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^ is the bitwise XOR operator, whereas ^^ is the logical XOR operator. If you don't know the difference, feel free to ask 
Joaquín M López Muñoz
Telefónica, Investigación y Desarrollo
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there is no logical XOR. at least not in VC6
-c
Smaller Animals Software, Inc.
You're the icing - on the cake - on the table - at my wake. Modest Mouse
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Oh yes, there ain't such an operator!  I guess I just extended the obvious analogy from & and && without thinking it too much. What strikes me is that all these years have passed without noticing before  Thanks for pointing out. My apologies to all.
Joaquín M López Muñoz
Telefónica, Investigación y Desarrollo
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i wouldn't have known, if i wasn't looking for the exact thing last night. didn't find it, of course.
-c
Smaller Animals Software, Inc.
You're the icing - on the cake - on the table - at my wake. Modest Mouse
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I found out that while answering a question to a VB programmer. He asked about XOR in C++, and I thought: "What a stupid question!", but I found out it was not that stupid after all. 
I vote pro drink ![Beer | [beer]](https://codeproject.global.ssl.fastly.net/script/Forums/Images/beer.gif)
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That's what I like about answering questions here -- you have to relearn the stuff in order to give accurate answers, even when you get to the basics.
Joaquín M López Muñoz
Telefónica, Investigación y Desarrollo
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VC++ spews at ^^ 
if( aCurrCPA[ref_c] ^^ aCurrCPA[temp_c] )
produces this error
D:\Pete\Capture\vss\Tracker.cpp(208) : error C2059: syntax error : '^'
is it non-standard C?
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GEEK OUT ALERT!!!!
XOR via AND and OR:
(a & !b) | (b & !a)
Tim Smith
I know what you're thinking punk, you're thinking did he spell check this document? Well, to tell you the truth I kinda forgot myself in all this excitement. But being this here's CodeProject, the most powerful forums in the world and would blow your head clean off, you've got to ask yourself one question, Do I feel lucky? Well do ya punk?
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Shouldn't it be
(a & ~b) | (b & ~a) not to mix logical and bitwise operators?
Joaquín M López Muñoz
Telefónica, Investigación y Desarrollo
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so for a logical operator this should work
(bool1 && !bool2) || (bool2 && !bool1)
right?
Thanks for the super-fast answers, BTW
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In fact, if we want get all clever about it we could use
#define XOR( a, b ) ((a) && !(b)) || ((b) && !(a))
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template <class T>
T XOR(const T val1, const T val2)
{
return ( ((val1)&(~val2)) | ((~val1)&(val2)) );
}
Hope that I am correct! 
"Dirty hands lead to important discovery..." - Thomas Edison
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For this to work, the class used has to implement the operators &,| and ~.
Otherwise you get comile errors (right?)
Good Luck,
Bill
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This is more efficient (maximum number of evaluations is 2, in the previous formula it is 4)
bool1?!bool2:bool2
Joaquín M López Muñoz
Telefónica, Investigación y Desarrollo
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nice 
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But that isn't done with AND and ORs.
Tim Smith
I know what you're thinking punk, you're thinking did he spell check this document? Well, to tell you the truth I kinda forgot myself in all this excitement. But being this here's CodeProject, the most powerful forums in the world and would blow your head clean off, you've got to ask yourself one question, Do I feel lucky? Well do ya punk?
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So much stuff from such a little subject
Another advantage of a?!b:b with respect to (a&&!b)||(!a&&b) is that the former does not have the "double evaluation" problem of the latter, which is important if one plans to make a macro out of it.
Joaquín M López Muñoz
Telefónica, Investigación y Desarrollo
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My version works better for circuits
Tim Smith
I know what you're thinking punk, you're thinking did he spell check this document? Well, to tell you the truth I kinda forgot myself in all this excitement. But being this here's CodeProject, the most powerful forums in the world and would blow your head clean off, you've got to ask yourself one question, Do I feel lucky? Well do ya punk?
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If you're doing TTL (which means only NANDs and negators), I think this is the way to go (my electronics are a little rusty, nevertheless, I could be wrong)
(a NAND !b) NAND (!a NAND b)
Joaquín M López Muñoz
Telefónica, Investigación y Desarrollo
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Yes that is logically the same thing. And since NAND is the primitive circuit that version is better.
Damn you Joaquin!!!!
(I just looked up half-adder in google and posted the first one I found.)
Tim Smith
I know what you're thinking punk, you're thinking did he spell check this document? Well, to tell you the truth I kinda forgot myself in all this excitement. But being this here's CodeProject, the most powerful forums in the world and would blow your head clean off, you've got to ask yourself one question, Do I feel lucky? Well do ya punk?
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brilliant. you are a genius. damn i am still cursing myself for reaching this post and not thinking of it.
god... only a couple of hours here and i already love this site.
impuzible
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Yup...
Tim Smith
I know what you're thinking punk, you're thinking did he spell check this document? Well, to tell you the truth I kinda forgot myself in all this excitement. But being this here's CodeProject, the most powerful forums in the world and would blow your head clean off, you've got to ask yourself one question, Do I feel lucky? Well do ya punk?
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Hello,
I want to access the Main dialog because of my program structure:
I have one dialog that has to be able to write data in two other dialogs.
All those three dialogs are used as controls of the main dialog.
Moreover, one of those dialogs has included two property sheets that need to send information to those other dialogs too.
I want to be able to access them and I've thought in those methods below:
----------------------------------------------
1)
Create a member function in the Application class that returns a pointer to the main dialog.
This allows me to use an extern call to the App object and to get directly the right type (returned from the function):
extern CMyApp MyApp;
MyApp.GetMainDialogPointer();
----------------------------------------------
2)
Get the main Window using AfxGetMainWnd():
((CMyMainDialog *)AfxGetMainWnd())->...
This is not the method I would like to use because of the fact that I need to use a type modifier each time I need to use the main dialog.
----------------------------------------------
3) Sending messages to the MainWindow and use the handler of those messages to interpret the parameters and to pass them to the right dialog?
----------------------------------------------
4)???
I don't know how to do it, do you know any other better method?
I'm sure that there is another method that could allow me to do that.
----------------------------------------------
As always thank you very much.
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I think the better way is to use Doc/View architechture support,so if you use FormView then you can access it easily from every where.
Mazy
"So,so you think you can tell,
Heaven from Hell,
Blue skies from pain,...
How I wish,how I wish you were here."
Wish You Were Here-Pink Floyd-1975
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There's a myriad of ways to achieve that. For one, create a member variable m_pMainDlg in CMyApp of type CMyMainDialog *. Set this member appropriately in CMyApp::OnInitInstance. Now these two little helpers get you that pointer from anywhere:
CMyApp * GetMyApp()
{
return static_cast<CMyApp *>(AfxGetApp());
}
CMyMainDialog * CMyApp::GetMyMainDialog() const
{
return m_pMainDlg;
}To access the main dialog, use GetMyApp()->GetMyMainDialog().
Joaquín M López Muñoz
Telefónica, Investigación y Desarrollo
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