x86 - Endianness inside CPU registers

Question

I need help understanding endianness inside CPU registers of x86 processors. I wrote this small assembly program:

section .data
section .bss

section .text
    global _start
_start:
    nop
    mov eax, 0x78FF5ABC
    mov ebx,'WXYZ'
    nop  ; GDB breakpoint here.
    mov eax, 1
    mov ebx, 0
    int 0x80

I ran this program in GDB with a breakpoint on line number 10 (commented in the source above). At this breakpoint, info registers shows the value of eax=0x78ff5abc and ebx=0x5a595857.

Since the ASCII codes for W, X, Y, Z are 57, 58, 59, 5A respectively; and intel is little endian, 0x5a595857 seems like the correct byte order (least significant byte first). Why isn't then the output for eax register 0xbc5aff78 (least significant byte of the number 0x78ff5abc first) instead of 0x78ff5abc?

Answer 1

Endianness inside a register makes no sense since endianness describes if the byte order is from low to high memory address or from high to low memory address. Registers are not byte addressable so there is no low or high address within a register. What you are seeing is how your debugger print out the data.