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Awesome!
Thanks for the update!
Cheers,
Mark
Mark Salsbery
Microsoft MVP - Visual C++
![Java | [Coffee]](https://www.codeproject.com/script/Forums/Images/coffee.gif)
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i have a dialog box which shows a video and some controls now i want to implement a graph. what is the procedure to create a graph dynamically? i also need to read data from the file and plot it into the graph. guess that is the second part.. not thinking about that.. priority is to create a graph.
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See if this[^] helps.
Nobody can give you wiser advice than yourself. - Cicero
.·´¯`·->Rajesh<-·´¯`·.
Codeproject.com: Visual C++ MVP
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1) by searching google
2) by searching codeproject articles
3) by searching again and again...
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well i was expecting something like this from somebody.. and yes i am doing what u have told me...  .. the last point has been really taken into consideration...
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yes.. actually rajesh send me that link.. i think that will be of great help to me... .. will get back if i need some help..
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if you need some help directly related to the use of that control, better ask in the forum at the bottom of the article itself, so that the author will be email notified, and will reply come back to you problem more easily.
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search for NTGraph Control in google.... i have used this control for drawing graph, i have successfully completed this project using this control...
Regards
Brahma
brahma reddy
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are you sure you wanted to say that to me ?
if you don't reply to the OP, he won't receive a notification email...
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hi,
i am opening a binary file in visual studio2005.so the file will be the active window.when i click the addin in tool menu i have to take the binary file as input.
if any one knows about this please give me the code .
pmr
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Hi,
I m using CColorDialog to display the color pallete dialog in my project.
its working fine..
but i wan my dialogs to be updated with the color selected by the color selected by the user in tat CColorDialog.
I m using SetCurrentColor().But the color s not updated in the dialog..
how to achieve tis..
suggestions reqd..
My snippet follows:
[ color c;
CColorDialog dlg;
dlg.m_cc.Flags |= CC_FULLOPEN | CC_RGBINIT;
dlg.m_cc.rgbResult = RGB(155, 0, 0);
dlg.DoModal();
c.SetCurrentColor(RGB(155, 0, 0));
c.DoModal(); ]
Gita
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color c; // is this a derived class from CColorDialog ?
c.SetCurrentColor(RGB(155, 0, 0));
c.DoModal();
"Call this function after calling DoModal to force the current color selection to the color value specified in clr." from msdn.
This function is called from within a message handler.
if you want before DoModal, you can do similar to dlg.m_cc.rgbResult = RGB(155, 0, 0);
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Hello everyone,
For the following code from Bjarne's book, it is stated that template parameter T for function g will be instantised as int other than double. My question is why there are not two instantiations for both int and double version of template function g?
template <class T> void f (T a) { g(a); }
void g(int);
void h()
{
extern g (double);
f (2);
}
Here is the related statement from Bjarne,
--------------------
Each use of a template for a given set of template arguments defines a point of instantiation.
That point is the nearest global or namespace scope enclosing its use,
just before the declaration that contains that use.
--------------------
Does it before g (double) is not global function or namespace scope enclosing its use? If yes, I do not know why g (double) is not a global function, since it is declared as extern and some other compile unit should expose it?
thanks in advance,
George
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Simply I cannot understand you point: you're calling f(2), i.e. template function with int argument. It is quite obvious that compiler will instantiate only the int specialization of the template function.
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
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Hi CPallini,
Why statement extern g (double); will not instantise a double version f, i.e. f (double)?
regards,
George
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And why should it do? The template function is f, not g.
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
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Hi CPallini,
Bjarne's point, void g(int) statement will instantise
template <class T> void f (T a) { g(a); }
by deducing the type of T int, since in g(a), type of a is T and we are declaring g(int).
My question is why extern g(double) does not have the same effect to deduct parameter T to double?
regards,
George
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OK, things are more clear, at least to me, changing the sample into
template <class t=""> void f (T a) { g(a); }
void g(int x) { return;}
void h()
{
extern void g (double);
f (2.5);
}
</class>
I made a test (defining void g (double)) into another source file. The result was that only int specialization of f was built by the compiler, confirming Bjarne point.
Now, while it is clear that g(double) is exposed by correnponding source file, maybe that its declaration make it scoped only into h() i.e. local.
I have to admit the point it is not so clear.
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
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Hi CPallini,
Thanks for your reply. Do you have any effective and efficient ways to debug template function to find out the type of template parameters?
regards,
George
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George_George wrote: extern g (double);
extern is a sort of forward declaration... it doesn't define a g(double) function at all.
and if anywhere in your code, such a function exist, the only function to get called will be g(int) (because it's the only one which actually exist...)
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Hi toxcct,
Sorry I do not agree with you. Please look at Rajkumar's reply which proves what matters is whether declaration is local or global, what not matters is whether using extern or whether a definition or delaration. I think both definition and declaration can make an instantiation.
Any comments? Please feel free to correct me if I am wrong.
regards,
George
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George_George wrote: f (2);
Are you missing here something like f<double> (2); or f(2.0); ?
George_George wrote: Does it before g (double) is not global function or namespace scope enclosing its use? If yes, I do not know why g (double) is not a global function, since it is declared as extern and some other compile unit should expose it?
yes, the definition of the g(double) must be in global scope of some other compilation unit to specify the linkage specifier extern. but its declaration is not in global scope or namespace scope at the point of function template instantiation for 'void f<double>(T)' being compiled.
In your sample declaration is in function scope of h();
try this,
template <class T>
void f (T a)
{ g(a);
}
void g(int);
extern g (double);
void h()
{
f <double>(2);
}
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Cool, Rajkumar!
1.
I think the original code does not work is because declaration of using g is local? Confirms to Bjarne'e comments -- "global or namespace scope enclosing its use, just before the declaration that contains that use"? Important points are global and declaration. Right?
--------------------
Each use of a template for a given set of template arguments defines a point of instantiation.
That point is the nearest global or namespace scope enclosing its use,
just before the declaration that contains that use.
--------------------
2.
You like explicitly assign parameter for template function, like f<double>(2), other than f(2). Is it for the good of reducing ambiguity? Any other benefits?
regards,
George
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George_George wrote: Important points are global and declaration. Right?
yes.
George_George wrote: You like explicitly assign parameter for template function, like f <double>(2), other than f(2).
in your argument you want to deduce double but you specified f(2) not f(2.0), if you have this tendency always go for explicit deducing template function like f<double>(2);
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