问 *************************************************************************/
//        ---------------
//          LED显示摇棒
//        ---------------
//    作者：nim    于 http://www.21icbbs.com    2005-1-7
//
//    email: linyige@yahoo.com
//        
//    本设计版权归作者所有，请勿擅自用于商业目的，转载请注明作者及出处
/*************************************************************************/
#include <reg51.h>
#define SEG 17        //每帧图片分成17列来显示
#define INTERVAL 20    //每幅图片在左右摇摆20次后换下一幅

typedef unsigned char uchar;
typedef unsigned int uint;

code char pattern[17][3]={    //3幅图片的字模
{0xff,0xff,0xff},
{0xff,0xff,0xff},
{0xff,0xff,0xff},
{0xff,0x9f,0xff},
{0xff,0x6f,0xf9},
{0xff,0x77,0x65},
{0xfe,0xbb,0x1e},
{0x7e,0xdd,0xfe},
{0x00,0xee,0xfe},
{0x7e,0xdd,0xfe},
{0xfe,0xbb,0x1e},
{0xff,0x77,0x65},
{0xff,0x6f,0xf9},
{0xff,0x9f,0xff},
{0xff,0xff,0xff},
{0xff,0xff,0xff},
{0xff,0xff,0xff},
};


uchar phase,th1,tl1,index,count;


main()
{
    EA=0;
    EX0=1;
    ET1=1;
    PX0=1;        
    IT0=1;     
    TMOD=0x11;
    index=0;
    EA=1;
    while(1)
    {
    }
}

void Int0_Handle(void) interrupt 0 using 2
{
    uint t0_time;
    TR0=0;
    TR1=0;
    TF1=0;
        t0_time=TH0<<8|TL0;
    TL0=0;
    TH0=0;
    TR0=1;
    t0_time=65535-t0_time/SEG;
    th1=t0_time/256;
    TH1=th1;
    tl1=t0_time % 256;
    TL1=tl1;
    if (count<3*INTERVAL){        //3幅图片循环
        count++;
    }
    else{
        count=0;
    }
    index=count/INTERVAL;
    if (th1!=0xff || tl1!=0xff){    //如果摆动特别慢，定时器溢出就不显示
        phase=0;
        TR1=1;
    }
    else{
        TR1=0;
    }
}

void Timer1(void) interrupt 3 using 3
{
    if (phase<SEG)            //17段依次显示
    {
        P1=pattern[phase][index];
        phase++;
        TH1=th1;
        TL1=tl1;
    }
}

一直没弄明白如何用定时器0计算两次外部中断的间隔的摇摆时间的？
答 1: 顶！2051中的定时器/计数器0的初值如何赋值的，事先并不知道摇摆一次所需的时间啊！而timer/counter0工作必须先赋值进去（是由2051是步增的，溢出产生中断的）