Assembly 为什么'；在我的asm代码中输入呼叫工作？

我用汇编语言尝试了以下数据块传输代码。代码在非重叠情况下运行良好，但我无法键入输入以获取“count”值

section .data       
    name db 'John'  ;initialize name array  
    name_len equ $-name ;initialize name length 
    surname db 'Smith'  ;initialize surname array
    sur_len equ $-surname   ;intialize surname length
    tot equ name_len + sur_len  ;initialize total length
    msg1 db "Initial name is: "     ;message 1
    msg1_len equ $-msg1         ;length of message
    msg2 db 0xa,"Non-overlapped transfer: " ;message2
    msg2_len equ $-msg2
    msg3 db "Overlapped String transfer: "
    msg3_len equ $-msg3
    endl db 0xa
    endlen equ $-endl
    steps db 0                   ;Stores the value length-count
    msg4 db "Enter 1,2 or 3"
    msg4_len equ $-msg4
    msg5 db "Enter the count "
    msg5_len equ $-msg5
section .bss
    choice resb 1                    ;for non-overlapped or overlapped
    a resb 1                         ;storing final length in overlapped case
    count resb 1                     ;Count is the number of letters which are overlapped
section .text
    global _start       ;linker
_start:
beg:
    mov eax,4
    mov ebx,1
    mov ecx,msg4
    mov edx,msg4_len
    int 0x80    

    mov eax,4
    mov ecx,endl
    mov edx,endlen
    int 0x80

    mov eax,3
    mov ebx,1
    mov ecx,choice
    mov edx,1
    int 0x80

    cmp byte[choice],0x33
    je exit
    cmp byte[choice],0x32
    je overcall
    cmp byte[choice],0x31
    je noncall
noncall:
    call non
    jmp exit
overcall:   
    mov eax,4
    mov ebx,1
    mov ecx,msg5
    mov edx,msg5_len
    int 0x80

    mov eax,3
    mov ebx,1
    mov ecx,count                ;I can't input any value for count in the terminal
    mov edx,1
    int 0x80

    sub byte[count],0x30
    mov al,name_len
    add al,sur_len
    sub al,[count]      
    mov [a],al
    call over
    jmp exit
exit:
    mov eax,1       ;syscall for exit
    int 0x80        ;call kernel

non:
    mov eax,4       ;syscall for print
    mov ecx,msg1        ;move address of msg1 to ecx
    mov edx,msg1_len    ;move msg1 length to edx
    int 0x80        ;call kernel    

    mov eax,4       ;syscall for print
    mov ecx,name        ;move address of name to ecx
    mov edx,name_len    ;move name length to edx
    int 0x80        ;call kernel        

    mov esi,name        ;move address of name to esi
    mov edi,surname     ;move surname address to edi
    add esi,name_len    ;add name_length to move to last byte
    mov ebx,sur_len     ;move length of surname to ebx
up:
    mov eax,[edi]       ;move character of surname to eax
    mov [esi],eax       ;move character to last of name
    inc esi         ;increment esi
    inc edi         ;increment edi
    dec ebx         ;decrement count
    jnz up          ;jumpif not zero to up

    mov eax,4       ;syscall for print
    mov ecx,msg2        ;move address of name to ecx
    mov edx,msg2_len    ;move name length to edx
    int 0x80        ;call kernel    

    mov eax,4       ;syscall for print
    mov ecx,name        ;move address of name to ecx
    mov edx,tot     ;move total length to edx
    int 0x80        ;call kernel    

    mov eax,4
    mov ecx,endl
    mov edx,endlen
    int 0x80
    ret

over:
    mov eax,4
    mov ecx,msg3
    mov edx,msg3_len
    int 0x80

    xor rax,rax
    mov rsi,name
    mov rdi,surname
    mov al,name_len
    sub al,[count]
    mov [steps],al
    add rsi,rax
    mov ecx,sur_len
    xor rax,rax 

upp:
    mov al,[rdi]
    mov [rsi],al
    inc rsi
    inc rdi
    dec ecx
    jnz upp

    mov eax,4
    mov ecx,name
    mov edx,[a]
    int 0x80

    mov eax,4
    mov ecx,endl
    mov edx,endlen
    int 0x80
    ret

在选择选项2后，程序只是打印一些垃圾

以下是输出：

Enter 1,2 or 3
2
Enter the count Overlapped String transfer: JohnSmithInitial name is: 
Non-overlapped Smithfer: Overlapped String transfer: 
*Enter 1,2 or 3Enter the count 2/�

---

如果我只是初始化计数值，而不是从用户处接受它，代码就可以正常工作了

您实际上键入了

2

，然后键入

enter

，然后将该enter（换行符）保留在输入缓冲区中，因此读取

计数将返回该值。可能不会导致您在该代码中出现问题，但是您标记了这个x86-64（实际上使用了64位寄存器好几次），但是代码的其余部分都使用32位系统调用，甚至32位寻址模式！！当您尝试以这种方式使用任何堆栈地址时，这将立即中断，因为（与静态数据不同），堆栈内存的地址高于2^32，超出了内存的低4GiB。寻址模式将出现故障，int 0x80将返回
-EFAULT
@Jester在这种情况下，我应该如何给出输入？@AnishKoulgi读取，直到遇到换行符或输入结束。