NASM 程序 - 合并两个数组

Question

我是 nasm 新手。 我对nasm有非常基本的了解 我的程序基本上将两个数组作为输入并使用冒泡排序对它们进行排序，然后将它们合并以获得一个新数组（排序） 但是在接受输入时，程序并没有按照我期望的方式工作。 我似乎找不到错误。 我的所有功能都工作正常。“合并”标签可能存在问题。我无法检查

请帮忙

section .data

msg1: db "Enter the size of first array ",10
len1: equ $-msg1

msg2: db "Enter the size of second array ",10
len2: equ $-msg2

 msg3: db "Enter the element ",10
len3: equ $-msg3

msg4: db "MERGED ARRAY IS  ",10
len4: equ $-msg4

comma: db ","
nline:db 10

section .bss

 arr1: resb 100
arr2: resb 100
array: resb 200
d1: resb 1
d0: resb 1
temp: resb 1
num: resb 1
size1: resb 1
size2: resb 1
size: resb 1
i: resb 1
j: resb 1
s: resb 1
r: resb 1
read: resb 1
section .text
global _start:


      ;-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------
 read_digit:
 mov eax, 3
 mov ebx, 0
 mov ecx, d1
 mov edx, 1
 int 80h

 mov eax, 3
 mov ebx, 0
 mov ecx, d0
 mov edx, 1
 int 80h

 mov eax, 3
 mov ebx, 0
 mov ecx, temp
 mov edx, 1
 int 80h

  sub byte[d1], 30h
  sub byte[d0], 30h  

  mov al, byte[d1]
  mov dl, 10
  mul dl
  add al, byte[d0]
  mov byte[read],al
 ret
    ;---------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------


     ;----------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------
 read_array:

  push rbx

  ;Printing the message to enter the numbers
  mov eax, 4
 mov ebx, 1
  mov ecx, msg3
 mov edx, len3
 int 80h

  ;Reading the number
  mov eax, 3
  mov ebx, 0
  mov ecx, d1
  mov edx, 1
  int 80h

 mov eax, 3
 mov ebx, 0
 mov ecx, d0
 mov edx, 1
 int 80h

 mov eax, 3
 mov ebx, 0
 mov ecx, temp
 mov edx, 1
 int 80h


 sub byte[d1], 30h
 sub byte[d0], 30h  
 mov al, byte[d1]
 mov dl, 10
 mul dl
 add al, byte[d0]

 ;al now contains the number
 pop rbx

 mov byte[ebx], al

 add ebx, 1

 dec byte[r]
 cmp byte[r], 0
 jg read_array
 jmp end

end:
ret
  ;-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------


       ;---------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------
   sort:
   mov byte[i], 0

  i_loop:
mov byte[j], 0
j_loop:
  mov ebx, ecx;needs ebx so given it in ecx
  movzx eax, byte[j]
  add ebx, eax
  mov eax, ebx
  add eax, 1
  mov dl, byte[ebx]
  mov dh, byte[eax]
  cmp dh, dl
  jl swap
  jmp no_swap

  swap:
    mov byte[ebx], dh
    mov byte[eax], dl

 no_swap:
  inc byte[j]
  mov al, byte[s]
  sub al, byte[i]
  sub al, 1
   cmp byte[j], al
  jl j_loop

  inc byte[i]
  mov al, byte[s]
 cmp byte[i], al
  jl i_loop
 jmp over

over:
ret
     ;--------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------

 _start:


 mov eax, 4
 mov ebx, 1
 mov ecx, msg1
 mov edx, len1
 int 80h

 call read_digit
 mov al,byte[read]
 mov byte[size1],al
 mov byte[r],al
 mov ebx,arr1
 call read_array


  mov eax, 4
  mov ebx, 1
  mov ecx, msg3
  mov edx, len3
  int 80h

  call read_digit
  mov al,byte[read]
  mov byte[size2],al
  mov byte[r],al
  mov ebx,arr2
  call read_array

  mov al,byte[size1]
  mov byte[s],al
   mov ecx,arr1
  call sort

  mov al,byte[size2]
   mov byte[s],al
   mov ecx,arr2
   call sort

  mov al,byte[size1]
  mov ah,byte[size2]
   add al,ah
   sub al,1
  mov byte[size],al


  mov byte[i],0
 mov eax,arr1
 mov ebx,arr2
 mov ecx,array

 merge:
mov dl,byte[eax]
mov dh,byte[ebx]

cmp dl,dh
jna A                                                                   ;if L[a]<=L[b] ;a[i]=L[a] a=a+1
jmp B


A: 
   mov byte[ecx],dl
   add eax,1
   add ecx,1
   inc byte[i]
   mov dl,byte[size]
   cmp byte[i],dl
   je done
   jmp merge

B:
    mov byte[ecx],dh
   add ebx,1
   add ecx,1
   inc byte[i]
   mov dl,byte[size]
   cmp byte[i],dl
   je done
   jmp merge



 done:

 mov eax, 4
 mov ebx, 1
 mov ecx, msg4
 mov edx, len4
 int 80h

  mov ebx,array
  mov al,byte[size]
  mov byte[temp],al

 print:
 push rbx

  mov al,byte[array]
 mov byte[num],al

 movzx ax,byte[num]
  mov bl,10
  div bl
  mov byte[d1],al
 mov byte[d0],ah

  add byte[d1],48
 add byte[d0],48

 mov eax, 4
 mov ebx, 1
 mov ecx, d1
 mov edx, 1
 int 80h

  mov eax, 4
 mov ebx, 1
 mov ecx, d0
 mov edx, 1
 int 80h

 mov eax, 4
 mov ebx, 1
 mov ecx, comma
  mov edx, 1
 int 80h

 pop rbx

 add ebx,1
  dec byte[temp]
 cmp byte[temp],0
 jg print

mov eax,1
mov ebx,0
int 80h

Answer 1

在组装过程中只有 2 个可能的错误：

• 每条指令旁边的 cmets 描述了一个不正确的算法
• 说明与旁边的 cmets 所描述的不同

为了找出您的错误，我首先检查每条指令旁边的 cmets 是否描述了正确的算法。它没有。然后我继续，并检查每条指令是否按照它的评论所说的那样做。他们都没有。您的代码是 100% 纯错误。

比如这段代码是100%错误的：

    mov al, byte[d1]
    mov dl, 10
    mul dl
    add al, byte[d0]

...而这段代码可能 100% 正确：

    mov al, byte[d1]    ;al = first digit read from standard input
    mov dl, 10          ;dl = 10
    mul dl              ;al = ax = first digit * 10
    add al, byte[d0]    ;al = al + second digit from standard input
                        ;   = first digit * 10 + second digit

同理，这段代码100%错误：

    mov eax, 3
    mov ebx, 0
    mov ecx, d1
    mov edx, 1
    int 80h

...而这段代码可能 100% 正确：

%define SYSAPI_READ   3      ;Linux kernel function number for "READ()"
%define STDIN         0      ;File descriptor number for standard input

    mov eax, SYSAPI_READ     ;eax = function number for Linux API
    mov ebx, STDIN           ;ebx = file descriptor
    mov ecx, d1              ;ecx = address of destination buffer
    mov edx, 1               ;edx = number of bytes to read
    int 80h                  ;Call Linux kernel API

如果您尝试修复所有“缺少 cmets”错误，那么您很可能会在不尝试的情况下找到其他问题。