| Re: simple question (I hope) [message #53282 is a reply to message #53276] |
Fri, 30 March 2007 10:00   |
David Fanning
Messages: 11724
Registered: August 2001
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Senior Member |
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Ryan. writes:
> I have one more question about it, but it is more about how IDL works
> than the REMOVE routine.
> Say for example I do this:
>
> group_array = huge_array[groupidx]
> indices_2_remove_in_group_array = [...]
>
> And If I call the REMOVE routine
> REMOVE, indices_2_remove_in_group_array, huge_array[groupidx]
>
> Will this call remove the elements from the *huge_array* or will it
> remove them from a temporary array created when calling the REMOVE
> routine?
>
> I know that IDL passes references as arguments, but in this will it
> actually remove the elements from the original *huge_array* or not.
Actually, IDL passes *variables* by reference. Everything
else, including expressions like "huge_array[groupidx]", it
passes by value. So if you called REMOVE like this, you
would get no error messages, since it would work, but
you wouldn't know about it. :-)
Cheers,
David
--
David Fanning, Ph.D.
Fanning Software Consulting, Inc.
Coyote's Guide to IDL Programming: http://www.dfanning.com/
Sepore ma de ni thui. ("Perhaps thou speakest truth.")
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