At 06:28 AM 9/5/2001 +0000, manish jaggi wrote:
>I am facing a adddress arror when i try executing this code
>please help me out
>#pragma pack (1)
> struct Record
>{
> char A[25];
> Int16 B;
> Int16 C;
> char T;
>
>};
[...]
> Record *rec = new Record;
[...]
> int x = rec->B;----------------here the problem is
I don't know why compiler vendors don't warn you about pragmas
that can get you in trouble.
Of course this causes an address error. Palms have Motorola
CPUs, which have two-byte alignment requirements, i.e., any
memory access for a two-byte or larger value (word, double word,
float, or pointer) must occur at an even address. In your
example, the "new Record" call has produced an even address for
"rec"; the #pragma directive has insured that no padding occurs
in struct Record, so the address of rec->B is exactly 25 greater
than the address of B itself. That puts rec->B at an odd address,
and you're fetching a 16-bit value from that address. Kablooey.
You can get rec->B with a little work:
UInt8* p = (UInt8*)&rec->B;
x = p[1] | (p[0] << 8);
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