怎么样才能用AT24C02保存数据(断电保存)？

只看该作者 · 2011-8-14 16:58

C语言编写程序：利用定时器产生一个0~99秒变化的秒针，显示在数码管上，每过一秒将变化的数写入AT24c02内部，关闭电源再打开时单片机从AT24c02中将原来写入的数读出来，接着此数继续变化并显示在数码管上。
谢谢
我写的程序如下:
#include <reg52.h>
#include<intrins.h>
#define uchar unsigned char
#define uint unsigned int
#define  Nop()  _nop_()
sbit sda=P1^2;          //数据线
sbit scl=P1^1;    //时钟线
sbit wp=P1^0;    //读写保护
uint num1,num,jia;
uchar code table[]={
0xc0,0xf9,0xa4,0xb0,0x99,0x92,0x82,0xf8,0x80,0x98};
uchar write;
void delay()       //时序延时
{ _nop_();
  _nop_();
_nop_();
_nop_();
_nop_();
}
void init()       //初始化
{
  sda=1;
  delay();
  scl=1;
  delay();
}
void start()    //启动
{
  sda=1;
  delay();
  scl=1;
  delay();
  sda=0;
  delay();
}
void stop()    //停止
{
  sda=0;
  delay();
  scl=1;
  delay();
  sda=1;
  delay();
}
void writebyte(uchar date) //写数据
{
  uchar i,temp;
  temp=date;
  for(i=0;i<8;i++)
  {
  temp=temp<<1;
  scl=0;
  delay();
  sda=CY;
  delay();
  scl=1;
  delay();
  }
  scl=0;
  delay();
  sda=1;
  delay();
}
void respons()    //应答
{
  uchar m;
  m=0;
  scl=1;
  delay();
  while((sda==1)&&(m<250))m++;
  scl=0;
  delay();
}          //读数据
uchar readbyte()
{
  uchar j,k;
  scl=0;
  delay();
  sda=1;
  delay();
  for(j=0;j<8;j++)
  {
  scl=1;
  delay();
  k=(k<<1)|sda;
  delay();
  scl=0;
  delay();
  }

  return k;
}
void writeadd(uchar address,uchar date)
{ // wp=0;
  start();
  writebyte(0xa0);
  respons();
  writebyte(address);
  respons();
  writebyte(date);
  respons();
  stop();

}
uchar readadd(uchar address)
{
  uchar n;
  start();
  writebyte(0xa0); //寻址、控制字节
  respons();    //应答
  writebyte(address);  //把要读的首地址发送到从机？
  respons();
  start();
  writebyte(0xa1);  //再次发送寻址、控制字节
  respons();
  n=readbyte(); //读的字节数
  stop();
  return n;
}
void delayms(uchar shu)    //数码管显示延时
{
  uchar x,y;
  for(x=shu;x>0;x--)
  for(y=100;y>0;y--);
}
void display(uchar shi,uchar ge)  //数码管显示
{
  //int baochi=10;
//while(baochi--)
  P2=0xff;
P2=0xfe;
  P0=table[ge];

  delayms(5);
  P2=0xfd;
  P0=table[shi];

  delayms(5);

}
void main()
{
  uchar num2;

  init();
  num=readadd(jia);    //???
  if(num>100)
  num=0;
  TMOD=0x10;
  ET1=1;
  EA=1;
  TH1=(65536-50000)/256;
  TL1=(65536-50000)%256;
  TR1=1;
  while(1)
  {
  display(num/10,num%10);
  if(write==1)
  {
  write=0;
  num2=num;
  writeadd(jia,num2);

  }

  }
}
void timer1() interrupt 3
{
  TH1=(65536-50000)/256;
  TL1=(65536-50000)%256;
  num1++;

  if(num1==20)
  {
  num1=0;
  jia++;
  num++;
  write=1;
  if(num==100)
  num=0;
  }
}