How to get the Executable name of a window

Issue

I try to get the name of executable name of all of my launched windows and my problem is that:

I use the method

UINT GetWindowModuleFileName(      
HWND hwnd,
LPTSTR lpszFileName,
UINT cchFileNameMax);

And I don’t understand why it doesn’t work.

Data which I have about a window are:
-HWND AND PROCESSID

The error is:
e.g:

HWND: 00170628 
ProcessId: 2336        
WindowTitle: C:\test.cpp - Notepad++
GetWindowModuleFileName():  C:\test.exe

HWND: 00172138 
ProcessId: 2543        
WindowTitle: Firefox
GetWindowModuleFileName():  C:\test.exe

HWND: 00120358 
ProcessId: 2436        
WindowTitle: Mozilla Thunderbird
GetWindowModuleFileName():  C:\test.exe

Note: test.exe is the name of my executable file, but it is not the fullpath of Notepad++… and it make this for Mozilla Thunderbird too… I don’t understand why

I use the function like this:

char filenameBuffer[4000];
if (GetWindowModuleFileName(hWnd, filenameBuffer, 4000) > 0)
{
    std::cout << "GetWindowModuleFileName(): " << filenameBuffer << std::endl;
}

Thank you for your response.

Solution

The GetWindowModuleFileName function works for windows in the current process only.

You have to do the following:

  1. Retrieve the window’s process with GetWindowThreadProcessId.
  2. Open the process with PROCESS_QUERY_INFORMATION and PROCESS_VM_READ access rights using OpenProcess.
  3. Use GetModuleFileNameEx on the process handle.

If you really want to obtain the name of the module with which the window is registered (as opposed to the process executable), you can obtain the module handle with GetWindowLongPtr with GWLP_HINSTANCE. The module handle can then be passed to the aforementioned GetModuleFileNameEx.

Example:

TCHAR buffer[MAX_PATH] = {0};
DWORD dwProcId = 0; 

GetWindowThreadProcessId(hWnd, &dwProcId);   

HANDLE hProc = OpenProcess(PROCESS_QUERY_INFORMATION | PROCESS_VM_READ , FALSE, dwProcId);    
GetModuleFileName((HMODULE)hProc, buffer, MAX_PATH);
CloseHandle(hProc);

Answered By – avakar

This Answer collected from stackoverflow, is licensed under cc by-sa 2.5 , cc by-sa 3.0 and cc by-sa 4.0

Leave a Reply

(*) Required, Your email will not be published