小弟我写的这个程序在未来汇编编译一点有关问题都没有，但在MASM6.15就说有有关问题～

;------------------------------------------------
;Description:To output a value to screen in the hexadecimal form
;Author: 
;creat date:2007-11-18
;modify date:
;Algorithms: 1 首先判断数据是否小于0，若小于0，首先在屏幕上输出一个字符'0'。
;            2 AX,BX同时保存这个数据，BX与1111B相与，结果是BX中只保留下最低4位，即我们要的数字
;              若数字大于9，则BL-10＋'a'，但到其16进制的字符，否则BL+'0'也得到其16进制最低位的字符
;            3 AX循环右移4位，再Mov bx,ax，重复2的步骤至所有的4个4 bits的数据都取到
;------------------------------------------------
    
data1 segment
      ; Please input here the code of data segment
      val  dw  -78
      msg  db  4 dup(?),'H',13,10,'$'
data1 ends

stack1 segment
    ;Please input here the code of data segment
stack1 ends

code1 segment 
    assume cs:code1,ds:data1,ss:stack1
start:
    mov   ax,data1
    mov   ds,ax
    ;Please input here the code of code segment
    mov   ax,[val]
    cmp   ax,0
    jl    prezero                         ; If the val is less than 0, output a zero first
    
contin:
    mov   ax,[val]                        ; If ax<0, the prezero will overwrite ax, so we reset it to make sure it is proper
    mov   cx,3                              

pro:
    mov   si,offset msg
    add   si,cx       
    mov   bx,ax
    and   bx,1111b                         ; logical and make the higher 4 bits zero, only leave the lower 4 bits
    cmp   bl,10
    jge   hexad                            ; if the lower 4 bits is great or equal to 10, jump to set letter
    
adcx:                                      ; if the lower 4 bits is not great or equal to 10, go down to set number's ASCII
    add   bl,'0'
    mov   byte ptr [si],bl
    jmp   shad                             ; jump to decrement cx, and shift right of ax
    
    
    
hexad:                                     ; The following fragment process the hexadecimal letter
    sub   bl,10
    add   bl,'a'
    mov   byte ptr [si],bl
    
shad:
    dec   cx
    cmp   cx,0
    jl    quit
    ROR   ax,4                             ;Shift ax 4 bits right
    jmp   pro

prezero:
    mov   dl,'0'
    mov   ah,2 
    int   21h
    jmp   contin   
    

quit:
    mov   ah,9
    mov   dx,offset msg
    int   21h
    mov   ah,4ch
    int   21h
code1 ends
    end start

 0031                shad:       ;**.lst里面的信息
 0031  49                dec   cx
 0032  83 F9 00                cmp   cx,0
 0035  7C 0A                jl    quit
                    ROR   ax,4                             ;Shift ax 4 bits right
val2hex.asm(60) : error A2070: invalid instruction operands
 0037  EB DA                jmp   pro

    ROR   ax,4        ;Shift ax 4 bits right