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But sir
The main thing how can i know about the namespace the dll provide ?
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If I remember correctly, the default namespace matches the name of the library.
Use the "OLE/COM Object Viewer" (you find it in the Tools menu of Visual Studio), browse to the according Type Library and open it. Somewhere you will find something like:
[
...
]
library SERVERLib
{
...
}
In this case SERVERLib will be the default namespace.
Regards,
Tim
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hye,
i m making a project in which i have to display video in an edit box....
i have a DVR card but i want to make a separate aaplication using the functions of that card
can anybody tell me the procedure or steps of displaying the video in general.i mean not specific to any software......
pls give me some hints
looking 4 help
NT
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You'd use DirectX to do this ( specifically DirectShow, which is now in the Platform SDK ). Plenty of articles on this on CP and elsewhere.
Christian Graus - Microsoft MVP - C++
Metal Musings - Rex and my new metal blog
"I am working on a project that will convert a FORTRAN code to corresponding C++ code.I am not aware of FORTRAN syntax" ( spotted in the C++/CLI forum )
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actually i want to use functions of my DVR card
the link u suggested is an application
i dnt need code
i need articles which explains how to display video in VisualC++
i want to knw the steps nd functions which i shld include in my code for dispplaying video
looking 4 reply...........
NT
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#include<iostream>
#include<string>
#include<cmath>
using namespace std;
class Complex{
public:
Complex();
Complex(double);
Complex(double,double);
void write() const ;
bool operator==(const Complex& ) const ;
bool operator!=(const Complex& ) const ;
Complex operator-();
friend ostream& operator<<(ostream& out,Complex& c);
private:
double real;
double imag;
};
Complex::Complex(){
real=imag=0.0;
}
Complex::Complex(double re){
real=re;
imag=0.0;
}
Complex::Complex(double re,double im){
real=re;
imag=im;
}
void Complex::write() const {
cout<<real<<"+"<<imag<<'i'<<endl;
}
bool Complex::operator==(const Complex& u ) const {
if(real==u.real&&imag==u.imag)
return true;
else return false;
}
bool Complex::operator!=(const Complex& u ) const {
if(real!=u.real||imag!=u.imag)
return true;
else return false;
}
Complex Complex::operator-(){
real=-real;
imag=-imag;
}
ostream& operator<<(ostream& out,const Complex& c)
{
return out<<c.real<<"+"<<c.imag<<'i'<<endl;
}
int main(){
Complex c1(2.3,5.6);
Complex c2(1.4,2.5);
if(c1==c2) {
c1.write();
c2.write();
cout<<c1<<endl;
cout<<c2<<endl;
}
return 0;
}
I have declared :friend ostream& operator<<(ostream& out,Complex& c);
but why the error says:
cannot access private member declared in class 'Complex'?
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Hi,
Your function declaration has different prototype,
one has const reference other is not const.
Best Regards
Raj
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But after I modified the declaration of the function to const,there are two errors which says that operator<< is ambiguous added ~
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It should work. What's the concrete message of the error?
--
=====
Arman
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No, It should work for the << operator
but shows error for
Complex Complex::operator-(){<br />
real=-real;<br />
imag=-imag;<br />
}<br />
because it doesnot return value.
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NO~~~after I delete the funtion Complex::operator-() from the class,the error is the same as before~~~
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Hi,
I simply copied your code and put in an empty project I compiles well would you post the exact error message
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there are four:
error C2248: 'real' : cannot access private member declared in
class 'Complex'
error C2248: 'imag' : cannot access private member declared in
class 'Complex'
error C2593: 'operator <<' is ambiguous
error C2593: 'operator <<' is ambiguous
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By the way,the two previous errors point to the location
"return out<
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Because-because-because you have two different operator << overloads;
friend ostream& operator<<(ostream& out, Complex& c);
ostream& operator<<(ostream& out, const Complex& c)
{
...
}
Their prototypes are different due to the difference inside their parameter list;
in first case; Complex &
in second case; const Complex &
So make them the same; put const in the friend declaration.
--
=====
Arman
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NO~no~no~~I have modified this problem~~but it do noting to the error~
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Chen-XuNuo wrote: I have modified this problem...
To what? By making both the declaration and the definition the same, the compiler error (C2248) will go away.
"A good athlete is the result of a good and worthy opponent." - David Crow
"To have a respect for ourselves guides our morals; to have deference for others governs our manners." - Laurence Sterne
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What is the size of windows message queue ? i.e how many messages can queue up before the next one being discarded?
And how long can a message stay in the message queue waiting to be processed?
I've been looking all over but haven't found an answer for this ?
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On Windows 2000 message queues defaulted to about 5000 entries which will be there until they get processed, however long it takes. I would guess that the sizes have gone up in XP and Vista and the implementation may have changed. If these limits are causing you problems then may I venture to suggest that you may have deeper problems in your design 
Nothing is exactly what it seems but everything with seems can be unpicked.
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thanks a lot,
i was wondering about this just for the sake of curiousity. it's really dangerous using a queue without knowing how much it can handle. it's wierd I couldn't found any information about this (which I thing is very important). now i'm so relieved  , nobody says a word about it because it's almost limitless
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GameProfessor wrote: it's really dangerous using a queue without knowing how much it can handle
It's also dangerous to make assumptions about the size of the windows message queue.
It's probably best to process the messages ASAP - for your sake and the user's sake 
Mark
"Posting a VB.NET question in the C++ forum will end in tears." Chris Maunder
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 I am attempt to make my software multilingual.In my application I have used CColorDialog. Now I didn't know how to translate these Common Dialog (CColorDialog / CFindReplaceDialog / CFontDialog / CPrintDialog / CPrintDialogEx) into all kinds of language version.
-- modified at 3:44 Monday 4th June, 2007
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my exe will modify a file, but just only the account which is created by IT dept. of my co. can do that, how can i run the exe with that account while the login account is another?
Thanks!!!
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use runas command .. u can shellexecute runas from your program.
if u want more help type just runas in command prompt window.
If u can Dream... U can do it
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