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Christian Graus wrote: What on earth is mupad ?
MuPad[^]...
A pretty cool tool, used it in a course once...As far as tying it into .NET, I am not too sure.
If you try to write that in English, I might be able to understand more than a fraction of it. - Guffa
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As looking at the web page, you can install on:
Windows, MacOS X, Linux
There is no way .NET framework is build for MacOS X, Linux.
It is pointless, to buil MuPad with .NET especaly for windows, but corp can use c++ cross-platform files, for maintaing is a lot easier, a lot cheaper, and the same techonlogy. And betwen them is very little diffrence in code. The Only big diffrence is Prebuild libraries for specific OS.
And you are defently in the wrong forum.
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bsaksida wrote: And you are defently in the wrong forum.
I'm not but maybe the OP is...
--- modified
I didn't think there was a way to tie MuPad into .NET 
Some people have a memory and an attention span, you should try them out one day. - Jeremy Falcon
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Hi friends. I'm a new programmer. In an application, when i click About menu, i want my About.h dialog form to display. In the MSDN library, i saw such an example to display user defined dialogs;
{
Form ^ f = gcnew Form();
f->ShowDialog();
}
And i changed this code to display my About.h form dialog such as;
{
Form ^ About = gcnew Form();
About->ShowDialog();
}
But, i still see the same empty dialog instead of my About.h form. So how can i display this dialog when i click About menu ?
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You're creating an empty form, an instance of the base Form class. You need to create an instance of the about dialog class, which will be derived from Form. If it's called AboutForm, then do this:
AboutForm ^ f = gcnew AboutForm();
f->ShowDialog();
Christian Graus - Microsoft MVP - C++
Metal Musings - Rex and my new metal blog
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I saw this sample in also MSDN and tried many times, but compiler says AboutForm and f are unknown identifiers. Is there anything else i missed ?
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I said 'if your about form is called 'AboutForm'. I have no idea what it's called, and obviously it needs to be in scope where you're trying to call it.
If you don't understand these things, you should stop playing with Managed C++, and also with any sort of GUI code, and work through a basic C++ book.
Christian Graus - Microsoft MVP - C++
Metal Musings - Rex and my new metal blog
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Of course, i modified the codes u send with respect to my Form's name, but there is still same error... Unknown identifiers...
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Then your problem is probably to do with your #include settings. It's impossible to say, based on what you've told me, I'd need to see the code.
Christian Graus - C++ MVP
'Why don't we jump on a fad that hasn't already been widely discredited ?' - Dilbert
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I am trying to run this packet logging/sending program for a game; which is coded within the Microsoft Visual Studio 6.0 C++ enviorment, however when I try executing it says's "This application has failed to start because the application is incorrect. Reinstalling the application may fix this problem.” any idea's?
-h7h7
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h7h7 wrote: which is coded within the Microsoft Visual Studio 6.0 C++ enviorment,
This means it was not written using Managed C++, and you're in the wrong forum
When you find the right one, no-one can help you based on what you've said, you need to set breakpoints or otherwise work out where the program is crashing, so we can see some code to comment on.
Christian Graus - Microsoft MVP - C++
Metal Musings - Rex and my new metal blog
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Can anyone please recommend a book that is a good guide for beginners to the STL Library. I purchased one book that was supposedly for beginners, the C++ Standard Template Library, and it is so confusing i cant understand it. I would appreciate someone's help with this.
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The book you've bought is the standard reference on the topic. However, there are lots of more elementary texts on the topic. There are also heaps of tutorials on the web ( I wrote a few on this site ). Work through the stuff on CP, then go back to your book.
And, in future, managed C++ means .NET C++, which does not include STL. Use the Visual C++ forum, if you're not using Managed C++ or C++/CLI.
Christian Graus - Microsoft MVP - C++
Metal Musings - Rex and my new metal blog
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thanks for the advice. Can you please reference an elementary text on the stl library. I cant seem to find a good tutorial on the web. So there are some on this sight? I would appreciate any help you can give me. Thanks much
Seth Isaacks
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Is there any way I coud manage a connection to an mdb database without using ODBC?Because I'm making a small program to store info like phone numbers and such, but I would rather not having to rely on the user having the ODBC for mdb databases, and I've been told that I can manage the connection by only adding a few things to the project, and no need to use ODBC.
I would appreciate any help you could give me.
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Well, as you're asking in the managed C++ forum, I guess you can use the .NET framework classes for database connections. If you posted in the wrong forum, then I'm not sure what your options are, but you should move the question where it belongs, in order to find out.
Christian Graus - C++ MVP
'Why don't we jump on a fad that hasn't already been widely discredited ?' - Dilbert
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No, I actually don't use .NET, I was told that I could access an mdb db from my app by only adding some library to it, and I'm pretty sure it was not a .NET library...
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can anyone walk me through the sequence of events in the following code execution? thank you.
main()
{
int x=20,y=35;
x=y++ + x++;
y= ++y + ++x;
printf(“%d%d\n”,x,y);
in my guess this is what happens...
x = 20; y = 35;
x = 35 + 20; y = 35+1;
x = 35 + (20+1);
x = 56; y = 36;
y = 37 + 57; x = 57;
x = 57; y = 94;
so output is: 5794
i got it printing 5794 but is my analysis correct? thank you
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As i see
x=20; y=35;
x = (35+1) + (20+1) // y has nev value 36
x = 57
y = (36+1) + (57+1);
y = 95;
y++ // this is add 1 to y;
++y // this 1+y, but doesn't increase y for 1.
ps: You are posting in wrong forum, you shoud Visual c++
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bsaksida wrote: ++y // this 1+y, but doesn't increase y for 1.
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int y=1;
x = ++y; //x is 2, y is 1;
x = y++//x is 2, y is 2;
y++; // increase y by 1
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bsaksida wrote: x = ++y; //x is 2, y is 1;
No ! y is 2 here,too. And 2 will be assigned to x.
bsaksida wrote: y++; // increase y by 1
y will become 3 here.
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j11Software wrote: x=y++ + x++;
Here, x(20) is added to y(35) and assigned to x(55). And then incremented(++) with 1 ie.e x is now 56. Y is incremented to 1. Y is 36 now.
j11Software wrote: y= ++y + ++x;
x(56) is incremented for 1 now x is 57. y is incremented for 1. Y is 37 now. Both are added to eqate to y. i.e. y is 94 now.
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Very close.
This is the C equivalent of what the Microsoft compiler produces:
x = 20; y = 35;
x = y + x; // 35 + 20 = 55
x = x+1; // 56
y = y+1; // 36
y = y+1; // 37
x = x+1; // 57
y = y + x; // 94
I would like to note that you should not write code like "y = y++ + x++" because the results are undefined by the language. That is the results of a statement like "x = x++" is considered undefined and it is up to the compiler vendor to decide how to implement it.
The following C equivalent code may be produce by a different compiler:
x = 20; y = 35;
tx = x+1; // 21
ty = y+1; // 36
x = y + x; // 35 + 20 = 55
x = tx; // 21 (unexpected result)
y = ty; // 36
y = y+1; // 37
x = x+1; // 22
y = y + x; // 59
One other thing I would like to point out is that other languages, that have the '++' operator, interpret it differently.
For furthur information, try searching the Net for "i=i++".
INTP
"Program testing can be used to show the presence of bugs, but never to show their absence."Edsger Dijkstra
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