修改cx时，TASM检查参数 [英] TASM checking arguments when cx is modified

问题描述

我需要检查是否给出了大量的参数.目前，我的程序知道什么时候没有参数，什么时候给出两个以上的参数.如果给出一个，那是不好的，但是我的程序无法识别它.我不能做简单的检查，因为有一种可能性，当它被允许时(当/?是参数时)，并且cx随循环而变化.例如，如果我将其命名为prog/?，则应打印描述，当我将其命名为prog arg1 arg2时，应允许使用它，但不允许prog arg1.我该如何检查?

I need to check whether good amount of arguments is given. Currently my program knows when there are no arguments and when there are more than two arguments given. If one is given, then it is bad, but my program doesn't recognise it. I can't do simple check because there is one possibility when it is allowed (when /? is argument) and cx changes with loop. For example, if I call it prog /?, it should print description, when I call it prog arg1 arg2, it should be allowed, but prog arg1 shouldn't. How can I check that?

  mov ch, 0h         
  mov cl, [es:0080h] 
  push cx
  cmp cx, 0         
  je print_description

  mov bx, 81h          
  jmp search_help

  jmp exit

search_help:
  cmp [es:bx], '?/'  
  je print_description 
  inc bx               
  loop search_help    


  pop cx
  cmp cx, 2
  jne print_description
  mov bx, 82h     
  mov si, offset input_filename 
  mov cx, 255                   

search_input_filename:
  mov dl, [es:bx]               
  inc bx                       
  cmp dl, 20h                  
  je search_output_filename_prep

  mov ds:[si], dl               
  inc si                       
  loop search_input_filename 

search_output_filename_prep:
  mov si, offset output_filename 

search_output_filename:
  mov dl, [es:bx]            
  inc bx                     
  cmp dl, 0Dh             
  je program               
  cmp dl, 20h             
  je print_description     

  mov ds:[si], dl         
  inc si                     
  loop search_output_filename 

推荐答案

两个错误:

1)我的TASM 5.0不喜欢[ES:xxxx].段覆盖必须写为ES:[xxxx].

1) My TASM 5.0 doesn't like [ES:xxxx]. The segment override has to be written as ES:[xxxx].

2)ES:[0080h]包含命令行的长度，不参数的数量.因此，这行不按照您的想法进行操作:

2) ES:[0080h] contains the length of the command line, not the count of the arguments. Thus this lines don't do what you think:

push cx
...
pop cx
cmp cx, 2
jne print_description

您可以改用我的get_argc函数.遵循以下测试用例(希望如此):

You can use my get_argc function instead. Following test case works (hopefully):

LOCALS @@
.MODEL small
.STACK 1000h

.DATA
    description db 'description',13,10,'$'
    input_filename db 80 DUP ('$')
    output_filename db 80 DUP ('$')

.CODE
main PROC
        mov ax, @data
        mov ds, ax

        mov ch, 0h
        mov cl, es:[0080h]
        cmp cx, 0
        je print_description

        mov bx, 81h
        jmp search_help

        jmp exit

    search_help:
        cmp word ptr es:[bx], '?/'
        je print_description
        inc bx
        loop search_help

        call get_argc
        cmp ax, 2
        jne print_description

        mov bx, 82h
        mov si, offset input_filename
        mov cx, 255

    search_input_filename:
        mov dl, es:[bx]
        inc bx
        cmp dl, 20h
        je search_output_filename_prep

        mov ds:[si], dl
        inc si
        loop search_input_filename

    search_output_filename_prep:
        mov si, offset output_filename

    search_output_filename:
        mov dl, es:[bx]
        inc bx
        cmp dl, 0Dh
        je program
        cmp dl, 20h
        je print_description

        mov ds:[si], dl
        inc si
        loop search_output_filename

    program:
        mov dx, OFFSET input_filename
        mov ah, 09h
        int 21h
        call crlf
        mov dx, OFFSET output_filename
        mov ah, 09h
        int 21h

    exit:
        mov ax, 4C00h
        int 21h

    print_description:
        lea dx, description
        mov ah, 09h
        int 21h
        mov ax, 4C01h
        int 21h

main ENDP

get_argc PROC
    mov bx, 0
    mov di, 80h
    mov cx, es:[di]

    @@L1:
        inc di
        mov al, es:[di]
        cmp al, 20h
        je @@L1
        cmp al, 09h
        je @@L1
        cmp al, 0Dh
        je @@done

        inc bx

    @@L2:
        inc di
        mov al, es:[di]
        cmp al, 20h
        je @@L1
        cmp al, 09h
        je @@L1
        cmp al, 0Dh
        je @@done
        jmp @@L2

    @@done:
        mov ax, bx
        ret
get_argc ENDP


crlf PROC
        mov ah, 2
        mov dl, 13
        int 21h
        mov ah, 2
        mov dl, 10
        int 21h
        ret
crlf ENDP

END main

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