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I like the message box Idea.... LOL
Unfortuately, back then the theory was that some other program (like the operating system would be the culprit.
Back when I first started looking into it, I was able to find the original KB article that highly recommended you do this, and also the later KB article that says it's a bad idea.
My more recent searches on msdn don't turn up the first. Perhaps it has been removed.
I am interesting it your Both! comment....
Why both?
Do use visual assist with VS6?
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By Both, I meant installing both environments. For the odd tweak, leave your code in VS6. If you have to do major changes to a module, take that chance to modernise it.
While having your code split into bits may give you headaches, it does have the advantage of being able to be converted module by module, instead of all or nothing.
I hope that's more clear.
In the process of moving to Sweden for love (awwww).
If you're in Scandinavia and want an MVP on the payroll (or happy with a remote worker), give me a job!
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OK, thanks for the clarification...
-Parker
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I have been searching the internet for days trying to figure this out and the only thing I can find is other people with this problem and no resolution.
I am writing a C++ application that does not require .NET of any kind.
I can get the currently logged in users information using the NetWkstaUserGetInfo() API call. I then pass the username and domain to the NetUserGetGroups() function and can get back the groups that the user is a member of, but it doesn't show nested groups.
Example:
Group 1
|____Group A
|____*User1*
Group 2
|____Group B
|____*User1*
If a user, *User1*, is a member of "Group A" and a member of "Group 2" I would like for the function to return:
Group A (explicit membership)
Group 1 (implicit membership thru Group A)
Group 2 (explicit membership)
Right now it is only showing:
Group A (explicit membership)
Group 2 (explicit membership)
Please tell me someone has done this before without .NET!
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See here.
"Old age is like a bank account. You withdraw later in life what you have deposited along the way." - Unknown
"Fireproof doesn't mean the fire will never come. It means when the fire comes that you will be able to withstand it." - Michael Simmons
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Dude I award you the Googling MOFO of the day trophy!
Of course not if you have that URL tattooed on the back of your hand.
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union u
{
char ch[2];
int i;
};
int main()
{
union u x={0,2};
cout<<x.ch<<"\n\n\n";
cout<<x.i<<endl;
return 0;
}
Why does this print 512??? What is this x={0,2}; exactly doing?
----------------------------
286? WOWW!
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The union looks something like this:
-----------------------
| 512 |
-----------------------
| 00000010 | 00000000 |
-----------------------
| 2 | 0 |
-----------------------
"Old age is like a bank account. You withdraw later in life what you have deposited along the way." - Unknown
"Fireproof doesn't mean the fire will never come. It means when the fire comes that you will be able to withstand it." - Michael Simmons
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Thanks.
----------------------------
286? WOWW!
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Is doing politely what you asked with your code.
The character array and the integer share the same memory space (at least the first two bytes), hence assigning one of the two will affect the other (you know that: it is a union , after all... ).
_8086 wrote: union u x={0,2};
Here the compiler initialise the ch member (this surpised a bit me) of the union with the characters having ASCII codes 0 and 2 . Incidentally 0 corrensponds to string terminator so ch eventually contains an empty string, this explains the output of the
_8086 wrote: cout<<x.ch<<"\n\n\n";
line.
Such a initialization affect also the integer (i ) member, and since you computer is a little endian one, you get 0 * 2^0 + 2 * 2 ^ 8 = 512 .
This explains the output of the
_8086 wrote: cout<<x.i<<endl;
line.
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]
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Such a small thing has got this much hidden! That's all about that crazy union . Thanks mate.
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286? WOWW!
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_8086 wrote: That's all about that crazy union
Farout
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lol
----------------------------
286? WOWW!
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Unions are powerful things, but until you realise that the parts share the same memory, you'll struggle. David's picture and Carlo's talk both help, I hope.
They are very powerful in their limited way. Here's a sample of my code (no real secrets here):
union __ChannelsOn
{
BYTE Mask;
struct {
BYTE On1 : 1;
BYTE On2 : 1;
BYTE On3 : 1;
BYTE On4 : 1;
BYTE OnTOF : 1;
BYTE Unused : 1;
BYTE MasterOn : 1;
BYTE ScanOn : 1;
} Bits;
} ChannelsOn;
I have some hardware that has a command I send to it to turn channels on and off. I send a byte made up of flag bits. I could say:
__ChannelsOn c;
c.Mask = 1 << 3 | 1 << 7;
SendChannels (c);
or I say:
__ChannelsOn c;
c.Mask = 0;
c.Bits.On3 = 1;
c.Bits.ScanOn = 1;
SendChannels (c);
Both do the same thing - but which is more readable?
They are also used to make the variant structure, used to talk with COM/VB.
It's equivalent to:
struct VARIANT
{
int nType;
union {
int nInt;
long lLong;
DWORD dwDword;
BSTR bstr;
} Var;
};
I hope that helps a bit,
Iain.
In the process of moving to Sweden for love (awwww).
If you're in Scandinavia and want an MVP on the payroll (or happy with a remote worker), give me a job!
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hi
what is the diffrence, if i write the interface declaration inside library in my .idl file . Are these two implementation serve the same purpose.
for Ex:
library XYZ
{
interface ABC
{
method 1 ;
}
[
uuid(________),
helpstring("_________")
]
coclass PQR
{
[default]interface ABC;
};
};
OR
interface ABC
{
method 1 ;
};
library XYZ
{
[
uuid(________),
helpstring("_________")
]
coclass PQR
{
[default] interface ABC ;
};
} ;
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pandit84 wrote: Are these two implementation serve the same purpose.
Yes, since I interpret that as a question...
"It's supposed to be hard, otherwise anybody could do it!" - selfquote "High speed never compensates for wrong direction!" - unknown
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I defined a ON_UPDATE_COMMAND_UI handler for save and every thing went right. But after making some modification on my project I noticed the menu item "Save" is always disabled and its OnUpdate handler is never called. I removed its handler and removed its declaration in MESSAGE MAP too. But it's disabled yet
any idea?
Thank you masters!
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Jusef Marzbany wrote: any idea?
No, sorry. you bet on the wrong horse.
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]
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Jusef Marzbany wrote: But after making some modification on my project
Jusef Marzbany wrote: any idea?
Yes, my idea is that some of your modification broke your project. Does that help you as much as it helps us?
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Jusef Marzbany wrote: ...and every thing went right.
Does this mean that the OnUpdateCmdUI() handler was getting called correctly?
"Old age is like a bank account. You withdraw later in life what you have deposited along the way." - Unknown
"Fireproof doesn't mean the fire will never come. It means when the fire comes that you will be able to withstand it." - Michael Simmons
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That should be a big clue then. Since it is no longer being called, the new code you added is at fault.
"Old age is like a bank account. You withdraw later in life what you have deposited along the way." - Unknown
"Fireproof doesn't mean the fire will never come. It means when the fire comes that you will be able to withstand it." - Michael Simmons
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Yes. Did you somehow think that once was sufficient? I fully expect to see a similar response to ours shortly.
"Old age is like a bank account. You withdraw later in life what you have deposited along the way." - Unknown
"Fireproof doesn't mean the fire will never come. It means when the fire comes that you will be able to withstand it." - Michael Simmons
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