摘要: divdw:push bxpush simov bx,ax;暂存低位mov ax,dx;高位除法mov dx,0div cxmov si,ax;暂存商mov ax,0add ax,bx;得到高位余数和低位之和div cx;ax中得到商,dx中是余数mov cx,dxmov dx,sipop sipop bxret 阅读全文
posted @ 2011-03-21 13:30 Mose 阅读(153) 评论(0) 推荐(0) 编辑