[Beowulf] fftw2, mpi, from 32 bit to 64 and fortran
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Lombard, David N dnlombar at ichips.intel.comWed Aug 6 14:56:01 PDT 2008
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On Tue, Aug 05, 2008 at 02:57:42AM -0700, Ricardo Reis wrote: > On Mon, 4 Aug 2008, Mark Kosmowski wrote: > > > So, why did the 32-bit test case work? Shouldn't the same problem > > crash both systems if it is a code issue? Not necessarily given the error described below. > I asked the same question myself... The function interface is: > > call rfftwnd_f77_mpi(plan_c2r, & > 1, local_data, work, use_work, FFTW_NORMAL_ORDER) > > where use_work is an integer, value 1 if you use the work temporary > array, 0 otherwise. This was the variable I wasn't passing. ... > The wrapper function for this is (from rfftw_f77_mpi.c): > > void F77_FUNC_(rfftwnd_f77_mpi,RFFTWND_F77_MPI) > (rfftwnd_mpi_plan *p, int *n_fields, fftw_real *local_data, > fftw_real *work, int *use_work, int *ioutput_order) > .... So it must be a pointer issue revealed by the 64 bit, no? When I > wasn't doing it "properly" the value of *ioutput_order wasn't set. The value of the first element of local_data was used for the n_fields scalar. The work array was being laid down starting at the location of the use_work scalar. The FFTW_NORMAL_ORDER value was being interpreted as use_work scalar. Finally, ioutput_order scalar was some random value. So, a lot was going wrong there. It's just one of life's little, um, pleasures that it looked like it was working for your 32-bit test case. Don't worry, you'll likely do this again, as likely *every* one of us on this list has, too. BTW, Fortran passes by reference; that's why all args are pointers. -- David N. Lombard, Intel, Irvine, CA I do not speak for Intel Corporation; all comments are strictly my own.
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