Display a "Debug" button?
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i made a error reporter and i want to display a "Debug" button on my form... I want to only show this button if theirs a debugger installed... basically i want to display the Just-In-Time debugger window. Anyway to do this? :-D
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i made a error reporter and i want to display a "Debug" button on my form... I want to only show this button if theirs a debugger installed... basically i want to display the Just-In-Time debugger window. Anyway to do this? :-D
There is a precompiler option #if DEBUG that could work for you. Otherwise a series of if-statements is the only choice.
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i made a error reporter and i want to display a "Debug" button on my form... I want to only show this button if theirs a debugger installed... basically i want to display the Just-In-Time debugger window. Anyway to do this? :-D
FocusedWolf wrote:
I want to only show this button if theirs a debugger installed...
Look at the
Debugger
class, specificallyIsAttached
,Attach()
andBreak()
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