x86汇编-利用栈反转内存数据.md

x86汇编-利用栈反转内存数据

assume cs:codeseg

codeseg segment

    dw 1h, 2h, 3h, 4h, 5h, 6h, 7h, 8h, 9h
    dw 9 dup(?)
    
start:
    ; 设置栈段
    mov ax, cs
    mov ss, ax
    ; 36 = 9 * 2 * 2
    mov sp, 36

    sub bx, bx
    mov cx, 9
s_push:
    mov ax, cs:[bx]
    push ax
    add bx, 2
    loop s_push

    sub bx, bx
    mov cx, 9
s_pop:
    pop cs:[bx]
    add bx, 2
    loop s_pop

    mov ah, 4ch
    int 21h

codeseg ends

end start