测试.S

.text
.global _start
    _start:
        xor %ax, %ax
        mov %ax, %ds
        mov %ax, %ss
        mov %ax, %es
        mov %ax, %fs
        mov %ax, %gs

我这样做得到了反汇编代码文件

$ x86_64-elf-gcc -g -c -O0 -m32 -fno-pie -fno-stack-protector -fno-asynchronous-unwind-tables .\test.S
$ x86_64-elf-ld .\test.o -m elf_i386  -Ttext=0x7c00 -o test.elf
$ x86_64-elf-objdump -x -d -S -m i386 ./test.elf > test_dis.txt

test_dis.txt

./test.elf:     file format elf32-i386
./test.elf
architecture: i386, flags 0x00000112:
EXEC_P, HAS_SYMS, D_PAGED
start address 0x00007c00

Program Header:
    LOAD off    0x00000000 vaddr 0x00007000 paddr 0x00007000 align 2**12
         filesz 0x00000c0d memsz 0x00000c0d flags r-x

Sections:
Idx Name          Size      VMA       LMA       File off  Algn
  0 .text         0000000d  00007c00  00007c00  00000c00  2**0
                  CONTENTS, ALLOC, LOAD, READONLY, CODE
  1 .debug_aranges 00000020  00000000  00000000  00000c10  2**3
                  CONTENTS, READONLY, DEBUGGING
  2 .debug_info   00000049  00000000  00000000  00000c30  2**0
                  CONTENTS, READONLY, DEBUGGING
  3 .debug_abbrev 00000014  00000000  00000000  00000c79  2**0
                  CONTENTS, READONLY, DEBUGGING
  4 .debug_line   0000003b  00000000  00000000  00000c8d  2**0
                  CONTENTS, READONLY, DEBUGGING
SYMBOL TABLE:
00007c00 l    d  .text  00000000 .text
00000000 l    d  .debug_aranges 00000000 .debug_aranges
00000000 l    d  .debug_info    00000000 .debug_info
00000000 l    d  .debug_abbrev  00000000 .debug_abbrev
00000000 l    d  .debug_line    00000000 .debug_line
00007c00 g       .text  00000000 _start
00008c0d g       .text  00000000 __bss_start
00008c0d g       .text  00000000 _edata
00008c10 g       .text  00000000 _end


Disassembly of section .text:

00007c00 <_start>:
.text
.global _start
    _start:
        xor %ax, %ax
    7c00:   66 31 c0                xor    %ax,%ax
        mov %ax, %ds
    7c03:   8e d8                   mov    %eax,%ds
        mov %ax, %ss
    7c05:   8e d0                   mov    %eax,%ss
        mov %ax, %es
    7c07:   8e c0                   mov    %eax,%es
        mov %ax, %fs
    7c09:   8e e0                   mov    %eax,%fs
    7c0b:   8e e8                   mov    %eax,%gs

我的问题
我想知道为什么我得到了像mov %eax,%ds这样的代码，它不适合我的原始汇编代码？
我的期望
我认为mov %eax,%ds应该是mov %ax,%ds，并且我认为%eax（32位）不适合%ds（16位）