stdcall, @s, and ld/dlltool...

[Matthew Brett <matthew dot brett at mrc-cbu dot cam dot ac dot uk>]

Hi,

I am stuck with a linking problem, and think this may be to do with
dlltool/dllwrap.

I have a third party dll, with no source, and stdcall calling
conventions, which exports @ decorated functions, like

foo@16

I have lots of code that I cannot modify, that calls the routines in
the library.  In fact it is in Fortran, but the same problem occurs in
c, if you use the -mrtd switch to enforce stdcall calling conventions.

Here's a simple replication of the problem: I made a dll like this:

/* cdll.c */
__stdcall void dllf1 () 
{
     printf("in dllf1.\n");
     return;
}

and compiled it with:
gcc -c -o cdll.o cdll.c
dllwrap --export-all --output-def cdll.def -o cdll.dll cdll.o

This exports (pexports.exe)

LIBRARY cdll.dll
EXPORTS
dllf1@0

I can generate an input library thus:

dlltool --input-def cdll.def -l libcdll.a 

and then try and call it from a little test function:

#include <stdio.h>
void dllf1();

int main () { 
  dllf1();
  return 0;
}

gcc -c -mrtd -o usedll.o usedll.c 
gcc -o usedll.exe usedll.o libcdll.a

The last line won't link with error:

usedll.o(.text+0xc):usedll.c: undefined reference to `dllf1'

I guess this is because of the difference between the decorated and
undecorated names.  The addition of --enable-stdcall-fixup to the last
call to gcc does not remove the error.  If I edit the .def file to
give:

LIBRARY cdll.dll
EXPORTS
dllf1=dllf1@0

then I can run the program usedll.exe, but get an error telling me
that the procedure entry point dllf1 could not be located in the dll
cdll.dll.

I could solve all this by adding the __stdcall attribute to the dllf1
function definition in usedll.c, rather than using the -mrtd switch,
but a) I can't do this in g77 AFAIK, and b) I can't edit the code
anyway.

So, is there any way of linking a dll function exported as "foo@0",
from an object file that is expecting "foo", without having the code
from the dll?  Why doesn't --enable-stdcall-alias work? - it seems as
though it should from the documentation.

Sorry if I have missed something very obvious (I have that feeling...)

Many thanks,

Matthew