The ABINIT Users Mailing List ( CLOSED )

["Anglade Pierre-Matthieu" <anglade@gmail.com>]

Hi,

The buffer size used by the program for stdout depends on the
compiler. In order to optimize computation speed some compilers tend
to increase this size a lot. Nevertheless Abinit log (and even output)
is a very big file and usually you get at least a few line after a few
seconds of run. Especially because Abinit startup is interactive
(unless you have chosen some special compilation options), meaning
that you must  get at least something in the log file.
1) So, if you really get nothing in the log and haven't use special
compilation options it is quite likely that this is a compiler (or
also OS) bug.
Do your version of Abinit stand the self tests ?
2) Otherwise if you just have to wait very long for the first sensible
information to be written in the output file this is not a bug. Just a
big computation.

regards

PMA

On 8/23/07, Hua Bao <hbao@purdue.edu> wrote:
> Dear Matthieu,
>
> Thank you for your reply. I changed the input file and use abinis to do
> the calculation. However, it just got running for a long time and no
> output was there. If I killed the job, there was even no output. The out
> file and log file were just nothing.
> It this a bug or something? Or is there any compiling problem? At least
> there should be some output in the log file.
>
> Thanks,
> Hua
>
>
> Matthieu Verstraete wrote:
> >
> > Hello Hua,
> >
> > 1) you should not set chkprim 0; use an irreducible unit cell unless
> > you are certain of what you are doing
> >
> > 2) The default kpt shift is 1 shift at 0.5 0.5 0.5. This could be what
> > breaks the symmetries, as you are in a hexagonal, not a cubic system.
> > Specify shiftk explicitly as 0.0 0.0 0.5 for hex systems, or even 0.0
> > 0.0 0.0. If you include  0.5 0.5 0.5, a single kpt shift is probably
> > not enough. See the tutorials and tests (in v1) on generating good kpt
> > grids.
> > Depending on what you are doing the code sometimes imposes that the
> > kpt grid has the full lattice symmetry. In any case this is always a
> > good idea, and is broken if your shift can be rotated out of the kpt
> > grid by a symop.
> >
> > In your case the full point group symmetry appears to be P1, which I
> > think is wrong for corundum (probably your xangst are not precise
> > enough - should probably try natrd and associated variables). For the
> > kpoints the bravais lattice symmetries are used, ie for the hexagonal
> > cell.
> >
> > Matthieu
> >
>
>


-- 
Pierre-Matthieu Anglade