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Do generatable array fields always belong to the same CFS? In case one of the list fields has a constraint, and another field has a different constraint and they are not connected. Will both the fields belong to the same CFS?

Yanir
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  • The question is not entirely clear. A simple code example would help clarify the question. – Kalev Jan 13 '16 at 10:21

2 Answers2

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If the fields aren't connected than each field will be on a different CFS.

Nadav
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The question is not totally clear but here is an attempt to answer:

  1. if this is a list of struct with several fields, the different fields will belong to the same CFS only if connected (e.g. l[0].x and l[0].y will belong to the same CFS only if there is a constraint connecting them).

  2. assuming the question referred to different indices of the same list path (e.g. l[0].x and l[1].x, or m[0] and m[1]), then we need to differentiate between the static and runtime considerations:

statically, both paths are considered to belong to the same CFS. For example, ICFS analysis assume that, so "keep x < f(m[0]); keep m[1] < g(x)" will create an ICFS.

on runtime, IntelliGen tries to solve the list items one-by-one (as if they were in different CFSs), for performance reasons. However, when the list items are connected (either directly or to different variables), which in IntelliGen's terms is called 'lace', they are indeed solved as one CFS, which can be very large.

for more details, I suggest to read section 4.3 ("Avoid Dependencies Between List Elements") in the IntelliGen user guide.

Reuven Naveh
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