51单片机电子钟程序数码管显示

如题所述

第1个回答  2021-08-04
#include<reg51.h>

#define uchar unsigned char

sbit dula=P2^6;
sbit wela=P2^7;
sbit beep=P2^3;
unsigned char j,k,a1,a0,b1,b0,c1,c0,s,f,m,key=10,temp,qq;
uchar shi20,shi10,fen20,fen10,miao20,miao10,new,ok=1,wei;
unsigned int pp;
unsigned char code table[]={0x3f,0x06,0x5b,0x4f,0x66,0x6d,0x7d,
0x07,0x7f,0x6f,0x77,0x7c,0x39,0x5e,0x79,0x71};

void delay(unsigned char i)
{
for(j=i;j>0;j--)
for(k=125;k>0;k--);
}
void display(uchar shi2,uchar shi1,uchar fen2,uchar fen1,uchar miao2,uchar miao1)
{
dula=0;
P0=table[shi2];
dula=1;
dula=0;

wela=0;
P0=0xfe;
wela=1;
wela=0;
delay(5);

P0=table[shi1]|0x80;
dula=1;
dula=0;

P0=0xfd;
wela=1;
wela=0;
delay(5);

P0=table[fen2];
dula=1;
dula=0;

P0=0xfb;
wela=1;
wela=0;
delay(5);

P0=table[fen1]|0x80;
dula=1;
dula=0;

P0=0xf7;
wela=1;
wela=0;
delay(5);

P0=table[miao2];
dula=1;
dula=0;

P0=0xef;
wela=1;
wela=0;
delay(5);

P0=table[miao1];
dula=1;
dula=0;

P0=0xdf;
wela=1;
wela=0;
delay(5);
}

void keyscan0()
{
P3=0xfb;
temp=P3;
temp=temp&0xf0;
if(temp!=0xf0)
{
delay(10);
if(temp!=0xf0)
{
temp=P3;
switch(temp)
{
case 0xbb:
ok=0;
break;

case 0x7b:
ok=1;
break;
}
}
}
}

void keyscan()
{
{
P3=0xfe;
temp=P3;
temp=temp&0xf0;
if(temp!=0xf0)
{
delay(10);
if(temp!=0xf0)
{
temp=P3;
switch(temp)
{
case 0xee:
key=0;
wei++;
break;

case 0xde:
key=1;
wei++;
break;

case 0xbe:
key=2;
wei++;
break;

case 0x7e:
key=3;
wei++;
break;
}
while(temp!=0xf0)
{
temp=P3;
temp=temp&0xf0;
beep=0;
}
beep=1;
}
}
P3=0xfd;
temp=P3;
temp=temp&0xf0;
if(temp!=0xf0)
{
delay(10);
if(temp!=0xf0)
{
temp=P3;
switch(temp)
{
case 0xed:
key=4;
wei++;
break;

case 0xdd:
key=5;
wei++;
break;

case 0xbd:
key=6;
wei++;
break;

case 0x7d:
key=7;
wei++;
break;
}
while(temp!=0xf0)
{
temp=P3;
temp=temp&0xf0;
beep=0;
}
beep=1;
}
}
P3=0xfb;
temp=P3;
temp=temp&0xf0;
if(temp!=0xf0)
{
delay(10);
if(temp!=0xf0)
{
temp=P3;
switch(temp)
{
case 0xeb:
key=8;
wei++;
break;

case 0xdb:
key=9;
wei++;
break;
}
while(temp!=0xf0)
{
temp=P3;
temp=temp&0xf0;
beep=0;
}
beep=1;
}
}
}
}

void main()
{
TMOD=0x01;

TH0=(65536-46080)/256;// 由于晶振为11.0592,故所记次数应为46080,计时器每隔50000微秒发起一次中断。
TL0=(65536-46080)%256;//46080的来历,为50000*11.0592/12
ET0=1;
EA=1;

while(1)
{ keyscan0();

if(ok==1)
{ TR0=1;
wei=0;

if(pp==20)
{ pp=0;
m++;
if(m==60)
{
m=0;
f++;
if(f==60)
{
f=0;
s++;
if(s==24) //为24h一个循环,若要12h,只需在此改为12即可。
{
s=0;
}
}
}
}

a0=s%10;
a1=s/10;
b0=f%10;
b1=f/10;
c0=m%10;
c1=m/10;
display(a1,a0,b1,b0,c1,c0);
}
else
{ TR0=0;
keyscan();
if(key!=10)
{

switch(wei)
{
case 1: if(key<3) //小时最高位为2
a1=key;
else
wei--;
break;
case 2: if(a1==1|a1==0)
a0=key;
else
if(key<5)
a0=key; //当小时最高位为2时,低位最高为4
break;
case 3: if(key<7) //分钟最高位为6
b1=key;
else
wei--;
break;
case 4: b0=key; break;
case 5: if(key<7) //秒最高位为6
c1=key;
else
wei--;
break;
case 6: c0=key; break;
}
key=10;
}
m=c1*10+c0;
f=b1*10+b0;
s=a1*10+a0;
display(a1,a0,b1,b0,c1,c0);
}
}
}

void time0() interrupt 1
{ TH0=(65536-46080)/256;
TL0=(65536-46080)%256;
pp++;
}