| Re: saving variables between calls to a procedure? [message #31596 is a reply to message #31595] |
Thu, 01 August 2002 07:45   |
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Paul Van Delst[1]
Messages: 1157
Registered: April 2002
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Senior Member |
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Mark Hadfield wrote:
>
> "David Fanning" <david@dfanning.com> wrote in message
> news:MPG.17b21214ff7d9b35989944@news.frii.com...
>
>> Paul van Delst (paul.vandelst@noaa.gov) writes:
>>
>>>> pro define,ptr
>>>> *ptr=[*ptr,10]
>>>> end
>>>
>>> Hmm. That seems like an extremely dangerous thing to do - couldn't
>>> you clobber something by concatenating like that? If IDL is smart
>>> enough to recognise that the next bit of memory may be used by
>>> something else it then seems that you would end up with a
>>> non-contiguous data structure (in the figurative).
>>
>> This doesn't seem dangerous to me (perhaps because I use the
>> construct all the time). It seems like one of those wonderful things
>> IDL occasionally does that makes you think to yourself "Now, by God,
>> that's how software *ought* to work!"
>>
>> In any case, it works, over and over and over. And it never occurred
>> to me that non-contiguous data storage could be involved, even
>> remotely.
>
> Extending an array a with the a = [a,b] syntax doesn't create a
> non-contiguous data structure.
I'm not worried about extending arrays like a=[a,b], but pointers ala *ptr=[*ptr,10]. If
ptr is pointing to some data structure (scalar or array), it seems feasible to me that the
next position in memory could be occupied by something else. If one then does a
*ptr=[*ptr,10], what happens to what was in that memory? Is it "protected" somehow so that
the value "10" is then placed in the next "free" spot in memory giving a non-contiguous
array [not efficient, but safe], or are the contents clobbered and replaced by the value
"10" [very very bad], or are the contents moved to some other memory location so that the
original position can be used to store the value "10" [relatively efficient(?) and safe].
> What is does is create a new array a,
> insert the elements from the existing a and b into it, then delete the
> old a.
So does this mean that
*ptr = [*ptr,10]
is equiavalent to:
a = *ptr
ptr_free, ptr
a=[a,10]
ptr=ptr_new(temporary(a))
?
(Which was my original question, just poorly phrased)
paulv
--
Paul van Delst
CIMSS @ NOAA/NCEP/EMC Beer is good.
Ph: (301)763-8000 x7274 My wife.
Fax:(301)763-8545
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