Comment 12 for bug 1083720

I think I got a slightly simpler idea based on your test/trial counter
idea. Just need to think it through. Will attempt to implement it tomorrow.

On 28 November 2012 20:03, Johan Hake <email address hidden> wrote:

> On 11/28/2012 07:15 PM, Martin Sandve Alnæs wrote:
> > I fixed this permanently by constructing both test and trial function
> > simultaneously on demand using Argument(V) and storing them to
> > V._argument_functions. Will the circular references "functionspace <->
> > testfunction" be a problem for garbage collection?
>
> Yes.
>
> I wonder how this can be implemented without some error prone
> heuristics. A robust, but quite intrusive way, would be to always
> instantiate a TestFunction together with a TrialFunction whenever one
> need the latter.
>
> Or one could introduce a trial and test space specific counter.
> Something like:
>
> class Argument(.):
> _test_and_trial_counters = [dict(test=False, trial=False)]
> ...
>
> class TestFunction(.):
> def __init__(.):
> .
> counters = Argument._test_and_trial_counters
> if counters[-1]["test"]:
> counters.append(dict(test=False, trial=False))
> counters[-1]["test"] = True
> Argument(self, ., count=len(counters)*2)
>
> class TrialFunction(.):
> def __init__(.):
> .
> counters = Argument._test_and_trial_counters
> if counters[-1]["trial"]:
> counters.append(dict(test=False, trial=False))
> counters[-1]["trial"] = True
> Argument(self, ., count=len(counters)*2-1)
>
> Johan
>
> --
> You received this bug notification because you are a member of DOLFIN
> Core Team, which is subscribed to DOLFIN.
> https://bugs.launchpad.net/bugs/1083720
>
> Title:
> replace() fails if the new function spaces have a different mesh
>
> Status in DOLFIN:
> Fix Committed
>
> Bug description:
> I submit this bug report following Marie's suggestions on the related
> DOLFIN question
> (https://answers.launchpad.net/dolfin/+question/215120)
>
> DOLFIN crashes in the solve() routine after using replace() to replace
> the function spaces of a form.
>
> While the solve succeeds if replaced function spaces have:
> - a different polynomial degree or
> - a different polynomial degree and a different mesh,
> it does not work if the new function spaces is based on a new mesh only.
>
> This is the error message I get:
> ***
> -------------------------------------------------------------------------
> *** Error: Unable to define linear variational problem a(u, v) = L(v)
> for all v.
> *** Reason: Expecting the solution variable u to be a member of the
> trial space.
> *** Where: This error was encountered inside
> LinearVariationalProblem.cpp.
> ***
> ------------------------------------------------------------------------
>
> The attached file demonstrates the problem.
>
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>