On 28 Jan 2009, at 2:24 am, Jeremy Pereira wrote:
Yes. That is correct, but since buffer is already a pointer to the
first byte of the array and then you are taking a reference to it,
key will end up containing the address of the buffer. You really
need:
uint key = *(uint*)buffer;
That's incorrect. Or rather, in this case it doesn't make any
difference, but the use of '&' is more general as it works whether
buffer is an array or not.
When you declare:
uint8 buffer[8];
You've reserved 8 bytes on the stack. Whether you use 'buffer' or
'&buffer' you obtain the same address, as C treats array variables as
pointers. But:
uint8 buffer;
You get very different results from 'buffer' and '&buffer' so in the
interests of defensive programming, using '&' is allowing your code to
tolerate a change to the buffer declaration without breaking. Because
of the additional cast *(int*) such a change would go unnoticed by the
compiler but probably have a very bad outcome at runtime. It's a good
habit IMO to always take the address in these cases to make it clear
in your code what your intentions were when you wrote it.
--Graham
_______________________________________________
Cocoa-dev mailing list (Cocoa-dev@lists.apple.com)
Please do not post admin requests or moderator comments to the list.
Contact the moderators at cocoa-dev-admins(at)lists.apple.com
Help/Unsubscribe/Update your Subscription:
http://lists.apple.com/mailman/options/cocoa-dev/archive%40mail-archive.com
This email sent to arch...@mail-archive.com
