功能說明:輸入兩組16位二進制代碼模擬兩為16位二進制整數(shù),對其求差并以16進制輸出.
基本算法:采用補碼,模擬二機制減法(加法類同),得到差(補碼)轉(zhuǎn)換成原碼后,轉(zhuǎn)化成十六進制并輸出.16為二進制轉(zhuǎn)換成4為十六進制數(shù)的方法是采用:將16位二進制分成4組,每組由連續(xù)4位二進制組成,模擬4位二進制到一位十六進制轉(zhuǎn)換,將每組轉(zhuǎn)換成16進制數(shù)并輸出.詳見代碼

本代碼在MASM.5調(diào)試通過
運行界面：
如
binary(YuanMa)   num1=1000000000001011  ;被減數(shù)原碼
binary(YuanMa)   num2=1000000000001101 ;減數(shù)原碼
binary(Buma) num1  BM=1111111111110101 ;被減數(shù)補碼
binary(Buma) num2  BM=1111111111110011 ;減數(shù)補碼
binary(Buma) diff sub=0000000000000010 ;差補碼
binary(YuanMa)diffsub=0000000000000010 ;差原碼
hex sumHEX=0002

;/************************************************/
;
;bsub.asm
;hbyufan@163.com
;

TSOUT MACRO dx0
        mov ah,09h
        mov dx,offset dx0
        int 21h
ENDM


data segment
num1 db 19 dup(''$'')
num2 db 19 dup(''$'')
sum  db 19 dup(''$'')
hexsum db 5 dup(''$'')

tsnum1  db 0ah,0dh,''binary(YuanMa)   num1=$''
tsnum2  db 0ah,0dh,''binary(YuanMa)   num2=$''
tssum   db 0ah,0dh,''binary(Buma) diff sub=$''
tsnum11 db 0ah,0dh,''binary(Buma) num1  BM=$''
tsnum22 db 0ah,0dh,''binary(Buma) num2  BM=$''
tssum33 db 0ah,0dh,''binary(YuanMa)diffsub=$''
tssumhex   db 0ah,0dh,''hex sumHEX=$''

data ends

code segment
       assume ds:data,cs:code
start:
        mov ax,data
        mov ds,ax

re:
       TSOUT tsnum1
        mov si,0
n1roat:
        mov ah,01h
        int 21h

        cmp al,''0''   ;對NUM1輸入錯誤簡單處理---重新輸入
        jl  re
        cmp al,''1''
        ja  re

        mov num1[si],al
        inc si
        cmp si,16
        jl  n1roat

re2:
        TSOUT tsnum2
        mov si,0
n2roat:
        mov ah,01h
        int 21h

        cmp al,''0''   ;對NUM1輸入錯誤簡單處理---重新輸入
        jl  re2
        cmp al,''1''
        ja  re2

        mov num2[si],al
        inc si
        cmp si,16
        jl  n2roat

        mov bx,offset num1
        call questBuma
        mov bx,offset num2
        call questBuma
                TSOUT tsnum11;//
                TSOUT num1;//
                TSOUT tsnum22;//
                TSOUT num2;//

        call bsubproc
                 TSOUT tssum;       
                 TSOUT sum ;//test

        lea bx,sum
        call questBuma
                TSOUT tssum33;//
                TSOUT sum;//
        lea bx,sum
        call BinaryToHexAndOut
                TSOUT tssumhex;//
                TSOUT hexsum;//
;******************************
        mov ah,4ch
        int 21h
;*******************************************
questBuma proc near   ;功能:求補碼
     ;算法:模擬求補碼運算,如二進制求補碼方法一致.
                   ;bx    ;bx作為參數(shù),傳入NUM1或NUM2內(nèi)存空間首地址
        mov al,[bx+0]
        cmp al,''0''
        jz qbret   ;如果是正整數(shù),函數(shù)返回;如果是負(fù)數(shù),符號位不變,其余位取反加1.
        mov si,1   
qbroat:
        mov al,[bx][si]
        cmp al,''0''
        jz  qb1
        mov al,''0''
        mov [bx][si],al
        jmp qb2
qb1:
        mov al,''1''
        mov [bx][si],al
qb2:
        inc si
        cmp si,16
        jl  qbroat

qnext:     
        mov dh,1      ;加1處理.方法類似減法
        mov si,15
qbincre:
        mov al,[bx][si]
        sub al,''0''
        add al,dh
        mov dh,0

        cmp al,2
        jnz qbi1
        mov al,0
        mov dh,1
qbi1:
        add al,''0''
        mov [bx][si],al
        dec si
        cmp si,0
        jnl qbincre
qbret:
        ret
questBuma endp

;*******************************************
bsubproc proc near   ;功能:16位二進制數(shù)減法模擬.
     ;算法:逐位相減,flag為借位標(biāo)志(BX)
        push ax
        push bx
        push cx
        push si

        mov bx,0 ;//flag
        mov si,15

broat:  mov al,num1[si]
        add al,bh
        mov bh,0   ;//clear bh
        cmp al,num2[si]
        jnl bnext
        add al,2
        dec bh
bnext:  
        sub al,num2[si]
        add al,''0''
        mov sum[si],al
        dec si
        cmp si,0
        jnl  broat     

        pop si
        pop cx
        pop bx
        pop ax
        ret
bsubproc endp
;*******************************************
BinaryToHexAndOut proc near  ;功能如函數(shù)名
                ;bx   ;bx作為參數(shù),傳入NUM1或NUM2內(nèi)存空間首地址
        mov dh,2
        mov si,0
bth1:
        mov al,0   ;分組轉(zhuǎn)換,模擬4為二進制到1為16進制數(shù)的轉(zhuǎn)換
        mov di,0
bth2:     
        mul dh
        add al,[bx][di]
        sub al,''0''
        inc di
        cmp di,4
        jl  bth2
        cmp al,9
        jna bth3
        sub al,10
        add al,''A''
        jmp bth4
bth3:
        add al,''0''
bth4:
        mov hexsum[si],al
        inc si
        add bx,4
        cmp si,4
        jl  bth1
        ret
BinaryToHexAndOut endp

;*******************************************
code ends
        end start


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