I'm assuming that a single write and read will suffice to send the value. Robust code will need to loop until completion, handle EINTR, etc. Or use some higher-level IPC mechanism.

  • At some size, stack-allocating the receive buffer will become a problem.

  • It's slow. Well, actually, I'm impressed with the speed of fork on Linux. My machine solves both backtracking problems in about a second, forking about 2000 processes along the way. You can speed it up more with static linking. But it's still far more overhead than the alternatives.

As usual, you can get the code from GitHub.

Categories: Offsite Blogs

FLTK bindings near completion. Need communityassistance ...

haskell-cafe - Sun, 01/11/2015 - 8:39pm
Hi all, I've been working on bindings [1] to the FLTK GUI toolkit [2] and I've come to the point where it's mostly complete but I could use some help from the community before I put it up on Hackage. My hope is to eventually fill the need in the Haskell ecosystem for a native GUI library that: - runs on all platforms - is easily installable - is small and fast - has a minimum of dependencies. Currently this library uses only base, stm, bytestring and c2hs. - is conservative with extensions. Currently the most recent extension required is GADTs. If any of this sounds good to you, I could use some help with: - testing the build on Linux and OSX. This should be as simple as installing FLTK, cloning the repo [1] and running `cabal build`. Better instructions are in included README. - getting it to build on Windows. I unfortunately don't have access to a Windows machine. - getting it to build on older GHC's. I used 7.8.3 because that is what I have, but it should work back to the first version of GHC that intro
Categories: Offsite Discussion