Re: Strangeness when moving data through an array.
Re: Strangeness when moving data through an array.
- Subject: Re: Strangeness when moving data through an array.
- From: Sherm Pendley <email@hidden>
- Date: Wed, 1 Dec 2004 03:46:10 -0500
On Dec 1, 2004, at 2:36 AM, Andrew Farmer wrote:
On 30 Nov 2004, at 23:23, Sherm Pendley wrote:
On Nov 30, 2004, at 11:55 PM, John Draper wrote:
Now, I take the data from "pkt" and I want to later stick it into a
16 bit (u_short), and would expect... short_test=258, but
after I execute this statement...
bcopy((u_short)&pkt[MSEQ], &short_test, 2);
short_test = 160
WHY? What's going on? How can I extract data from an array in
any arbitrary
index into it (even or odd), and still get back that 2 byte int
correctly?
You're typecasting a pointer (&pkt[MSEQ]) to an integer (u_short),
which smells fishy.
I'd say you probably need to cast to a different pointer type:
bcopy((u_short *)&pkt[MSEQ], &short_test, 2);
Or do something rather simpler.
pkt[MSEQ] = short_test;
(or is there a reason you aren't doing it this way?)
That's backwards - he's assigning *to* short_test.
short_test = pkt[MSEQ];
But regardless, that kind of straightforward assignment would only work
if pkt[] happens to be an array of u_shorts. If it's an array of chars
(for example), it would take the char at index MSEQ, and assign the
value of that char to short_test.
But I don't think that's what John wants to do - I think he wants to
assign the two bytes pkt[MSEQ] and pkt[MSEQ+1] to a u_short. To do that
with a direct assignment, you'd need to take the address of the array
element you want, typecast that pointer to a u_short*, and then
dereference it:
short_test = *((u_short*)&pkt[MSEQ]);
That's a *little* simpler than the function call, but not a whole heck
of a lot.
sherm--
Cocoa programming in Perl: http://camelbones.sourceforge.net
Hire me! My resume: http://www.dot-app.org
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