具有段和偏移量的推入寄存器的含义

Question

i want to create MBR with assembly language . But i have code , give by somebody that i don't understand . When you write

push ax
push bx

What is the meaning , when we use segment and offset

For more details, read this code :

    [BITS 16]
[ORG 0x0]

mov ax, 0x07C0
mov ds , ax
mov es , ax
mov ax , 0x8000
mov ss , ax
mov sp , 0xf000

mov si ,msgDebut
call afficher

end :
    jmp end

msgDebut db "HelloWorld!!!!",13,10,0

afficher :
    push ax
    push bx

.debut :
    lodsb
    cmp al ,0
    jz .fin
    mov ah , 0x0E
    mov bx , 0x07
    int 0x10
    jmp .debut

.fin :
    pop bx
    pop ax
    ret

times 510-($-$$) db 144
    dw 0xAA55

can you explain me this code line by line using memory addressing, segmentation, and offset abstraction

Answer 1

The push instruction places its operand on the stack for conservation.
The stack is an area of memory set aside for this purpose.

The stackpointer determines where the data will be stored in memory. When the stackpointer SS:SP is equal to 0x8000:0xEFFE, push ax will first lower the stackpointer by 2 so it becomes 0x8000:0xEFFC and then write the contents of the AX register in that memory address.
Hereafter push bx will again lower the stackpointer by 2 so it now becomes 0x8000:0xEFFA and then write the contents of the BX register in that memory address.

The pop instruction reverses the actions of push . You use these to restore the registers to their original values.

afficher :
    push ax   ; Preserve AX
    push bx   ; Preserve BX
    ...
    pop bx    ; Restore BX
    pop ax    ; Restore AX
    ret