这是一个基于定时器的电子时钟,以前总是觉得数码管和LED挺简单的,谁知道组合起来还是有些困难的,但还是一一的解决了程序中的bug,新手也可以借鉴一下,感觉还挺有趣的。如果发现程序中有什么bug可以回帖相互探讨一下。
单片机源程序如下:
#include
#include
#define uchar unsigned char
#define uint unsigned int
uchar code huayang1[]={0x7f,0xbf,0xdf,0xef,0xf7,0xfb,0xfd,0xfe,0xfd,0xfb,0xf7,0xef,0xdf,0xbf}; //花样1
uchar code huayang2[]={0x7f,0xfe,0xbf,0xfd,0xdf,0xfb,0xef,0xf7,0xef,0xfb,0xdf,0xfd,0xbf,0xfe}; //花样2
uchar code huayang3[]={0x7f,0x3f,0x1f,0x0f,0x07,0x03,0x01,0x00,0x80,0xc0,0xe0,0xf0,0xf8,0xfc,0xfe,0xff}; //花样3
uchar code huayang4[]={0x55,0xaa,0xcc,0x33,0x99,0x66,0x0f,0xf0}; //花样4
uchar code Segcode[13]={0xC0,0xF9,0xA4,0xB0,0x99,0x92,0x82,0xF8,0x80,0x90,0xff,0xbf, 0xC7};
uchar Dispbut[8]={10,10,10,10,10,10,10,10};
uchar Bitselect[8]={0x01,0x02,0x04,0x08,0x10,0x20,0x40,0x80};
uchar Displayer[8]={0x00};
uint i,a,b;
sbit DU=P2^0;
sbit WE=P2^1;
#define Led_rod_DU 11
uchar Sec=0,Min=30,Hour=12;
void Display(void)
{
static uchar Num=0;
DU=1;
P0=0xff;
DU=0;
WE=1;
P0=Bitselect[Num];
WE=0;
DU=1;
P0=Segcode[Dispbut[Num]];
DU=0;
Num++;
if(Num>=8)
Num=0;
}
void Timer0_int()interrupt 1
{
TH0=(65536-5000)/256;
TL0=(65536-5000)%256;
Display();
}
void Timer1_int()interrupt 3
{
static uint Timer1_Count=0;
TH1=(65536-50000)/256;
TL1=(65536-50000)%256;
Timer1_Count++;
if(Timer1_Count>=20)
{
Timer1_Count=0;
Sec++;
if(Sec>59)
{
Sec=0;
Min++;
if(Min>59)
{
Min=0;
Hour++;
if(Hour>23)
Hour=0;
}
}
Dispbut[0]=Hour/10;
Dispbut[1]=Hour%10;
Dispbut[2]=Led_rod_DU;
Dispbut[3]=Min/10;
Dispbut[4]=Min%10;
Dispbut[5]=Led_rod_DU;
Dispbut[6]=Sec/10;
Dispbut[7]=Sec%10;
}
}
void delay(uint i)
{
while(i--)
for(b=0;b<100;b++)
;
}
void liushui()
{
a=0xfe;
for(i=0;i<8;i++)
{
P1=a;
a=_crol_(a,1);
delay(200);
}
a=0xfc;
for(i=0;i<8;i++)
{
P1=a;
a=_crol_(a,1);
delay(200);
}
for(i=0;i<14;i++)
{
P1=huayang1[i];
delay(200);
}
for(i=0;i<14;i++)
{
P1=huayang2[i];
delay(200);
}
for(i=0;i<14;i++)
{
P1=huayang3[i];
delay(200);
}
for(i=0;i<14;i++)
{
P1=huayang4[i];
delay(200);
}
}
……………………
『本文转载自网络,版权归原作者所有,如有侵权请联系删除』