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How DLL (Dynamic library) is linked during run time?

+2 votes

For an example:
I have a dll file liba.dll,
While compiling i am giving the path of the library file. It is compiling properly.

Now i have to load this binary/.exe file and DLL file to other system/device,
Then what about the DLL file?
How our program gets to know the path of DLL during run time?
I mean how DLL file will be linked?

posted Aug 26, 2015 by anonymous

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Is it a Linux case or Windows, in Linux we don't have dll we have .so, please clarify
In Linux....
Do i have to place .so file on same path
Is it compulsary that i have to pass .so file as command line argument?
Ex: ./a.out
Not necessary, it can be anywhere pointed out by the LD_LIBRARY_PATH.

Also no need to pass the .so in the command line.
Then how our exe/binary will get to know the path of .so file?

1 Answer

0 votes

GetModuleFileName() works fine from inside the DLL's codes. Just be sure NOT to set the first parameter to NULL, as that will get the filename of the calling process. You need to specify the DLL's actual module instance instead. You get that as an input parameter in the DLL's DllEntryPoint() function, just save it to a variable somewhere for later use when needed.

A complete example:

CStringW thisDllDirPath()
    CStringW thisPath = L"";
    WCHAR path[MAX_PATH];
    HMODULE hm;
                            (LPWSTR) &thisDllDirPath, &hm ) )
        GetModuleFileNameW( hm, path, sizeof(path) );
        PathRemoveFileSpecW( path );
        thisPath = CStringW( path );
        if( !thisPath.IsEmpty() && 
            thisPath.GetAt( thisPath.GetLength()-1 ) != '\\' ) 
            thisPath += L"\\";
    else if( _DEBUG ) std::wcout << L"GetModuleHandle Error: " << GetLastError() << std::endl;

    if( _DEBUG ) std::wcout << L"thisDllDirPath: [" << CStringW::PCXSTR( thisPath ) << L"]" << std::endl;       
    return thisPath;
answer Aug 27, 2015 by Amit Kumar Pandey
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