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Posted: **Fri Oct 03, 2014 9:51 pm**

Hi guys!
I'm trying to write functions that set up gdt in c.
My definition of segment descriptor is:

typedef struct
{
    uint16 limit_low;
    uint16 base_low;
    uint8 base_middle;
    uint8 access;
    uint8 limit_mid_flag;
    uint8 base_high;
} gdt_descriptor;

My problem is about the first uint16 field.
I know 80386 is little-endian. So when writing a 16-bit value to mem(let's say 0x1234), the actual data in ram would be

Code: Select all

Addr Data
0x0 0x34
0x1 0x12

Suppose I have a limit of 0x123456, I assign 0x1234 to limit_low and if you look at the binary of bits 0-15 of segment descriptor, it should be

Code: Select all

Bit   Data
0-7   0x34 
8-15  0x12.

But intel seems to expect this:

Code: Select all

Bit   Data
0-7   0x12 
8-15  0x34.

I've seen fair amount of tutorials. Some of them do 0x12 0x34 and the others do 0x34 0x12.
So I'm confused here:
Should it be 0x12 0x34 or 0x34 0x12(little-endian form)?
Thanks!

Posted: **Fri Oct 03, 2014 10:10 pm**

Hi,

HyperAssembler wrote:Should it be 0x12 0x34 or 0x34 0x12(little-endian form)?

Um, what?

The lowest 16-bits of 0x123456 would be 0x3456 (not 0x1234); which would be stored in little-endian as the bytes 0x56, 0x34.

Cheers,

Brendan

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Segment descriptor and endianness

Segment descriptor and endianness

Re: Segment descriptor and endianness