[Lazarus] C Integer types

Bernd K. prof7bit at gmail.com
Sun Jul 8 21:39:46 CEST 2012

On 08.07.2012 17:18, Hans-Peter Diettrich wrote:
> Bernd schrieb:
>> 2012/7/7 Marco van de Voort <marcov at stack.nl>:
> IMO all these members should be pointers, of size 64 bit according to 
> the index increment (8 bytes?). Typecasts will be needed for 32 bit 
> pointers, as demonstrated in above code.

Wait a monment, I have again looked at it. I think I was right and my
other email admitting that I was wrong was wrong.

seqence_crc = *((int64_t*)data);
seqence_key = *((int64_t*)&data[8]);

from the inside out:
data is a pointer, say it points to a buffer
(int64_t*)data still points to that buffer
*((int64_t*)data) are bytes 0..7 of that buffer as int64.

now the second line

seqence_key = *((int64_t*)&data[8]);

data[8] is the same as *(data+8)
data[8] represents the 8th byte of that buffer (already dereferenced!)
&data[8] is the address of that byte
&data[8] would be the same as (data+8)
((int64_t*)&data[8]) is still the same pointer
*((int64_t*)&data[8]) are bytes 8..17 of that buffer as int64

this should be equivalent:
seqence_crc = *((int64_t*)&data[0]);
seqence_key = *((int64_t*)&data[8]);
compr_crc   = *((int64_t*)&data[16]);
compr_len   = *((int32_t*)&data[24]);

and this also:
seqence_crc = *((int64_t*)data);
seqence_key = *((int64_t*)(data+8));
compr_crc   = *((int64_t*)(data+16));
compr_len   = *((int32_t*)(data+24));

Its really just a simple record with 4 numbers in it.


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