
请专业解答!!非常感谢! 郭天祥单片机 I方C总线 接E方PROM 的记忆问题 求解! 感激耐心看完 好人一生平安.
下面程序想实现倒计时100-0在数码管显示,并每一秒都往E方PROM中输入,以此来记忆,每次重开单片机都会在原来数的基础上继续倒计时.但是下面我做的程序怎么不能实现啊.....
下面程序想实现倒计时100-0 在数码管显示,并每一秒都往E方PROM 中输入,以此来记忆,每次重开单片机 都会在原来数的基础上继续倒计时. 但是下面我做的程序怎么不能实现啊....百思不得其解.求好人耐心解答!!!谢谢!
#include<reg52.h>
#include<intrins.h>
#define uchar unsigned char
#define uint unsigned int
sbit sda=P2^0;
sbit scl=P2^1;
sbit wela=P2^7;
sbit dula=P2^6;
uchar code table[]={0x3f,0x06,0x5b,0x4f,0x66,0x6d,0x7d,0x07,0x7f,0x6f,0x77,0x7c,0x39,0x5e,0x79,0x71};
uchar a=0xfe,num=0,num1=100;
void delayms(uint xms)
{
uint j;
for(;xms>0;xms--)
for(j=110;j>0;j--);
}
void delay()
{ ;; }
void start() //开始信号
{ scl=1;
delay();
sda=1;
delay();
sda=0;
delay();
}
void stop() //停止
{
scl=1;
delay();
sda=0;
delay();
sda=1;
delay();
}
void respons() //应答
{
uchar i;
scl=1;
delay();
while((sda==1)&&(i<250))i++;
scl=0;
delay();
}
void init() //订计时初始化
{
TMOD=0x01;
TH0=(65536-45872)/256;
TL0=(65536-45872)%256;
EA=1;
ET0=1;
TR0=1;
}
void write_byte(uchar date) //往"记忆"芯片中写入一字节
{
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() //从记忆芯片中读一字节
{
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) //写
{
start();
write_byte(0xa0);
respons();
write_byte(address);
respons();
write_byte(date);
respons();
stop();
}
uchar read_add(uchar address) //读 address为记忆芯片中的储存地址
{
uchar date;
start();
write_byte(0xa0); //0xa0为记忆芯片在总线中的地址
respons();
write_byte(address);
respons();
start();
write_byte(0xa1);
respons();
date=read_byte();
stop();
return date;
}
void display(uchar num)
{
P0=0xfe;
wela=1;
wela=0;
P0=table[num/100];
dula=1;
dula=0;
delayms(1);
P0=0xfd;
wela=1;
wela=0;
P0=table[num%100/10];
dula=1;
dula=0;
delayms(1);
P0=0xfb;
wela=1;
wela=0;
P0=table[num%10];
dula=1;
dula=0;
delayms(1);
}
void main()
{
init();
delayms(100);
num1=read_add(23);
while(1)
{
display(num1);
}
}
void T0_time() interrupt 1
{
TH0=(65536-45872)/256;
TL0=(65536-45872)%256;
num++;
if(num==20)
{
num=0;
num1--;
if(num1==0)num1=100;
write_add(23,num1);
}
} 展开
#include<reg52.h>
#include<intrins.h>
#define uchar unsigned char
#define uint unsigned int
sbit sda=P2^0;
sbit scl=P2^1;
sbit wela=P2^7;
sbit dula=P2^6;
uchar code table[]={0x3f,0x06,0x5b,0x4f,0x66,0x6d,0x7d,0x07,0x7f,0x6f,0x77,0x7c,0x39,0x5e,0x79,0x71};
uchar a=0xfe,num=0,num1=100;
void delayms(uint xms)
{
uint j;
for(;xms>0;xms--)
for(j=110;j>0;j--);
}
void delay()
{ ;; }
void start() //开始信号
{ scl=1;
delay();
sda=1;
delay();
sda=0;
delay();
}
void stop() //停止
{
scl=1;
delay();
sda=0;
delay();
sda=1;
delay();
}
void respons() //应答
{
uchar i;
scl=1;
delay();
while((sda==1)&&(i<250))i++;
scl=0;
delay();
}
void init() //订计时初始化
{
TMOD=0x01;
TH0=(65536-45872)/256;
TL0=(65536-45872)%256;
EA=1;
ET0=1;
TR0=1;
}
void write_byte(uchar date) //往"记忆"芯片中写入一字节
{
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() //从记忆芯片中读一字节
{
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) //写
{
start();
write_byte(0xa0);
respons();
write_byte(address);
respons();
write_byte(date);
respons();
stop();
}
uchar read_add(uchar address) //读 address为记忆芯片中的储存地址
{
uchar date;
start();
write_byte(0xa0); //0xa0为记忆芯片在总线中的地址
respons();
write_byte(address);
respons();
start();
write_byte(0xa1);
respons();
date=read_byte();
stop();
return date;
}
void display(uchar num)
{
P0=0xfe;
wela=1;
wela=0;
P0=table[num/100];
dula=1;
dula=0;
delayms(1);
P0=0xfd;
wela=1;
wela=0;
P0=table[num%100/10];
dula=1;
dula=0;
delayms(1);
P0=0xfb;
wela=1;
wela=0;
P0=table[num%10];
dula=1;
dula=0;
delayms(1);
}
void main()
{
init();
delayms(100);
num1=read_add(23);
while(1)
{
display(num1);
}
}
void T0_time() interrupt 1
{
TH0=(65536-45872)/256;
TL0=(65536-45872)%256;
num++;
if(num==20)
{
num=0;
num1--;
if(num1==0)num1=100;
write_add(23,num1);
}
} 展开
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