| Re: size of a structure [message #7977 is a reply to message #7905] |
Wed, 29 January 1997 00:00  |
Peter Mason
Messages: 145
Registered: June 1996
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Senior Member |
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On Mon, 27 Jan 1997, Phil Williams wrote:
> 1) What is the "length" of a struture in the help,/st mean? It's not
> the "size" which is what I thought it was. i.e.
> IDL> help,/st,info
< ... structure with several string fields & some other fields, with
< help,/struct showing length=124 >
< ... code showing that structure written to disk is 162 bytes long >
It seems that the structure's "length" is correct except when it comes to
structure members which are strings. A string member contributes 16
bytes to the structure's "length", regardless of string length.
This is probably because strings are dynamic, while everything else about
an IDL structure's size (#members, member datatypes and dimensions) is
static: the 16 bytes shown for a string member is probably some sort of
string descriptor with a pointer to the actual string.
> 2) Is there a way, besides brute force, to determine the size of a
> structure w/in a procedure?
Not that I know.
But brute force is often OK, especially for a computer (which is by nature
a brute :)
e.g.,
; Return the size of variable V in bytes
forward_function sizeof ;make sure our recursion will work
function sizeof,v
j=size(v) &t=j(j(0)+1) &n=j(j(0)+2) &j=0
case t of
0:rv=0L
1:rv=1L
2:rv=2L
3:rv=4L
4:rv=4L
5:rv=8L
6:rv=8L
7:begin ;string
rv=0L &for i=0L,n-1L do rv=rv+strlen(v(i)) &n=1L
end
8:begin ;structure
rv=0L &nt=n_tags(v(0))-1L
for ii=0L,n-1L do begin ;loop here because of string members
vv=v(ii)
for i=0,nt do rv=rv+sizeof(vv.(i)) ;loop thru members
endfor
n=1L
end
9:rv=16L
else:rv=0L
endcase
return,rv*n
end
Peter Mason
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