在内存中存储负数

Question

我试图从用户输入负数并将它们存储在寄存器中。这是我的代码 -

%include        'functions.asm'
SECTION .bss
b2:      RESB    4
SECTION .text
global  _start

_start:

    mov edx,4
    mov ecx,b2
    mov ebx,0
    mov eax,3
    int 80h
    mov eax,b2
    call atoi
    mov [b2],eax

这是我的atoi功能 -

;------------------------------------------
; int atoi(Integer number)
; Ascii to integer function (atoi)
atoi:
    push    ebx             ; preserve ebx on the stack to be restored after function runs
    push    ecx             ; preserve ecx on the stack to be restored after function runs
    push    edx             ; preserve edx on the stack to be restored after function runs
    push    esi             ; preserve esi on the stack to be restored after function runs
    mov     esi, eax        ; move pointer in eax into esi (our number to convert)
    mov     eax, 0          ; initialise eax with decimal value 0
    mov     ecx, 0          ; initialise ecx with decimal value 0

.multiplyLoop:
    xor     ebx, ebx        ; resets both lower and uppper bytes of ebx to be 0
    mov     bl, [esi+ecx]   ; move a single byte into ebx register's lower half
    cmp     bl, 48          ; compare ebx register's lower half value against ascii value 48 (char value 0)
    jl      .finished       ; jump if less than to label finished
    cmp     bl, 57          ; compare ebx register's lower half value against ascii value 57 (char value 9)
    jg      .finished       ; jump if greater than to label finished
    cmp     bl, 10          ; compare ebx register's lower half value against ascii value 10 (linefeed character)
    je      .finished       ; jump if equal to label finished
    cmp     bl, 0           ; compare ebx register's lower half value against decimal value 0 (end of string)
    jz      .finished       ; jump if zero to label finished

    sub     bl, 48          ; convert ebx register's lower half to decimal representation of ascii value
    add     eax, ebx        ; add ebx to our interger value in eax
    mov     ebx, 10         ; move decimal value 10 into ebx
    mul     ebx             ; multiply eax by ebx to get place value
    inc     ecx             ; increment ecx (our counter register)
    jmp     .multiplyLoop   ; continue multiply loop

.finished:
    mov     ebx, 10         ; move decimal value 10 into ebx
    div     ebx             ; divide eax by value in ebx (in this case 10)
    pop     esi             ; restore esi from the value we pushed onto the stack at the start
    pop     edx             ; restore edx from the value we pushed onto the stack at the start
    pop     ecx             ; restore ecx from the value we pushed onto the stack at the start
    pop     ebx             ; restore ebx from the value we pushed onto the stack at the start
    ret  

但是当我尝试显示b2中存储的值时，我得到以下值 - 1717986918。我该如何解决这个问题？