DLL export __stdcall without DEF file and decorated function names with VS

Question

If I have a DLL file...how can I see functions are built as __stdcall or __cdecl?
I want to create a DLL to export functions as __stdcall without DEF file and decorated function names with Visual Studio like:

__declspec(dllexport) int __stdcall TestFunktion(void)
{
  return 0;
}

The result looks like this:

Is there any easy way to create the functions without decoration without DEF file in Visual Studio?

It look good. But this is a bad idea to mix __stdcall and __cdecl. Thats is the point. The main programm expect __stdcall without decoration. — Aleksej, Sep 06 '21 at 10:28

Daniel Kleinstein · Accepted Answer · 2021-09-06T10:45:59.320

The standard approach to this problem is to define your export's undecorated name in a .def file - and this is the approach you should use.

But if for whatever reason you don't want to use a .def file, you can write code like:

__declspec(dllexport) int __stdcall TestFunktion(void)
{
    #pragma comment(linker, "/EXPORT:" __FUNCTION__ "=" __FUNCDNAME__)
    ...
}

__FUNCDNAME__ will evaluate to your decorated function name, while __FUNCTION__ will evaluate to your undecorated function name - this is effectively a replacement for the equivalent line in the .def file.

As for your other question:

If I have a DLL file...how can I see functions are built as __stdcall or __cdecl?

this is not strictly possible from the DLL alone, much less if you deliberately omit hints from the export table (i.e. you might be able to see a decorated export symbol and infer that it's __stdcall, but you're choosing to export them as undecorated symbols) - perhaps if you perform some sort of analysis on the function's disassembly and try to understand its calling convention, but this is non-hermetic and very hacky.

This is why well-documented calling conventions and well-documented API signatures are important.

DLL export __stdcall without DEF file and decorated function names with VS

1 Answers1