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Re: MOD operator [message #22034]

Tue, 17 October 2000 00:00

Phillip David
Messages: 36
Registered: April 1999

Member

Theo Brauers wrote:
>
> I was wondering if the result of the MOD operator in IDL changed
> from previous versions (4.x, 5.0) to the current version (5.3.1).
> Now the output is:
>
> IDL> PRINT, (FINDGEN(8)-4.) MOD 3
> -1.00000 0.000000 -2.00000 -1.00000 0.000000
> 1.00000 2.00000 0.000000
>
> When I programmed a function long ago I used the MOD operator
> expecting the output
>
> IDL> PRINT, (FINDGEN(8)-4.) MOD 3
> 2.00000 0.000000 1.00000 2.00000 0.000000
> 1.00000 2.00000 0.000000
>

To the best of my knowledge, this is always the way IDL (and most other
languages providing a MOD operator) has worked. If you wish to take the
mod of a negative number , the result will be a negative number between
-(n+1) and 0. A positive input yields a positive result. So when we're
trying to limit a value to between 0 and 2*pi, we often end up with code
that looks something like:

x = ((y mod (2*pi)) + 2*pi) mod (2*pi)

Hope this helps.

Phillip

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[Message index]

		Re: MOD operator By: Phillip David on Tue, 17 October 2000 00:00
		Re: MOD operator By: arkos on Tue, 17 October 2000 00:00
		Re: MOD operator By: Alex Schuster on Tue, 17 October 2000 00:00

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