Thread: Need urgently help!

Hello!

I'd like to update my former message that remained unanswered.
As I'm really stuck and can't keep on developing my program, I installed Visual C++ 6.0 on another computer and have been re-building up my project (from scratch, yes) for hours and hours, when I got, once more, that terrific type of message :

"Unable to open the file (C:\Programmi\Radiate\RadiateView.cpp). The file is either corrupted or inexistent in the Wizard Data Base etc."

So, I must stop working again. Should I give up programming with Visual C++ 6.0 because it seems to me that it 's unsecure ?

Can really nobody help me ??

I can't imagine that 2 computers are infected by who knows what type of virus that causes the same disaster.

Hope that a 'good soul' will think a while about my problem and tell me what I should do.

Thanks a lot.

Best regards.

Frandy

There is no virus, you are just making the same mistake twice. There could be a number of things wrong, so here is what I suggest you do.

Create a new project in a new workspace. Compile and run it to make sure everything is OK. Then slowly make changes to the project until you get the point you are at now. Keep compiling and running the project so you know the exact change that leads you to the error you are getting. Doing this will show you what your problem is.

Many thanks Bithub for the suggestion.
Before going that way, as my Visual C++ 6.0 has been designed for Windows 98, I'll first try to get an update for XP.
Thanks again.
Kind regards

Frandy

Or try moving to Visual Studio .NET. It offers several neat features - you can even use functions from the .NET Framework if you're too preoccupied with things to come up with your own.

Thanks to arti_valekar for his suggestion that will be useful later on.
For the moment I 'd like to build up a brand new but simplified project .
Therefore, I need to know what's the meaning of 'protect:' for a class. Have I to fear problems if I change 'protect:' with 'public:' ?

Any idea ?

Thanks a lot.

Kind regards.

Frandy