| Re: Could you explain why this happen? [message #70686] |
Thu, 29 April 2010 03:57 |
Carsten Lechte
Messages: 124
Registered: August 2006
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Senior Member |
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David Fanning wrote:
> An IDL start-up file is simply a text file containing
> IDL commands of the sort you would type at the IDL
> command line. You point to it via an environment
> variable named IDL_STARTUP.
Note however, that COMPILE_OPT only affects the local scope, i.e. you have
to put it into every procedure and function definition, not just into the
startup file, or you will be wondering if the sky is falling. Example:
IDL Version 6.4 (linux x86 m32). (c) 2007, ITT Visual Information Solutions
IDL> print, 32000*6
-4608
IDL> compile_opt defint32
IDL> print, 32000*6
192000
IDL> .comp
- pro test
- print, 32000*6
- end
% Compiled module: TEST.
IDL> test
-4608
IDL> print, 32000*6
192000
IDL>
chl
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| Re: Could you explain why this happen? [message #70704 is a reply to message #70686] |
Wed, 28 April 2010 12:26  |
David Fanning
Messages: 11724
Registered: August 2001
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Senior Member |
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Bryan writes:
> Thank you for your kind response. Then, could you let me know how to
> set the IDL start-up file like you?
An IDL start-up file is simply a text file containing
IDL commands of the sort you would type at the IDL
command line. You point to it via an environment
variable named IDL_STARTUP.
Here is how you might define it in a UNIX environment:
http://www.dfanning.com/misc_tips/idlsetup.html
In a Windows environment, you define it via a Windows
Preference dialog.
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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| Re: Could you explain why this happen? [message #70705 is a reply to message #70704] |
Wed, 28 April 2010 10:54 |
bryan.s.hong
Messages: 12
Registered: November 2008
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Junior Member |
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On 4월28일, 오후12시06분, David Fanning <n...@dfanning.com> wrote:
> Bryan writes:
>> This may be a stupid question, but I really want to know why.
>> Please, see below and explain. Thanks.
>
>> IDL> print, 132*30
>> 3960
>> IDL> print, 132*30*10
>> -25936
>> IDL> print, 132*300
>> -25936
>> IDL> data1 = 132
>> IDL> data2 = 300
>> IDL> data3 = data1*data2
>> IDL> print, data3
>> -25936
>
> Here is a clue. These are the same IDL commands in my IDL
> session:
>
> IDL> print, 132*30
> 3960
> IDL> print, 132*30*10
> 39600
> IDL> print, 132*300
> 39600
> IDL> data1 = 132
> IDL> data2 = 300
> IDL> data3 = data1*data2
> IDL> print, data3
> 39600
>
> Probably the only difference between your IDL session and mine
> is that I have this command in my IDL startup file:
>
> compile_opt defint32
>
> Why would that matter?
>
> 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.")
Thank you for your kind response. Then, could you let me know how to
set the IDL start-up file like you?
Thanks.
Bryan
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| Re: Could you explain why this happen? [message #70706 is a reply to message #70705] |
Wed, 28 April 2010 10:06 |
David Fanning
Messages: 11724
Registered: August 2001
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Senior Member |
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Bryan writes:
> This may be a stupid question, but I really want to know why.
> Please, see below and explain. Thanks.
>
> IDL> print, 132*30
> 3960
> IDL> print, 132*30*10
> -25936
> IDL> print, 132*300
> -25936
> IDL> data1 = 132
> IDL> data2 = 300
> IDL> data3 = data1*data2
> IDL> print, data3
> -25936
Here is a clue. These are the same IDL commands in my IDL
session:
IDL> print, 132*30
3960
IDL> print, 132*30*10
39600
IDL> print, 132*300
39600
IDL> data1 = 132
IDL> data2 = 300
IDL> data3 = data1*data2
IDL> print, data3
39600
Probably the only difference between your IDL session and mine
is that I have this command in my IDL startup file:
compile_opt defint32
Why would that matter?
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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| Re: Could you explain why this happen? [message #70707 is a reply to message #70706] |
Wed, 28 April 2010 10:04 |
Michael Galloy
Messages: 1114
Registered: April 2006
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Senior Member |
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On 4/28/10 10:46 AM, Bryan wrote:
> This may be a stupid question, but I really want to know why.
> Please, see below and explain. Thanks.
>
> IDL> print, 132*30
> 3960
> IDL> print, 132*30*10
> -25936
> IDL> print, 132*300
> -25936
> IDL> data1 = 132
> IDL> data2 = 300
> IDL> data3 = data1*data2
> IDL> print, data3
> -25936
The default integer in IDL is a 16-bit signed integer (i.e., INT), range
-32768 to 32767. Your calculation involved two of these 16-bit integers,
so IDL calculated a result which was a 16-bit integer, wrapping around
to negative values in the process:
IDL> help, 132
<Expression> INT = 132
IDL> help, 300
<Expression> INT = 300
IDL> help, 132 * 300
<Expression> INT = -25936
The way to fix this to get the "correct" result is to make one or both
of the operands in the operation be of type long, making the result of
type long (range -2^31 to 2^31 - 1):
IDL> help, 132L * 300L
<Expression> LONG = 39600
Mike
--
www.michaelgalloy.com
Research Mathematician
Tech-X Corporation
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| Re: Could you explain why this happen? [message #70708 is a reply to message #70707] |
Wed, 28 April 2010 10:01 |
jeanh
Messages: 79
Registered: November 2009
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Member |
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On 28/04/2010 12:46 PM, Bryan wrote:
> This may be a stupid question, but I really want to know why.
> Please, see below and explain. Thanks.
>
> IDL> print, 132*30
> 3960
> IDL> print, 132*30*10
> -25936
> IDL> print, 132*300
> -25936
> IDL> data1 = 132
> IDL> data2 = 300
> IDL> data3 = data1*data2
> IDL> print, data3
> -25936
Hi,
you are working with integers, the maximum value being 32 767. When you
go over this limit, IDL goes back to the minimum value (-32 768). So 32
767 + 1 = -32 768
Use long integers instead, and you will have the value you are expecting!
Jean
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