Hello mailing list subscribers, hi Steven,
From the Meep reference:
(at-time T step-functions...)
"Given zero or more step functions, evaluates them only once, after a T
time units have elapsed from the start of the run."
It seems that this is not really what happens: the step functions actually
get called twice, as you can see by running meep, printing the meep-time in
the step function:
meep>
(run-until 10
(at-time 5
(lambda (x) (display "time: ") (display (meep-time)) (newline))
)
)
This results in
time: 5.0
time: 10.0
run 0 finished at t = 10.0 (200 timesteps)
Apparently the step function is called twice, first at the desired time T,
and then once more at the end of the run.
I guess this is related to the fact that, for the last time step, the run
functions call the step functions twice: once with 'step and once more with
'finish. Apparently this confuses the (at-time T ...) "step function
modifier": it lets the last time step slip by, and it doesn't work quite as
expected.
(Other time step modifiers do work as expected, for instance (at-beginning
..) only executes its step function once, at the beginning.)
Of course several straightforward workarounds come to mind right away;
but perhaps it would still be useful to modify the (at-time T ...) in the
next release,
to make it conform to the manual pages?
Or is there something I am overlooking?
Best regards,
Mischa
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