Question

我一直在想用haskell创建midori插件会很好，但似乎几乎不可能。问题在于通过ffi导出haskell函数，因为ghc编译器使用了大量的-u开关。

有没有人见过haskell在类似的环境中使用，而不必为ghc替换gcc？如果是这样，它是如何运作的，以及它们经历了什么环节？

编辑：请求了一些示例，所以在这里：

export.hs

{-# LANGUAGE ForeignFunctionInterface #-}
module Export where
import Foreign.C
import Foreign.C.Types

foo :: IO Int
foo = return 2

foreign export ccall foo :: IO Int

test.c（ifdefs snipped）

#include <stdio.h>
#include "HsFFI.h"
#include "export_stub.h"

extern void __stginit_Export(void);

int main(int argc, char **argv)
{
    int i;
    hs_init(&argc, &argv);
    hs_add_root(__stginit_Export);
    i = foo();
    printf("%d\n", i);
    hs_exit();
    return 0;
}

使用ghc --make -no-hs-main export.hs test.c进行编译会创建一个a.out 可执行的工作。 ghc使用以下命令进行链接：

collect2 --build-id --eh-frame-hdr -m elf_i386 --hash-style=both -dynamic-linker /lib/ld-linux.so.2 -o a.out -z relro -u ghczmprim_GHCziTypes_Izh_static_info -u ghczmprim_GHCziTypes_Czh_static_info -u ghczmprim_GHCziTypes_Fzh_static_info -u ghczmprim_GHCziTypes_Dzh_static_info -u base_GHCziPtr_Ptr_static_info -u base_GHCziWord_Wzh_static_info -u base_GHCziInt_I8zh_static_info -u base_GHCziInt_I16zh_static_info -u base_GHCziInt_I32zh_static_info -u base_GHCziInt_I64zh_static_info -u base_GHCziWord_W8zh_static_info -u base_GHCziWord_W16zh_static_info -u base_GHCziWord_W32zh_static_info -u base_GHCziWord_W64zh_static_info -u base_GHCziStable_StablePtr_static_info -u ghczmprim_GHCziTypes_Izh_con_info -u ghczmprim_GHCziTypes_Czh_con_info -u ghczmprim_GHCziTypes_Fzh_con_info -u ghczmprim_GHCziTypes_Dzh_con_info -u base_GHCziPtr_Ptr_con_info -u base_GHCziPtr_FunPtr_con_info -u base_GHCziStable_StablePtr_con_info -u ghczmprim_GHCziBool_False_closure -u ghczmprim_GHCziBool_True_closure -u base_GHCziPack_unpackCString_closure -u base_GHCziIOziException_stackOverflow_closure -u base_GHCziIOziException_heapOverflow_closure -u base_ControlziExceptionziBase_nonTermination_closure -u base_GHCziIOziException_blockedIndefinitelyOnMVar_closure -u base_GHCziIOziException_blockedIndefinitelyOnSTM_closure -u base_ControlziExceptionziBase_nestedAtomically_closure -u base_GHCziWeak_runFinalizzerBatch_closure -u base_GHCziTopHandler_runIO_closure -u base_GHCziTopHandler_runNonIO_closure -u base_GHCziConc_ensureIOManagerIsRunning_closure -u base_GHCziConc_runSparks_closure -u base_GHCziConc_runHandlers_closure /usr/lib/gcc/i486-linux-gnu/4.4.3/../../../../lib/crt1.o /usr/lib/gcc/i486-linux-gnu/4.4.3/../../../../lib/crti.o /usr/lib/gcc/i486-linux-gnu/4.4.3/crtbegin.o -L/usr/lib/ghc-6.12.1/base-4.2.0.0 -L/usr/lib/ghc-6.12.1/integer-gmp-0.2.0.0 -L/usr/lib/ghc-6.12.1/ghc-prim-0.2.0.0 -L/usr/lib/ghc-6.12.1 -L/usr/lib/gcc/i486-linux-gnu/4.4.3 -L/usr/lib/gcc/i486-linux-gnu/4.4.3 -L/usr/lib/gcc/i486-linux-gnu/4.4.3/../../../../lib -L/lib/../lib -L/usr/lib/../lib -L/usr/lib/gcc/i486-linux-gnu/4.4.3/../../.. -L/usr/lib/i486-linux-gnu export.o export_stub.o test.o -lHSbase-4.2.0.0 -lHSinteger-gmp-0.2.0.0 -lgmp -lHSghc-prim-0.2.0.0 -lHSrts -lm -lffi -lrt -ldl -lgcc --as-needed -lgcc_s --no-as-needed -lc -lgcc --as-needed -lgcc_s --no-as-needed /usr/lib/gcc/i486-linux-gnu/4.4.3/crtend.o /usr/lib/gcc/i486-linux-gnu/4.4.3/../../../../lib/crtn.o

删除-u开关（只留下-l，-L和一些额外的标志）上一个命令不编译并返回（大约50个左右）线）

/usr/lib/ghc-6.12.1/libHSrts.a(RtsAPI.o): In function `rts_mkFunPtr':
(.text+0x5a9): undefined reference to `base_GHCziPtr_FunPtr_con_info'
/usr/lib/ghc-6.12.1/libHSrts.a(RtsAPI.o): In function `rts_mkString':
(.text+0x60f): undefined reference to `base_GHCziPack_unpackCString_closure'

Answer 1

我能够解决问题。

我正在使用以下文件：

的main.c

#include <stdio.h>
#include "lib_stub.h"
int main(int argc, char **argv)
{
    puts("foo");
    printf("%d\n", hsfun(5));
    return 0;
}

lib.hs

{-# LANGUAGE ForeignFunctionInterface #-}
module Test where
import Foreign.C.Types

hsfun :: CInt -> IO CInt
hsfun x = do
  putStrLn "Hello from haskell"
  return (42 * x)

foreign export ccall hsfun :: CInt -> IO CInt

module_init.c

#include <HsFFI.h>
extern void __stginit_Test(void);

static void library_init(void) __attribute__((constructor));
static void library_init(void)
{
    static char *argv[] = { "libtest.so", 0 }, **argv_ = argv;
    static int argc = 1;

    hs_init(&argc, &argv_);
    hs_add_root(__stginit_Test);
}

static void library_exit(void) __attribute__((destructor));
static void library_exit(void)
{
    hs_exit();
}

我正在用ghc --make -shared -dynamic -fPIC -o libtest.so lib.hs module_init.c -lHSrts-ghc6.12.1 -optl-Wl,-rpath,/usr/lib/ghc-6.12.1/ -L/usr/lib/ghc-6.12.1编译库可执行文件gcc -c main.c -I/usr/lib/ghc-6.12.1/include -L. -ltest -Wl,-rpath=$PWD

重要的是让库共享并拥有rts库链接也没有默认。 rpath使它成为可能在没有LD_LIBRARY_PATH的情况下运行。

http://weblog.haskell.cz/pivnik/building-a-shared-library-in-haskell/

http://www.well-typed.com/blog/30

Answer 2

这可能是手册的相关部分：

8.2. Using the FFI with GHC

特别参见8.2.1.2。您可以创建一个用Haskell编写的库，并可以从C代码中调用。然后你只需要在C中写一些胶水代码就可以变成插件或其他什么。但我自己没有这样做，请等待foreign export更有经验的用户回答。

连接haskell与c

2 个答案: