C / C++ / MFC

11-Dec-09 15:03

Hi,

I use the following code to create a Named Mutex, to ensure that only One Instance of an App is running on a Machine.

<br />
     CString AppName=GetAppNameRoot();<br />
     CString MutexName=AppName+"_60D44FC0_07ED_11D8_A55E_444553540000_";<br />
<br />
     HANDLE exclusion = CreateMutex(NULL, FALSE, MutexName);<br />
     DWORD err = ::GetLastError();<br />
     // Check exclusion and err, Get Out Gracefully if CreateMutex Failed<br />
     <br />

All worked fine as advertised. Then I placed the module containing the Above Code in a DLL, and CreateMutex never Fails, allowing Multiple copies to Run.

I Don't understand why, or, even How to Debug this.

Regards, Smile | :)

Bram van Kampen

«_Superman_»11-Dec-09 18:37

«_Superman_»

11-Dec-09 18:37

You're clearly doing something wrong.
It works perfectly even if the named mutex is inside a DLL.

«_Superman_»
I love work. It gives me something to do between weekends.

Microsoft MVP (Visual C++)

Rajesh R Subramanian11-Dec-09 22:29

11-Dec-09 22:29

Bram van Kampen wrote:
Then I placed the module containing the Above Code in a DLL, and CreateMutex never Fails

Well, the call won't fail (the return value won't be NULL). It rather returns a handle to the existing mutex object. However, if you make an immediate call to GetLastError() (which you should do), you'd see that the error code has been set to ERROR_ALREADY_EXISTS.

I've always thought this is kinda weird, but that's how it is.

“Follow your bliss.” – Joseph Campbell

krmed12-Dec-09 4:16

krmed

12-Dec-09 4:16

Calling GetLastError when the call did not fail is not recommended - the error code that you get will probably be from the last failed action.

You should only call GetLastError if the CreateMutex call failed - then it will give you a reliable error code.

Karl - WK5M
PP-ASEL-IA (N43CS)
PGP Key: 0xDB02E193
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Rajesh R Subramanian12-Dec-09 4:52

12-Dec-09 4:52

krmed wrote:
Calling GetLastError when the call did not fail is not recommended - the error code that you get will probably be from the last failed action.
You should only call GetLastError if the CreateMutex call failed - then it will give you a reliable error code.

That statement is wrong and misleading. Please read the documentation[^]:

Return Value

If the function succeeds, the return value is a handle to the newly created mutex object.

If the function fails, the return value is NULL. To get extended error information, call GetLastError.

If the mutex is a named mutex and the object existed before this function call, the return value is a handle to the existing object, GetLastError returns ERROR_ALREADY_EXISTS, bInitialOwner is ignored, and the calling thread is not granted ownership. However, if the caller has limited access rights, the function will fail with ERROR_ACCESS_DENIED and the caller should use the OpenMutex function.

I also mentioned in my original post that the call to GetLastError() must be made immediately (before anything else).

“Follow your bliss.” – Joseph Campbell

krmed12-Dec-09 5:14

krmed

12-Dec-09 5:14

Yes of course - I do read the documentation, however I overlooked that detail. Many users call GetLastError after calling other methods that don't set it unless there's an error.

Sorry about that.

Karl - WK5M
PP-ASEL-IA (N43CS)
PGP Key: 0xDB02E193
PGP Key Fingerprint: 8F06 5A2E 2735 892B 821C 871A 0411 94EA DB02 E193

Rajesh R Subramanian12-Dec-09 6:15

12-Dec-09 6:15

krmed wrote:
Many users call GetLastError after calling other methods that don't set it unless there's an error.

Yes, but there are those crafty API calls that will do stuff like this.

krmed wrote:
Sorry about that.

No problem. Smile | :)

“Follow your bliss.” – Joseph Campbell

Bram van Kampen12-Dec-09 15:46

12-Dec-09 15:46

Been There, (actualy, Is part of the function the details of which I spared the Details, and covered by a comment)
GetLastError() returns 0!

