| renaming a variable without making a copy [message #68935] |
Tue, 08 December 2009 03:51  |
David Higgins
Messages: 13
Registered: August 2009
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Junior Member |
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Can I rename a variable to a more approrpriate name, instead of making
a copy of the variable
I'm using /OVERWRITE to save memory, through a number of operations on
a large multi-dimensional array. I'd like to keep my code readable and
rename the variable to reflect what the data currently are.
Thanks,
Dave
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| Re: renaming a variable without making a copy [message #68975 is a reply to message #68935] |
Thu, 10 December 2009 08:56  |
lecacheux.alain
Messages: 325
Registered: January 2008
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Senior Member |
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On 10 déc, 17:30, alx <lecacheux.al...@wanadoo.fr> wrote:
> On 10 déc, 16:36, Reimar Bauer <R.Ba...@fz-juelich.de> wrote:
>
>> alx schrieb:
>
>> IDL> a=findgen(2,2)
>> IDL> b=dblarr(4)
>> IDL> reads,a,b
>> IDL> print,b
>> 0.0000000 1.0000000 2.0000000 3.0000000
>
>> btw. just seen that reads, 1 seems to be valid, but what does it do?
>
>
Unfortunately, that does not work as you said: "reads" (cf. IDL doc)
converts
your 'a' array to an array of strings, then 'read' it to 'b' by using
implicit formatting.
That is actually not a "memory read" and, likely, does not work with
structures.
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| Re: renaming a variable without making a copy [message #68976 is a reply to message #68935] |
Thu, 10 December 2009 08:30 |
lecacheux.alain
Messages: 325
Registered: January 2008
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Senior Member |
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On 10 déc, 16:36, Reimar Bauer <R.Ba...@fz-juelich.de> wrote:
> alx schrieb:
>
> IDL> a=findgen(2,2)
> IDL> b=dblarr(4)
> IDL> reads,a,b
> IDL> print,b
> 0.0000000 1.0000000 2.0000000 3.0000000
>
> btw. just seen that reads, 1 seems to be valid, but what does it do?
>
I did not realize so far that 'reads' could actually mean "read from
memory". Thanks for the tip !
Here is an immediate application: you want to import some structure
from a COM object, written in C or whatsoever.
But the IDL COM bridge does not support passing of structure. So you
need to import it as a typecasted (variant) array
of simple type. Then, "reads" will allow to recover and use the
structure in IDL...
alain.
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| Re: renaming a variable without making a copy [message #68977 is a reply to message #68935] |
Thu, 10 December 2009 07:36 |
R.Bauer
Messages: 1424
Registered: November 1998
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Senior Member |
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alx schrieb:
> On 9 d�c, 18:06, Reimar Bauer <R.Ba...@fz-juelich.de> wrote:
>
>> read about reads or execute
>>
>
> Reimar,
> I understand that 'reads' might allow a string to be read out as a
> structure by using a binary format, but such a typecast method seems
> to me quite a bit involved.
the "s" it is not only about strings
IDL> a=findgen(2,2)
IDL> b=dblarr(4)
IDL> reads,a,b
IDL> print,b
0.0000000 1.0000000 2.0000000 3.0000000
btw. just seen that reads, 1 seems to be valid, but what does it do?
> I do not understand the usage of 'execute' for such a thing.
> alain.
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| Re: renaming a variable without making a copy [message #69146 is a reply to message #68935] |
Fri, 11 December 2009 08:38 |
penteado
Messages: 866
Registered: February 2018
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Senior Member
Administrator |
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On Dec 11, 11:22 am, alx <lecacheux.al...@wanadoo.fr> wrote:
> Look at the following:
>
> define some structure:
> IDL> a={n:1L,x:2.0}
>
> build its binary content as a byte chain:
> IDL> b=[byte(a.n,0,4),byte(a.x,0,4)]
> IDL> print,b
> 1 0 0 0 0 0 0 64
>
> define a void structure to get the result:
> IDL> aa={n:0L,x:0.0}
>
> then :
> IDL> reads,b,aa
> IDL> print,aa
> { 1 0.000000}
> does not provide the right result.
Good example. In case anybody is wondering why that happens, the reads
is getting the first field of a from the string "1", and the second
field from the string "0" (the string representations of the first and
second elements of the byte array b).
To properly obtain aa from the bytes in b, it is necessary to know
exactly how the fields are aligned, as David mentioned. Which in this
particular case would be:
IDL> aa.n=long(b,0)
IDL> aa.x=float(b,4)
IDL> print,aa
{ 1 2.00000}
Somebody also talked about using (real) pointers in IDL. Ronn Kling's
DLM book has a useful trick (by Nigel Wade), to use a memory copy to
hold a pointer's value in an IDL byte array of the proper size.
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| Re: renaming a variable without making a copy [message #69158 is a reply to message #68935] |
Fri, 11 December 2009 05:22 |
lecacheux.alain
Messages: 325
Registered: January 2008
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Senior Member |
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On 11 déc, 12:51, Reimar Bauer <R.Ba...@fz-juelich.de> wrote:
> what is that doing then?
> IDL> a=bindgen(3)
> IDL> b=ptr_new({s:intarr(3)})
> IDL> reads,a,*b
> IDL> print,*b
> { 0 1 2
Still the same: 'a' is implicitely converted to some string array by
'reads', then read out to *b.
Look at the following:
define some structure:
IDL> a={n:1L,x:2.0}
build its binary content as a byte chain:
IDL> b=[byte(a.n,0,4),byte(a.x,0,4)]
IDL> print,b
1 0 0 0 0 0 0 64
define a void structure to get the result:
IDL> aa={n:0L,x:0.0}
then :
IDL> reads,b,aa
IDL> print,aa
{ 1 0.000000}
does not provide the right result.
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| Re: renaming a variable without making a copy [message #69160 is a reply to message #68975] |
Fri, 11 December 2009 03:51 |
R.Bauer
Messages: 1424
Registered: November 1998
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Senior Member |
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alx schrieb:
> On 10 d�c, 17:30, alx <lecacheux.al...@wanadoo.fr> wrote:
>> On 10 d�c, 16:36, Reimar Bauer <R.Ba...@fz-juelich.de> wrote:
>>
>>> alx schrieb:
>>> IDL> a=findgen(2,2)
>>> IDL> b=dblarr(4)
>>> IDL> reads,a,b
>>> IDL> print,b
>>> 0.0000000 1.0000000 2.0000000 3.0000000
>>> btw. just seen that reads, 1 seems to be valid, but what does it do?
>>
> Unfortunately, that does not work as you said: "reads" (cf. IDL doc)
> converts
> your 'a' array to an array of strings, then 'read' it to 'b' by using
> implicit formatting.
> That is actually not a "memory read" and, likely, does not work with
> structures.
>
>
what is that doing then?
IDL> a=bindgen(3)
IDL> b=ptr_new({s:intarr(3)})
IDL> reads,a,*b
IDL> print,*b
{ 0 1 2
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