I would assume because it is easier and safest to implement. Then you always know you have the same parallel pattern aso.
Johan On Mon, May 26, 2014 at 1:59 PM, Miroslav Kuchta <[email protected]> wrote: > Thanks for clarification. I thought that that Function.assign simply > fills the vector of expansion coefficients > with new values, but I see now, that it creates a new vector with updated > values. Just out of curiosity, why > is it necessary to recreate the vector? > > Miro > > > On 05/26/2014 01:52 PM, Johan Hake wrote: > > And I do get the same values. It is just the third printed value of the > function using assign(foo) that differs. That value comes from the vector > of the original Function, which never gets it values updated. > > Johan > > > On Mon, May 26, 2014 at 1:48 PM, Martin Sandve Alnæs > <[email protected]>wrote: > >> I don't see how the id of a vector has anything to do with its value. The >> norm that Miro prints should be the same. >> >> Martin >> >> >> On 26 May 2014 13:43, Johan Hake <[email protected]> wrote: >> >>> To be honest, I would also assume these two to print the same. But I >>> have been bit by the Function::assign-recreate-the-vector behavior quite >>> many times now. The point is that the vector in a Function gets recreated >>> after an assign. >>> >>> from dolfin import * >>> mesh = UnitSquareMesh(2,2) >>> V = FunctionSpace(mesh,"CG",1) >>> u0 = Function(V) >>> u1 = Function(V) >>> >>> U = u0.vector() >>> print "Same:", U.id()==u0.vector().id() >>> u0.assign(u1) >>> print "Same:", U.id()==u0.vector().id() >>> >>> Johan >>> >>> >>> On Mon, May 26, 2014 at 1:32 PM, Martin Sandve Alnæs <[email protected] >>> > wrote: >>> >>>> Johan, I don't understand what you mean by that. I would also expect >>>> these to do the same, i.e. result in an exact copy of ut.vector()? >>>> >>>> Martin >>>> >>>> >>>> On 26 May 2014 13:27, Johan Hake <[email protected]> wrote: >>>> >>>>> Function.assign re-create the FOOvector holding the actual values. I >>>>> assume this explains the difference in behavior. >>>>> >>>>> Johan >>>>> >>>>> >>>>> >>>>> >>>>> On Mon, May 26, 2014 at 1:09 PM, Miroslav Kuchta >>>>> <[email protected]>wrote: >>>>> >>>>>> Hi, >>>>>> >>>>>> please consider the attached snippet. I would assume foo() and bar() >>>>>> to do the same thing but that is not the case. Is this a bug in >>>>>> Function.assign() or am I missing something about the method's >>>>>> behaviour? >>>>>> Thanks for answer. >>>>>> >>>>>> Regards, Miro >>>>>> _______________________________________________ >>>>>> fenics mailing list >>>>>> [email protected] >>>>>> http://fenicsproject.org/mailman/listinfo/fenics >>>>>> >>>>>> >>>>> >>>>> _______________________________________________ >>>>> fenics mailing list >>>>> [email protected] >>>>> http://fenicsproject.org/mailman/listinfo/fenics >>>>> >>>>> >>>> >>> >> > >
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