Regards Smile | :)

Bram van Kampen

Bram van Kampen12-Dec-09 16:54

12-Dec-09 16:54

Rajesh R Subramanian wrote:
Well, the call won't fail (the return value won't be NULL). It rather returns a handle to the existing mutex object. However, if you make an immediate call to GetLastError() (which you should do), you'd see that the error code has been set to ERROR_ALREADY_EXISTS.

Well, for you who needs an explanation, By 'Not Fail' I mean Did as Documented.

Why do you respond twice to this post!

Mad | :mad:

Bram van Kampen

Rajesh R Subramanian12-Dec-09 18:54

12-Dec-09 18:54

Bram van Kampen wrote:
Why do you respond twice to this post!

Who responded twice to any post? I see everyone has responded only once (except you have responded twice to my answer)!

I'd like to know if that's your vote on my post?

“Follow your bliss.” – Joseph Campbell

Bram van Kampen14-Dec-09 12:25

14-Dec-09 12:25

Hi,
Appologies,
My Fault!
Blush | :O

Bram van Kampen

Richard MacCutchan12-Dec-09 1:46

Richard MacCutchan

12-Dec-09 1:46

Bram van Kampen wrote:
CString AppName=GetAppNameRoot();

Are you sure this function always returns the same value, as you are using it as part of the mutex name?

Bram van Kampen12-Dec-09 16:08

12-Dec-09 16:08

Yes, It does!
It Calls GetModuleHandle(NULL)
ASSERTS (at this Stage that the ModuleHandle is 0x00400000)
Cals GetModuleFileName()
The Result is Stripped to remove up to the 'last '\', and the first '.' thereafter. This function is Also in the same DLL, I got it to print it's name to a DBG Text File,The printed name is always 'MyApp'.
When I trace into the DLL, the Name is Correct, and a New Mutex is created, regardless of the old one. When I move the file back to the App, it works a advertised.

Thanks Smile | :)

Bram van Kampen

Richard MacCutchan12-Dec-09 22:55

Richard MacCutchan

12-Dec-09 22:55

I noticed yesterday some remarks in the MSDN document about creating the MUTEX in local or global space. You may want to check that you are doing it according to those rules.

Migounette13-Dec-09 2:36

Migounette

13-Dec-09 2:36

Take care if you place it in a DLL you have to distinguish action done in the DLL:

BOOL WINAPI DllMain(
HINSTANCE hinstDLL, // handle to DLL module
DWORD fdwReason, // reason for calling function
LPVOID lpReserved // reserved
)
{
BOOL fInit = FALSE;
WORD wVersionRequested;
WSADATA wsaData;

// Register the log
switch( fdwReason )
{
case DLL_PROCESS_ATTACH:
...
case DLL_THREAD_ATTACH:
...
case DLL_THREAD_DETACH:
...
case DLL_PROCESS_DETACH:
...
};

What is the goal of your mutex ?
Prevent multiple loads of the DLL or block for security reasons ?

Tell me more...

Bram van Kampen13-Dec-09 21:34

Im just new in c++ [modified]

13-Dec-09 21:34

Migounette wrote:
What is the goal of your mutex ?
Prevent multiple loads of the DLL or block for security reasons ?

Tell me more...

The Goal is to ensure that at most one copy each of a suite of four apps is running per machine. A function 'OnceOnly()' is exported from the DLL. So, instead of the function 'OnceOnly()' being separately included in each of the Apps, it was moved to the DLL which contains the 'Common' Code for the Apps.

