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Re: memory allocation for structure arrays [message #11724 is a reply to message #11586]

Sat, 02 May 1998 00:00

mallors
Messages: 76
Registered: November 1997

Member

In article <3549F646.51FF035E@ngdc.noaa.gov>,
Ian Sprod <ian@ngdc.noaa.gov> writes:
>
> On a related point, when you do a :
>
> help,data_structure,/structure
>
> There is a field called "length". Is this the amount of memory in bytes
> allocated to the structure? I can't find this field defined in the
> hardcopy or online documentation.
> .
> .
> .
> test = {str1:'a'}
> help,test,/structure
> ** Structure <8194ba4>, 1 tags, length=8, refs=1:
>
> test = {str1:'aa',str2:'b'}
> help,test,/structure
> ** Structure <8194e0c>, 2 tags, length=16, refs=1:
>
> IDL seems to allocate 8 bytes per string field - no matter how
> long the string is initialized. I guess this is the "padding byte"
> problem Peter talks about in his email.

I was looking at the IDL Advanced Development Guide, and under the section
"Mapping of Basic Types", an IDL_STRING is defined as follows:

typedef struct {
unsigned short slen; /* Length of string */
short stype; /* Type of string */
char *s; /* Pointer to string */
} IDL_STRING;

I would guess this is how IDL defines its strings internally,
so your structure example above would make sense, that is, each
string would take 8 bytes: two for the length "slen", two for type
"stype", and then the 4-byte pointer to the string (on most machines,
but these are all machine-dependent).

For the other case you mention

> test = {long1:0l,byte2:0b,byte3:0b}
> help,test,/structure
> ** Structure <8194e5c>, 3 tags, length=8, refs=1:
>
> The length is 8 - I would expect 6 (4 + 1 + 1).

I think this has to do with how the structure is aligned in memory, as
mentioned by Peter. Some hint of this can be seen as follows:

1. test = {long: 0L, b1: 0B, b2: 0B} => length = 8
2. test = {long: 0B, b1: 0B, b2: 0L} => length = 8
3. test = {long: 0B, b1: 0L, b2: 0B} => length = 12

On a 32-bit machine (4-byte words), case 1 uses one
word to hold the long, and then another word to
hold both of the byte variables. Similarly, in case
2, first word for the two bytes, another word for
the long. But for case 3, the first word holds the
first byte. Then we need another word to hold the
long, because there is not enough room left in the
first word. Then a third word is needed for the
last byte, for a total of 3 words = 12 bytes.

-bob mallozzi

--
Robert S. Mallozzi
http://cspar.uah.edu/~mallozzir/

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[Message index]

		memory allocation for structure arrays By: Ian Sprod on Wed, 29 April 1998 00:00
		Re: memory allocation for structure arrays By: mallors on Sat, 02 May 1998 00:00
		Re: memory allocation for structure arrays By: Ian Sprod on Fri, 01 May 1998 00:00
		Re: memory allocation for structure arrays By: korpela on Thu, 30 April 1998 00:00
		Re: memory allocation By: Essa Yacoub on Tue, 27 July 1999 00:00
		Re: memory allocation By: Liam Gumley on Tue, 27 July 1999 00:00
		Re: memory allocation By: Craig Markwardt on Tue, 27 July 1999 00:00
		Re: memory allocation By: davidf on Mon, 26 July 1999 00:00
		Re: memory allocation By: Essa Yacoub on Mon, 26 July 1999 00:00
		Re: memory allocation By: davidf on Mon, 26 July 1999 00:00

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