按键按下一次1602显示字符内容，断电后送电显示器1602继续显示停电前的字符内容

ID:488631 · 发表于 2019-12-19 19:48

按对应按钮时1602显示对应字符内容，断电在送电又恢复初始状态，不能实现利用24c02读写达到断电保护的目的。请各位大佬指点一下小弟错在何处。

单片机源程序如下:

#include<reg52.h>
#include<intrins.h>
#define uchar unsigned char
#define uint unsigned int
uchar code table[]="I LOVE DIANZI";
uchar code table1[]=" ABCD ";
uchar num,sec;
sbit key1 = P3^0;
sbit key2 = P3^1;
sbit key3 = P3^2;
sbit key4 = P3^3;
/* sbit key5 = P3^4;
sbit key6 = P3^5;
sbit key7 = P3^6;
sbit key8 = P3^7;*/
sbit sda = P2^0;
sbit scl = P2^1;
bit write = 0;
bit k1 = 0;
bit k2 = 0;
sbit lcdrs = P2^6;
sbit lcdwr = P2^5;
sbit lcden = P2^7;
void delay1s(uint z)
{
uint x,y;
for(x=z;x>0;x--)
for(y=110;y>0;y--);
}
void delay()
{;;}
void at_start()
{
sda = 1;
delay();
scl = 1;
delay();
sda = 0;
delay();
}
void at_stop()
{
sda = 0;
delay();
scl = 1;
delay();
sda = 1;
delay();
}
void at_respons()
{
uchar i;
scl = 1;
delay();
while((sda==1)&&(i<255))i++;
scl = 0;
delay();
}
void at_init()
{
sda = 1;
delay();
scl = 1;
delay();
}
void at_write_byte(uchar date)
{
uchar i,temp;
temp =date;
for(i=0;i<8;i++)
{
temp=temp<<1;
scl = 1;
delay();
sda = CY;
delay();
scl = 1;
delay();
}
scl = 0;
delay();
sda = 1;
delay();
}
uchar at_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 at_write_add(uchar address,uchar date)
{
at_start();
at_write_byte(0xa0);
at_respons();
at_write_byte(address);
at_respons();
at_write_byte(date);
at_respons();
at_stop();
}
uchar at_read_add(uchar address)
{
uchar date;
at_start();
at_write_byte(0xa0);
at_respons();
at_write_byte(address);
at_respons();
at_start();
at_write_byte(0xa1);
at_respons();
date=at_read_byte();
at_stop();
return date;
}
bit lcd_busy()
{
bit result;
lcdrs = 0;
lcdwr = 1;
lcden = 1;
delay();
delay();
delay();
result=(bit)(P0&0x80);
lcden = 0;
return(result);
}
void lcd_send_com(uchar com)
{
while(lcd_busy());
lcdrs = 0;
lcdwr = 0;
P0 = com;
delay1s(5);
lcden = 1;
delay1s(5);
lcden = 0;
}
void lcd_send_data(uchar date)
{
while(lcd_busy());
lcdrs = 1;
lcdwr = 0;
P0 = date;
delay1s(5);
lcden=1;
delay1s(5);
lcden=0;
}
void lcd_init()
{
lcden = 0;
lcdrs = 0;
lcdwr = 0;
lcd_send_com(0x38);
lcd_send_com(0x0c);
lcd_send_com(0x06);
lcd_send_com(0x01);
}
uchar keyscan()
{
if(key1==0)
{
delay1s(5);
if(key1==0)
{
k1 = 1;
lcd_send_com(0x80);
for(num = 0;num<13;num++)
{
lcd_send_data(table[num]);
delay1s(5);
}
}
while(!key1);
}
else if(key2==0)
{
delay1s(5);
if(key2==0)
{
k1 = 2;
lcd_send_com(0x80);
for(num = 0;num<13;num++)
{
lcd_send_data(table[num]);
delay1s(5);
}
}
while(!key2);
}
return k1;
}
void displaychar()
{ uchar p1;
p1=at_read_add(8);
delay1s(20);
if(p1==1)
{
lcd_send_com(0x80);
for(num = 0;num<13;num++)
{
lcd_send_data(table[num]);
delay1s(5);
}
}
else if(p1==2)
{
lcd_send_com(0x80);
for(num = 0;num<13;num++)
{
lcd_send_data(table1[num]);
delay1s(5);
}
}
}
void main()
{
at_init();
lcd_init();
displaychar();
while(1)
{
keyscan();
at_write_add(8,k2);
}
}

复制代码

ID:584814 · 发表于 2019-12-20 08:24

全程无备注看得太眼花，提供点思路：
没存进的可能，没读出的可能，被初始化的可能。

ID:661787 · 发表于 2019-12-21 10:03

建议写程序时给程序添加解释说明

ID:664105 · 发表于 2019-12-21 14:54

有条理的写程序，每一段要明白自己要干什么

ID:488631 · 发表于 2019-12-25 14:48

好的，谢谢各位大佬指教，我从新添加注释又发了，谢谢！

ID:437615 · 发表于 2019-12-25 17:09

你的板子有没有问题，有没有延时断电功能

ID:475247 · 发表于 2019-12-25 20:54

思路是这样的，你按键按下后让24c02记下按键值，开机后用24c02读出的值来执行显示任务，这样能起到开机恢复断电前画面。

帐号		自动登录	找回密码
密码			立即注册