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[Conditional "DEBUG"]

abelo

abelo

20y
1.6k
1
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
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