The Full Code for OnceOnly() is:-

<br />
DLLEXPORT void OnceOnly(){<br />
// Protect against Two Instances on the Same Computer!<br />
        CString AppName=GetAppRootName();<br />
        CString MutexName=AppName+SG_UID;<br />
	<br />
        HANDLE exclusion = CreateMutex(NULL, FALSE, MutexName);<br />
     DWORD err = ::GetLastError();<br />
     if(err != 0){<br />
		 if(err == ERROR_ALREADY_EXISTS){<br />
			 CString Message;<br />
			 Message.Format("There is already an Instance of '%s' Running on This Computer\n\n"<br />
					 "Activate the Existing Instance instead",(LPCSTR)AppName);<br />
	    		 AfxMessageBox(Message);<br />
				 exit(0);<br />
		 }<br />
		 else{<br />
			 AfxMessageBox("An Error occurred whilst checking that No other instance of\n"<br />
				 "this program is running\n");<br />
			 exit(-1);<br />
		} // error <br />
	 }<br />

Thanks

Bram van Kampen

Amr M. K.11-Dec-09 9:32

Amr M. K.

11-Dec-09 9:32

Dear All,
I am still new in c++

i would like to implement this class
how??
void main()
}
;(A a1(1,1
;(A a3(2,2
;(A a3(0,0
a3=a1+a2;
++ a1
() a3.out
a3=a1-a3
{() a3.out

Yes. CreatiVity withOuT limiTs

modified on Friday, December 11, 2009 3:57 PM

«_Superman_»11-Dec-09 9:53

«_Superman_»

11-Dec-09 9:53

This doesn't look like C++ to me.

«_Superman_»
I love work. It gives me something to do between weekends.

Microsoft MVP (Visual C++)

Amr M. K.11-Dec-09 9:58

Amr M. K.

11-Dec-09 9:58

its like procedure must be implemnt in c++

Yes. CreatiVity withOuT limiTs

EliottA11-Dec-09 9:58

EliottA

11-Dec-09 9:58

I think he want's to port it over to C++, or maybe it's some sort of class diagram outlining functionality...

Check out the CodeProject forum Guidelines[^]
The original soapbox 1.0 is back![^]

CPallini11-Dec-09 11:30

CPallini

11-Dec-09 11:30

Do you want to convert ASCII-art into C++ code?
Big Grin | :-D

If the Lord God Almighty had consulted me before embarking upon the Creation, I would have recommended something simpler.
-- Alfonso the Wise, 13th Century King of Castile.

This is going on my arrogant assumptions. You may have a superb reason why I'm completely wrong.
-- Iain Clarke

[My articles]

Luc Pattyn11-Dec-09 11:53

Luc Pattyn

11-Dec-09 11:53

yes, and without limits please.

Smile | :)

Luc Pattyn [Forum Guidelines] [Why QA sucks] [My Articles]

I only read code that is properly indented, and rendered in a non-proportional font; hint: use PRE tags in forum messages

Amr M. K.11-Dec-09 12:31

Amr M. K.

11-Dec-09 12:31

#include <iostream>

using namespace std;
class A
{

private:
int x;
int y;

public:

A()
{
x = 0;
y = 0;
}

A(int First, int Second)
{
x=First;
y=Second;
}

// Opertator
// A operator= ( A &oA);
A operator- ( A oA);
A operator+ ( A oA);

A& operator++(int unu); // prefix

int GetMyX() { return x; }
int GetMyY() { return y; }

void SetMyX(int Xvalue)
{
x=Xvalue;
}

void SetMyY(int Yvalue)
{
y=Yvalue;
}

void Out()
{
cout<<"Value Of x:" << GetMyX()<<" "<<"Value Of y:"<<GetMyY()<<endl ;
}
};

A A:: operator- (A oA)
{
oA.SetMyX (oA.GetMyX ()-x);
oA.SetMyY (oA .GetMyY ()-y);

return oA;
}

A A::operator+ (A oA)
{

oA.SetMyX (oA.GetMyX ()+x);
oA.SetMyY (oA .GetMyY ()+y);
return oA;
}

A& A::operator++(int temp)
{
SetMyX (GetMyX ()+1);
SetMyY (GetMyY ()+1);
return *this;
}

int main()
{
using namespace std;
A a1(2,4);
A a2(3,6);
A a3(0,0);

a1++;

a1.Out ();
a3=a1+a2;
a3.Out ();

a3=a1-a3;

a3.Out ();

int x;

cin>>x;
return 0;
}

Yes. CreatiVity withOuT limiTs