采用對(duì)符號(hào)位單獨(dú)處理的辦法，處理步驟如下：

1. 單獨(dú)處理被乘數(shù)和乘數(shù)的符號(hào)位，辦法是單獨(dú)取出被乘數(shù)符號(hào)位并與乘數(shù)符號(hào)位進(jìn)行異或操作，因?yàn)榉e的符號(hào)位的產(chǎn)生規(guī)則是同號(hào)相乘為正，異號(hào)相乘為負(fù)。

2. 求被乘數(shù)和乘數(shù)的絕對(duì)值，并使兩絕對(duì)值相乘從而獲得積的絕對(duì)值，方法是分別判斷被乘數(shù)和乘數(shù)的符號(hào)位：若它為正，則其本身就是絕對(duì)值；若它為負(fù)，則對(duì)它求補(bǔ)。

3. 對(duì)積進(jìn)行處理。若積為正，則對(duì)積不進(jìn)行處理；若積為負(fù)，則對(duì)積求補(bǔ)，使之變?yōu)檠a(bǔ)碼的形式。

R0和R1中為兩個(gè)補(bǔ)碼形式的帶符號(hào)數(shù)，R3R2為積 R0*R1=R3R2

?

ORG 0600H’

SBIT BIT 20H.0

SBIT1 BIT 20H.1

SBIT2 BIT 20H.2

?

MOV A,R0???????????? ;被乘數(shù)送A

RLC A?????????????????? ;被乘數(shù)符號(hào)送CY

MOV SBIT1,C??????? ;送入SBIT1

MOV A,R1???????????? ;乘數(shù)送A

RLC A?????????????????? ;乘數(shù)符號(hào)送CY

MOV SBIT2,C??????? ;送入SBIT2

ANL C,/SBIT1??????? ;SBIT1非 與 SBIT2 送CY

MOV SBIT,C????????? ;送入SBIT

MOV C,SBIT1??????? ;SBIT1送CY

ANL C,SBIT2????????? ;SBIT1 與 SBIT2非 送CY

ORL C,SBIT??????????? ;積的符號(hào)位送CY

MOV SBIT,C????????? ; 送入SBIT

MOV A,R0???????????? ;處理被乘數(shù)

JNB SBIT1,NCH1??? ;若她為正，則轉(zhuǎn)NCH1

CPL A?????????????????? ;若她為負(fù)，則求補(bǔ)的絕對(duì)值

INC A?????????????????? ;

NCH1:??? MOV B,A????????????? ;被乘數(shù)絕對(duì)值送B

????????????? MOV A,R1???????????? :處理乘數(shù)

JNB SBIT2,NCH2??? ;若它為正，則轉(zhuǎn)NCH2

CPL A?????????????????? ;若她為負(fù)，則求補(bǔ)的絕對(duì)值

ADD A,#01H???????? ;

NCH2:??? MUL AB??????????????? ;求積的絕對(duì)值

JNB SBIT,NCH3???? ;若積為正，則轉(zhuǎn)NCH3

CPL A?????????????????? ;若積為負(fù)，則低字節(jié)求補(bǔ)

ADD A,#01H???????? ;

NCH3:??? MOV R2,A???????????? ;積的低字節(jié)存入R2

MOV A,B????????????? ;積的高字節(jié)送A

JNB SBIT,NCH4???? ;若積為正，則轉(zhuǎn)NCH4

CPL A?????????????????? ;若積為負(fù)，則高字節(jié)求補(bǔ)

ADDC A,00H???????? ;

NCH4:??? MOV R3,A???????????? ;積的高字節(jié)存入R3

SJMP $

END