| assignment inside boolean expression [message #20613] |
Mon, 10 July 2000 00:00  |
Patrick Broos
Messages: 27
Registered: December 1996
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Junior Member |
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I was wondering if it's common knowlege that one can put an IDL
assignment inside
a boolean expression (like in the C language). For example
if (v = 0) then ... assigns v and does not execute the "then"
statement, while
if (v = 1) then ... assigns v and does execute the then.
Just as in C I find this leads to really nasty bugs.
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Patrick S. Broos, Systems Analyst/Programmer patb@astro.psu.edu
Department of Astronomy and Astrophysics My office 814-863-7947
Penn State University
525 Davey Lab FAX 814-863-8686
University Park, PA 16802-6305
http://www.astro.psu.edu Group office 863-9550
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| Re: assignment inside boolean expression [message #20709 is a reply to message #20613] |
Tue, 18 July 2000 00:00  |
Martin Schultz
Messages: 515
Registered: August 1997
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Senior Member |
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Patrick Broos wrote:
>
> I was wondering if it's common knowlege that one can put an IDL
> assignment inside
> a boolean expression (like in the C language). For example
>
> if (v = 0) then ... assigns v and does not execute the "then"
> statement, while
> if (v = 1) then ... assigns v and does execute the then.
>
> Just as in C I find this leads to really nasty bugs.
>
> --
One application where I am actually using this is token search in
strings. It
can save you a few keystrokes and (in my opinion) make code more
readable, because
the temporary variable p is only used within the IF construct:
if ((p=StrPos(test,'Martin')) ge 0) then $
tail = StrMid(test,p,999)
as opposed to
p = StrPos(test,'Martin')
if p ge 0 then tail = StrMid(test,p,999)
--
[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[ [[[[[[[
[[ Dr. Martin Schultz Max-Planck-Institut fuer Meteorologie [[
[[ Bundesstr. 55, 20146 Hamburg [[
[[ phone: +49 40 41173-308 [[
[[ fax: +49 40 41173-298 [[
[[ martin.schultz@dkrz.de [[
[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[ [[[[[[[
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| Re: assignment inside boolean expression [message #20710 is a reply to message #20613] |
Tue, 18 July 2000 00:00 |
Martin Schultz
Messages: 515
Registered: August 1997
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Senior Member |
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Ben Tupper wrote:
>
> Craig Markwardt wrote:
>
>> Not to undercut you, but will (X AND 1) do the trick?
>>
>
> Thanks to Ken and Craig. I think for my purposes the following should suffice (I
> guess as long as I make sure that I'm working with an integer/long/byte type.)
>
> X = Indgen(6) - 2
>
> For i = 0, N_elements(X)-1 Do $
> If X then Print, X[i], ': Odd' Else print, X[i], ': Even'
>
> -2: Even
> -1: Odd
> 0: Even
> 1: Odd
> 2: Even
> 3: Odd
>
> Thanks again,
>
> Ben
but if you start increasing the number of elements of X to say 1000000,
you are
certainly better off with:
answer=['even','odd']
print,answer[ (x and 1) ]
no loop ;-)
Example:
IDL> x=lindgen(20)-5
IDL> answer=['even','odd']
IDL> print,answer[x and 1]
odd even odd even odd even odd even odd even odd even odd even odd even
odd even odd even
BTW: X MOD 2 does not work for negative numbers !!!
IDL> print,answer[x mod 2]
even even even even even even odd even odd even odd even odd even odd
even odd even odd even
Cheers,
Martin
--
[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[ [[[[[[[
[[ Dr. Martin Schultz Max-Planck-Institut fuer Meteorologie [[
[[ Bundesstr. 55, 20146 Hamburg [[
[[ phone: +49 40 41173-308 [[
[[ fax: +49 40 41173-298 [[
[[ martin.schultz@dkrz.de [[
[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[ [[[[[[[
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