Question

我的程序非常简单。它需要用户提供两个输入：

一个数组（只有2位数的正数）
要搜索的2位数字

输出：给定号码的次数。发生在数组

样品

输入数组的大小
  03
  输入数字
  10个
  输入数字
  12个
  输入数字
  10个
  输入要搜索的号码
  10

您的电话号码 02 次

0分段错误（核心转储）

请帮我解决这个问题

enter code here

  section .data

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

msg2: db "Enter a number",10    
len2: equ $-msg2     

msg3: db "Enter the number to be searched :"     
len3: equ $-msg3      

msg4: db "Your entered number appeared "    
len4: equ $-msg4    

section .bss      
temp: resb 1     
array: resb 100     
d1: resb 1      
d0: resb 1     
size: resb 1     
num: resb 1     
ele: resb 1     
time: resb 1     

section .text     

 global _start:     

 _start:      
 ;Getting the size of array     

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

 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,2          
 int 80h            


 sub byte[d1],48           
 sub byte[d0],48          

 mov al,byte[d1]          
 mov bl,10           
 mul bl          
 add byte[d0],al             
 mov al,byte[d0]           
 mov byte[size],al           
 mov byte[temp],al             

 mov ebx,array          

 reading:           
 push rbx           ;preserves the value of ebx     


  mov eax,4   
  mov ebx,1  
 mov ecx,msg2   
 mov edx,len2   
 int 80h   

 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,2    
 int 80h    

 sub byte[d1],48     
 sub byte[d0],48     

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

 pop rbx         

 mov byte[ebx],al          
 add ebx,1          
 dec byte[temp]          
 cmp byte[temp],0            
 jg reading              

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

 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,2          
 int 80h             

 sub byte[d1],48                 
 sub byte[d0],48              

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

 mov byte[ele],al              

 mov cl,byte[size]                
 mov byte[temp],cl              

 mov byte[time],0              

 search:             

 mov al,byte[ebx]             
 cmp byte[ele],al           
 jne nf               
 jmp f             
 mov cl,0


 f:   add ebx,1            
   add byte[time],1             
    dec byte[temp]           
   cmp byte[temp],cl           
    jg search          
    jmp next


 nf:        
     add ebx,1         
     add byte[time],0           
     dec byte[temp]       
     cmp byte[temp],cl          
     jg sea rch        
     jmp next


  next: 

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

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

   add byte[d0],48            
   add byte[d1],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,1      
    mov ebx,0        
    int 80h

Answer 1

global _start
section .text
_start:

mov rax,1
mov rdi,1
mov rsi,msg
mov rdx,msglen
syscall

mov rax,0
mov rdi,0
mov rsi,no1
mov rdx,5
syscall

mov rax,1
mov rdi,1
mov rsi,msg1
mov rdx,msg1len
syscall

mov rax,0
mov rdi,0
mov rsi,no2
mov rdx,5
syscall

mov rax,[no1]
sub rax,30h
mov rbx,[no2]
sub rbx,30h
add rax,rbx
add rax,30h
mov [res],rax
mov rdx,2
syscall

mov rax,1
mov rdi,1
mov rsi,res
mov rdx,2
syscall

mov rax,60h
mov rdx,0
syscall

section .data
msg: db "Enter 1st Number",0x0a
msglen equ $ -msg
msg1: db "Enter 2st Number",0x0a
msg1len equ $ -msg1
msg2:db "Addition is==",0x0a
msg2len equ $ -msg2

section .bss
no1: resb 5
no2: resb 5
res: resb 15

Answer 2

看看来源，我注意到了一件事。我认为你的'搜索'循环超出界限，因为你无法初始化寄存器cl;指令mov cl,0目前是死代码;它在无条件跳跃之下，所以永远不会被执行。

但正如我之前所说，如果有疑问，请使用调试器。

NASM编程 - 分段故障（核心转储）

2 个答案: