at2402程序的问题，

szgaoju

at2402程序的问题， [复制链接]

最近在学习《郭天祥老师10天学会51单片机和C语言》视频第9课最后留个课后作业：让两位数码管自动增长，当断电时自动存入AT24C02A，上电时还能自动在数码管上显示断电前的数字，并继续自动增长。以下是我写的程序，当上电时先显示“00”后在继续增长，不知为何，还请高手指教。[br]

#include  //52单片机头文件
#include
#define uint unsigned int //宏定义
#define uchar unsigned char //宏定义
sbit sda=P3^7;
sbit scl=P3^6;
uchar numwe,ge,shi,temp,temp1,aa,t1,a;
uchar code table[]={
0xc0,0xf9,0xa4,0xb0,
0x99,0x92,0x82,0xf8,
0x80,0x98,0x88,0x80,
0xc6,0xc0,0x86,0x8e};  //0,1,2,3,4,5,6,7,8,9,a,b,c,e,f

uchar code tablewe[]={
0xfe,0xfd,0xfb,0xf7};

void delay()
{ ;; }
void start()  //开始信号
{
sda=1;
delay();
scl=1;
delay();
sda=0;
delay();
}

void stop() //停止
{
sda=0;
delay();
scl=1;
delay();
sda=1;
delay();
}

void respons()  //应答
{
uchar i;
scl=1;
delay();
while((sda==1)&&(i<250))i++;
scl=0;
delay();
}

void write_byte(uchar date)
{
uchar i,temp1;
temp1=date;

for(i=0;i<8;i++)
{
temp1=temp1<<1;
scl=0;
      delay();
sda=CY;
delay();
scl=1;
delay();
// scl=0;
   // delay();
}
scl=0;
delay();
sda=1;
delay();
}

uchar read_byte()
{
uchar i,k;
scl=0;
delay();
sda=1;
delay();
for(i=0;i<8;i++)
{
scl=1;
delay();
k=(k<<1)|sda;
scl=0;
delay();
}
return k;
}

void write_add(uchar address,uchar date)
{
start();
write_byte(0xa0);
respons();
write_byte(address);
respons();
write_byte(date);
respons();
stop();
}

uchar read_add(uchar address)
{
uchar date;
start();
write_byte(0xa0);
respons();
write_byte(address);
respons();
start();
write_byte(0xa1);
respons();
date=read_byte();
stop();
return date;
}
void delay1(uint z)    //延时子函数
{
uint x,y;
for(x=z;x>0;x--)
for(y=110;y>0;y--);
}
void display(uchar ge,uchar shi)
{
if(numwe==2)
numwe=0;
P2=tablewe[numwe];
P0=table[shi];
numwe++;
delay1(5);
P2=tablewe[numwe];
P0=table[ge];
numwe++;
delay1(5);
}
void init()
{
TMOD=0x01;
TH0=(65536-50000)/256;
TL0=(65536-50000)%256;
EA=1;
ET0=1;
TR0=1;
sda=1;
delay();
scl=1;
delay();
temp=read_add(23);

}
void timer0() interrupt 1
{
TH0=(65536-50000)/256;
TL0=(65536-50000)%256;
aa++;
}

void main()       //主函数
{
init();//初始化子程序
while(1)
{
if(aa==20)
   {
aa=0;

temp++;
write_add(23,temp);
delay1(100);
if(temp==99)
   {
temp=0;
   }
shi=temp%100/10;
ge=temp%10;
   }
display(ge,shi);
}
}
[br]
连接图[br]
[br]

yuzejun2321

自己先顶一下

inter32

中断服务程序中得把当时的时间值放在EEPROM，然后再主函数的初始化程序上读EEPROM放时间的地址用显示代码刷一遍时间！

renrenren

引用 2 楼 huxinhui 的回复:
中断服务程序中得把当时的时间值放在EEPROM，然后再主函数的初始化程序上读EEPROM放时间的地址用显示代码刷一遍时间！

看着有点复杂，能提供点代码，看看吗

majiansong

留个邮箱

dnvtaje

好，hanyouxi3@163.com

ggw001

都没有人会吗？？

jonystar

EmF63D2as

NMYSvUIv7

yjx356657845

实在不想动手, 帮你顶吧

nishiyitouniu

收到你的邮件了，不知道怎么给你分

tunersys

呵呵，点“结贴”……建议一下ＬＺ（５２内置了ＥＥＰＲＯＭ，可以设置寄存器来使用它）不需要买个外围芯片，比较经济，学得也多

liyananliy

求24C02页写和连续读的函数

lijiamin11

我把正确答案给大家共享一下，谢谢huxinhui的指教。
#include
#define uchar unsigned char
char shu[]={0xc0,0xf9,0xa4,0xb0,0x99,0x92,0x82,0xf8,0x80,0x90,0xff};
char wei[4]={0xfe,0xfd,0xfb,0xf7};
char dis[4]={0x00,0x00,0x00,0x00};
char timedata[2]={0x00,0x00};
sbit sda=P3^7;
sbit scl=P3^6;
char ms50=0x00;
void delay1ms(int t)
{
  int i,j;
  for(i=0;i   for(j=0;j<100;j++);
}
scan()
{
  char k;
    dis[1]=timedata[0]%10;
dis[0]=timedata[0]/10;
//    dis[2]=timedata[1]%10;
// dis[3]=timedata[1]/10;
  for(k=0;k<2;k++)
  {
P0=shu[dis[k]];
P2=wei[k];
delay1ms(1);
// P2=0xfd;
  }
}
void delay()
{ ;; }
void start()
{
  sda=1;
  delay();
  scl=1;
  delay();
  sda=0;
  delay();
}

void stop()
{
  sda=0;
  delay();
  scl=1;
  delay();
  sda=1;
  delay();
}

void respons()
{
  uchar i;
  scl=1;
  delay();
  while((sda==1)&&(i<250))i++;
  scl=0;
  delay();
}
void write_byte(uchar date)
{
  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();
}

uchar read_byte()
{
uchar i,k;
scl=0;
delay();
sda=1;
delay();
for(i=0;i<8;i++)
{
   scl=1;
   delay();
   k=(k<<1)|sda;
   scl=0;
   delay();
}
return k;
}

void delay1(uchar x)
{
uchar a,b;
for(a=x;a>0;a--)
for(b=100;b>0;b--);
}

void write_add(uchar address,uchar date)
{
  start();
  write_byte(0xa0);
  respons();
  write_byte(address);
  respons();
  write_byte(date);
  respons();
  stop();
}

uchar read_add(uchar address)
{
  uchar date;
  start();
  write_byte(0xa0);
  respons();
  write_byte(address);
  respons();
  start();
  write_byte(0xa1);
  respons();
  date=read_byte();
  stop();
  return date;
}
uchar shuju()
{
uchar Time;
Time=timedata[0]+timedata[1]*100;
return Time;
}
void time(void) interrupt 1
{

ET0=0;TR0=0;TH0=0x3c;TL0=0xb0;TR0=1;ms50++;
if(ms50==20)
{
   ms50=0x00;timedata[0]++;
   if(timedata[0]==100)
   {
   timedata[0]=0;timedata[1]++;
   if(timedata[1]==100)
   {
      timedata[1]=0;
   }
   }
   write_add(12,shuju());
}
ET0=1;
}

void init()
{
  uchar x;
  sda=1;
  delay();
  scl=1;
  delay();
  x=read_add(12);
  timedata[0]=x%100;
  timedata[1]=x/100;
  scan();
  delay1(1);
}
main()
{
init();
shuju();
TMOD=0x01;
TH0=0x3c;
TL0=0xb0;
TR0=1;
ET0=1;
EA=1;
while(1)
{scan();}
}

at2402程序的问题， [复制链接]

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