Check If Application Is Running

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Is there a way to run a loop that would check to see if an application is running? I have this so far. It's in a windows console program. I'm using Visual Studio 2010. Its saying define _AFXDLL.

C++

#include "stdafx.h"
#include <afx.h>
int main()
{
HANDLE hd;
LPCWSTR mu = (LPCWSTR)"Notepad";
hd = CreateMutex( NULL, FALSE, mu);
if ( GetLastError() == ERROR_ALREADY_EXISTS )
{
//MessageBox::Show("The program is already running.");
return 1;
}
CloseHandle( hd );

Posted 8-Sep-11 15:01pm

Member 7766180

Updated 8-Sep-11 20:20pm

Emilio Garavaglia

v6

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Sergey Alexandrovich Kryukov 8-Sep-11 21:27pm

Why loop?
--SA

Member 7766180 8-Sep-11 21:29pm

Ok no loop! Any suggestions?

Sergey Alexandrovich Kryukov 8-Sep-11 21:38pm

Sure. See the article I referenced. Good luck.
--SA

Emilio Garavaglia 9-Sep-11 2:20am

removed spurious html tags

2 solutions

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Arun Parthasarathy · Accepted Answer · 2011-09-08T20:35:00

Use this:

AM just hardcoded and comparing whether internet.exe is running!
like this you do it dynamically to determing any service running. also you may hav arguments passed from command line.

CSS

#include "afxwin.h"
# include <wtsapi32.h>
# include <iostream>

#pragma comment(lib,"wtsapi32.lib")

using namespace std;

DWORD pCount = 0; 
	PWTS_PROCESS_INFO ppProcessInfo,pProcess;
	WTSEnumerateProcesses(WTS_CURRENT_SERVER_HANDLE,0,1,&ppProcessInfo,&pCount);
	pProcess = ppProcessInfo;
        CString internet = "iexplore.exe";

	for(int i = 0; i < pCount; i++)
	{
		pProcess->pProcessName;
		if (internet.CompareNoCase ( pProcess->pProcessName) == 0)
		{
			cout<< "iexplore running";
		}
		pProcess++;
	}</iostream></wtsapi32.h>

Sergey Alexandrovich Kryukov · Accepted Answer · 2011-09-08T15:38:00

Solution 1

You can use WMI to watch any process and get notification on system events related to processes like creation and termination. See this CodeProject article: Process Information and Notifications using WMI[^].

—SA

Posted 8-Sep-11 15:38pm

Sergey Alexandrovich Kryukov

Comments

Member 7766180 9-Sep-11 1:02am

Thank You.

Sergey Alexandrovich Kryukov 9-Sep-11 13:04pm

You are welcome.
If you agree that it makes sense, please formally accept this answer (green button) -- thanks.
Good luck, call again.
--SA