| Re: Float procedure [message #13742] |
Wed, 02 December 1998 00:00 |
eddie haskell
Messages: 29
Registered: September 1998
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Junior Member |
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Charlie Solomon wrote:
>
> Can anyone shed some light on how this byte array is converted into a
> floating point number?
> two_words = bytarr(4)
> two_words = [244, 232, 165, 64]
first off, the second assignment statement makes two_words an intarr,
not a bytarr as intended.
to keep two_words as a bytarr do something like:
two_words = byte([244,232,165,64]) -or maybe-
two_words[*] = [244, 232, 165, 64]
> IDL> print, float(two_words, 0)
> 2.13062e-038
when you use the offset in the float procedure, IDL takes the first 32
bits it finds from the point
specified in the offset, in this case, the beginning of the array.
you can see this if you look at the binary representation of the numbers
(i.e. using kevin ivory's
binary program):
IDL> print,binary(float(two_words,0))
0 0 0 0 0 0 0 0 1 1 1 1 0 1 0 0
0 0 0 0 0 0 0 0 1 1 1 0 1 0 0 0
IDL> print,binary(244),binary(232)
0 0 0 0 0 0 0 0 1 1 1 1 0 1 0 0
0 0 0 0 0 0 0 0 1 1 1 0 1 0 0 0
as to the difference between machines, that could be a big/little endian
thing but i really don't
know for sure (can anybody verify this?)
cheers,
eddie
----- ---- --- ---- --- --- --- --- -- -- -- - - - -
|\ A. G. Edward Haskell
|\ Center for Coastal Physical Oceanography
|\ Old Dominion University, Norfolk VA 23529
|\ Voice 757.683.4816 Fax 757.683.5550
|\ e-mail haskell@ccpo.odu.edu
----- ---- --- ---- --- --- --- --- -- -- -- - - - -
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| Re: Float procedure [message #13745 is a reply to message #13742] |
Wed, 02 December 1998 00:00  |
Liam Gumley
Messages: 473
Registered: November 1994
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Senior Member |
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Charlie Solomon wrote:
> Can anyone shed some light on how this byte array is converted into a
> floating point number?
> two_words = bytarr(4)
> two_words = [244, 232, 165, 64]
> IDL> print, two_words, format = '(z)'
> f4
> e8
> a5
> 40
> IDL> print, float(two_words, 0)
> 2.13062e-038
>
> On my computer at work (NT4, x86) I get a different value...5.18469
> Here at home (Win98, x86) I get this really small number....
> IDL> print, !version
> { x86 Win32 Windows 5.1.1 Jul 20 1998}
In addition to Alex Schuster's comments, the only suggestion I have is
to use the HELP routine to keep track of variable types, e.g.
IDL> two_words = bytarr(4)
IDL> help, two_words
TWO_WORDS BYTE = Array[4]
IDL> two_words = [244, 232, 165, 64]
IDL> help, two_words
TWO_WORDS INT = Array[4]
One of the most important things to learn in IDL programming is the
transient nature of variable types. Frequent use of HELP both on the
command line *and* within your IDL programs is a very effective way to
help ensure your IDL code works the way you intended.
Cheers,
Liam.
---
Liam E. Gumley
Space Science and Engineering Center, UW-Madison
1225 W. Dayton St., Madison WI 53706, USA
Phone (608) 265-5358, Fax (608) 262-5974
http://cimss.ssec.wisc.edu/~gumley
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| Re: Float procedure [message #13747 is a reply to message #13742] |
Wed, 02 December 1998 00:00 |
Alex Schuster
Messages: 124
Registered: February 1997
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Senior Member |
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Charlie Solomon wrote:
> Can anyone shed some light on how this byte array is converted into a
> floating point number?
> two_words = bytarr(4)
> two_words = [244, 232, 165, 64]
Caution! This statements creates an intarr(4). You could define
two_words this way:
IDL> two_words = [244b, 232b, 165b, 64b]
> IDL> print, float(two_words, 0)
> 2.13062e-038
I get this result:
IDL> f = float( two_words, 0 )
IDL> print, f
-1.47457e+32
This still isn't the value you expect, because your PC uses another byte
ordering scheme than my workstation. Changing this with the swap_endian
routine corrects this:
IDL> print, swap_endian( f )
5.18469
Alex
--
Alex Schuster Wonko@weird.cologne.de PGP Key available
alex@pet.mpin-koeln.mpg.de
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