assembly - 当我从命令行（ubuntu nasm 32bit）获取字符串时，如何将字符串转换为整数？

Question

%include "asm_io.inc"

                 segment .data
                 segment .bss
argument         resb 32                    ; argument[32]


                 segment .text
                 global main
main:

                 enter 0,0

                 mov ebx, 4
                 mov ecx, dword [ebp + 4 * ebx]
                 mov eax, dword [ecx + 4]

                 mov [argument], eax

                 mov al, byte [argument + 0]
                 sub al, 48

                 call print_int          




end:             leave
                 mov eax, 0
                 ret

我正在尝试将字符串从命令行转换为整数（例如，当我在命令行键入 $./Hello 30 时，'30' 必须是程序（过程）的整数参数）。在找到 [argument + 0] == '30', [argument + 1] = bin/bash 之后，我想我可以用 [argument + 0] 得到正确的数字。但结果就像-1075579555。

即使是一点点评论，也会很有帮助。谢谢你

标签： assemblyx86nasm

Answer 1

当您的程序启动时，堆栈的布局如下

       0
       Address of last environment string
       ...
       Address of 2nd environment string
       Address of 1st environment string
       0
       Address of last program argument
       ...
       Address of 2nd program argument
       Address of 1st program argument
       Address of program path
ESP -> Number of arguments

对于您的任务，您需要获取第一个程序参数的地址。
您会在dword [esp + 8].

在执行之后enter 0,0并且因为该指令推送了EBP寄存器，你会在dword [esp + 12]==找到它dword [ebp + 12]。

enter   0,0                   ; Same as PUSH EBP : MOV EBP, ESP

mov     ebx, dword [ebp + 12] ; Address of first argument
movzx   eax, byte [ebx]       ; First character, is e.g. "3"
sub     al, "0"               ; Convert from "3" -> 3
imul    eax, 10               ; EAX now holds 3 * 10 == 30
mov     dl, byte [ebx + 1]    ; Second character, is e.g. "5"
sub     dl, "0"               ; Convert from "5" -> 5
add     al, dl                ; EAX now holds 3 * 10 + 5 == 35

call    print_int             ; Prints "35"