TECHNOLOGIES
FORUMS
JOBS
BOOKS
EVENTS
INTERVIEWS
Live
MORE
LEARN
Training
CAREER
MEMBERS
VIDEOS
NEWS
BLOGS
Sign Up
Login
No unread comment.
View All Comments
No unread message.
View All Messages
No unread notification.
View All Notifications
Answers
Post
An Article
A Blog
A News
A Video
An EBook
An Interview Question
Ask Question
Forums
Monthly Leaders
Forum guidelines
abelo
NA
2
0
[Conditional "DEBUG"]
Jun 18 2004 10:49 AM
Hello, I have developed an Exception Controller for showing Exceptions in a Form. This controller is in an ExceptionForm.dll file. To show an exception, I call the method ExceptionForm.showException(). Now is the problem: If the calling application was build in Debug-configuration, i want to show some additional data for developers in ExceptonForm. How can I determine inside of ExceptionForm that the calling application was built with Debug? I created a conditional method in ExceptionForm: [Conditional("DEBUG")] private void showAdditionalData() { ...some code... } It does not work! This method will be executed if the ExceptionForm.dll was built with Debug configuration. But I don't need to check the build mode of ExceptionForm.dll, I need the information of build mode of the calling application. I hope some people can understand what I mean (my english is not perfect:-) thanks for help Fasimba
Reply
Answers (
1
)
Remote Components Setup
Converting from VB6 to C#