The run function is not a loop
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(print "Hello world!\n")) | (print "Hello world!\n")) | ||
- | '''This is wrong'''. It will output "Hello world!" ''once'', then give an error. What is going on? The problem is that you are thinking of <code>run-until</code> in the wrong way, as if it is a loop: | + | '''This is wrong'''. It will output "Hello world!" ''once'', then give an error. What is going on? The problem is that you are thinking of <code>run-until</code> in the wrong way, as if it were a loop: |
for time ≤ 200 do | for time ≤ 200 do |
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In Meep, there are functions run-until
and similar that run the simulation, and take arguments allowing custom actions to be performed on every time step, or on some subset of the time steps. Many users misunderstand this, however, and make the same mistake: they think the run
function is a "looping" construct of some kind, and that you can just put any code you want into it and it will get executed for every time step. This mistake and how to correct it are described in this article.
Hello world
Let's consider a "Hello world" example. Suppose we start with a control file that runs for 200 time units and outputs Ez on each time step:
(run-until 200 output-efield-z)
and now we want to modify it to also print "Hello world!" for every time step, as it is running.
The wrong way
Many users will naively write:
(run-until 200 output-efield-z (print "Hello world!\n"))
This is wrong. It will output "Hello world!" once, then give an error. What is going on? The problem is that you are thinking of run-until
in the wrong way, as if it were a loop:
for time ≤ 200 do output-efield-z (print "Hello world!\n")
This is not what is happening. Instead, run-until
is just a function that runs the simulation, and its arguments should be functions that are called for each time step. That is, it is really doing something like:
evaluate the arguments: 200: a number output-efield-z: a function (print "Hello world!\n"): prints output and returns #<unspecified> call run-until: time-step until t=200 at each time step, call the arguments: call (output-efield-z) call (#<unspecified>)
Two things went wrong. First, the arguments are evaluated before calling the function, which means that the print
statement is executed before run-until
even starts! Second, run-until
then tries to call the result of (print ...)
as if it were a function, which causes an error because (print ...)
does not return a function. (The print
returns a special Scheme code #<unspecified>
that means it doesn't really return anything at all.)
The right way
What we should have passed to run-until
, instead of the result of calling (print ...)
, is a function that calls (print ...)
. There are two ways to do it.
First, we could explicitly define a function, call it my-hello
, that does what we want:
(define (my-hello) (print "Hello world!\n")) (run-until 200 output-efield-z my-hello)
Notice two things. First, my-hello
is a function of no arguments, which means that it is just called at every time step (another, more complicated, possibility is described in the Meep Reference). Second, when we call run-until
, we just pass the name of the function my-hello
, and not the result of calling the function (my-hello)
.
A second possibility is that we could use Scheme's lambda
construct to define our function in-line. The lambda
syntax in Scheme allows you to define "anonymous" functions without assigning them a name via define
, and to stick the function definition right into another expression. It works like this:
(run-until 200 output-efield-z (lambda () (print "Hello world!\n")))
Here, the (lambda () ...)
defines a function of no arguments ()
that, when called, executes the ...
statements.