Interesting replies to my original question, but they got diverted into
<<uses>> versus <<include>>.
I was asking, "Does <<include>> on a use case dependency imply that ALL
of the included behavior of the use case is executed?"
YES, YES, YES !!!!!
I'm pretty sure the answer is no, because a use case is a flow of
control, which could be represented by an activity diagram, containing
branches in control. The use case does not have to take every branch
for every incoming event -- it only takes appropriate branches for the
existing conditions.
If a use case inserts behavior from another use case into its own
behavior, the inserted behavior only has to be that part of the
<<include>>'ed use case that is needed.
A use case is a collection of activities, or activity steps that must be
performed in order fulfil the over all goal of the use case. The individual
steps can be arranged as sequences, iterations, selections or branches and
any combination hereof, depending on domain and the goal of the use case.
This results in a number of paths from start to end of the use case. These
are typically called scenarios (one path equals one scenario) and an
execution of a use case means running through one scenario. So, when you
include a use case you include it all, but this does not mean that all
scenarios are run through each time the use case is executed.
You can make an analogy to including one source fine into another. If the
included file contains one function that's called from the code in the
including file, the "entire" function is executed, but if the function
contains braches and selections, this does not mean that all statements in
the function are executed in call.
Erik
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