include irvine32.inc;
.data
i dword 0;
sum qword 0;
str1 byte "請輸入16進制的(32位整數)乘數和被乘數",0
str2 byte "乘積為：",0;
j dword 0;

.code

main proc

L1:
?? mov edx,offset str1;
?? call writestring;

?? mov j,0;
?? mov eax,0;eax為（低32位）乘積值；
?? mov ebx,0;edx為（低32位）乘積值的中間值
?? mov edx,0;
?? mov esi,0;dsi為（高32位）乘積值的中間值
?? mov edi,0;edi為（高32位）乘積值；
?? mov cl,0;cl為移位的位數

?? call readhex;
?? cmp eax,80000000h;

?? jna L4;
?? neg eax;
?? inc j;
L4:
?? mov ebx,eax;??? ebx存儲乘數（來移位的）
?? ;call writeint;

?? call readhex;
?? cmp eax,80000000h;

?? jna L5;
?? neg eax;
?? inc j;
L5:
?? mov i,eax;
?? ;call writeint;
?? mov eax,0;
?? cmp ebx, 0;
?? jne L2;
?? cmp i,0;
?? je? finally;

? L2: cmp ebx,0;
????? jz quit;
????? shr ebx,1;
????? jnc next;if(ZF==0)

????? mov esi,0;每次都要清零；
????? mov edx,i;

????? shld esi,edx,cl;
????? shl edx,cl;

????? add eax,edx;
????? jnc L3;if(ZF==0)
????? add edi,1;

?L3:
????? add edi,esi;
next:
????? inc cl;
????? jmp L2;


quit:
???? call crlf;
???? mov edx,offset str2;
???? call writestring;
???? mov ebx,eax;

???? cmp j,1;
???? jne L6;
???? add edi,80000000h
???? ;call writestring
?L6:

???? mov eax,edi;
???? call writehex;

???? mov eax,ebx;
???? call writehex;


???? call crlf;

???? jmp L1;

finally:

exit
main endp;
end main

轉載于:https://www.cnblogs.com/hqu-ye/archive/2013/02/25/2932588.html

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