| Re: renaming a variable without making a copy [message #68920 is a reply to message #68919] |
Wed, 09 December 2009 01:42   |
lecacheux.alain
Messages: 325
Registered: January 2008
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Senior Member |
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On 8 déc, 23:39, David Fanning <n...@dfanning.com> wrote:
> Kenneth P. Bowman writes:
>> In article <MPG.2587fff16b170da3989...@news.giganews.com>,
>> David Fanning <n...@dfanning.com> wrote:
>
>>> newName = Temporary(oldName)
>
>> Can anyone explain to me what TEMPORARY actually does? The documentation
>> says
>
>> The TEMPORARY function returns a temporary copy of a variable, and sets
>> the original variable to "undefined".
>
>> which makes no sense to me at all. Doesn't making a "temporary copy
>> of a variable" occupy memory? Perhaps I am confused by the use of the name
>> "TEMPORARY".
>
>> My concept of an IDL variable (which could easily be wrong) is: some
>> metadata that describes the variable (what you get with the SIZE function)
>> and the actual data that comprises the variable. These things could be
>> in different places in memory, and the metadata could contain, for example,
>> a pointer to the actual data. Most of the time, I don't need to know.
>
>> Does TEMPORARY wipe out the old metadata (replacing it with
>> "undefined") and create new "unnamed" metadata that points to the data part
>> of the destroyed variable?
>
>> The example in the Docs is not very revealing.
>
> Here is how I wave my hands around this when explaining it
> in an IDL class. Remember, I am speaking metaphorically here.
> I have *no* idea what actually happens. ;-)
>
> You are right, a variable in IDL is composed of some metadata,
> one part of which is the variable's name, and some machine
> memory, where the variable lives. I like to say the variable
> is "attached" to the machine memory. When you issue a command
> like this:
>
> newVar = Temporary(oldVar) + 1
>
> You are saying to IDL, "Take that machine memory that is attached
> to oldVar and temporarily use it to perform whatever operation
> you are doing." Then, when you are finished, make another variable,
> newVar, and attach this temporary memory permanently to this variable.
> In IDL there is a rule that only one variable at a time can be
> permanently attached to machine memory, so the act of attaching this
> memory to newVar is to remove it from oldVar. A variable that has
> no machine memory attached to it is, by definition, an undefined
> variable.
>
>> Why does
>
>> A = TEMPORARY(A) + 1
>
>> use less memory than
>
>> A = A + 1
>
>> I suppose there is a good reason that the latter example "creates a new
>> array for the result of the addition, places the sum into the new array,
>> assigns it to A, and then frees the old allocation of A", although it
>> just seems to me like the interpreter is being obtuse.
>
> I'm sure there is a good reason. And if I think about it long
> enough, I'm sure it will come to me. Meantime, you may have
> to take it on faith that IDL just works that way. :-)
>
> 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.")- Masquer le texte des messages précédents -
>
Here is a question, maybe related to the previous one, regarding
variable typecast in IDL.
Changing the type of one scalar variable, even a vector, is fairly
easy by using the FIX function.
But let suppose that I get a variable (for instance from an external
routine, by reading a shared memory,
a socket, etc...) that is described as a vector of byte (or integer or
float or anything else).
I want to further consider this variable as a (elsewhere defined)
structure (in the IDL sense).
In other words, I want to typecast an untyped variable to a structured
one : what is the way in IDL ?
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