9.15学习笔记

数码管的动态显示：精简代码版,要注意关灯，否则有问题，还有时间消影，这个例子虽然很简单，但是精简代码百分之九十的人会写错

#include<reg52.h>

#define uint unsigned int

#define uchar unsigned char



void delay(uint z);

uchar code table[]={ 0x3f,0x06,0x5b,0x4f,

					0x66,0x6d,0x7d,0x07,

					0x7f,0x6f,0x77,0x7c,

					0x39,0x5e,0x79,0x71};



uchar code tablew[]={0xfe,0xfd,0xfb,0xf7,

					0xef,0xdf,0xbf,0x7f};



uchar aa,bb;

uint tt;

sbit dula =P2^6;

sbit wela = P2^7;



main()

{

	aa = 0;

	bb = 6;

	

	while(1)

	{

		wela = 1;

		P0 = tablew[aa];

		wela = 0;

		P0 = 0xff;

		//delay(1);



	

		dula = 1;

		P0 = table[bb];

		dula = 0;

		P0 = 0xff;

		tt=300;

		while(tt--);



		dula = 1;

		P0 = 0x0;//要关灯，或者这个程序有问题，

		dula = 0;



		delay(1);



		

		aa ++;

		bb--;

	

	

		if(aa == 6)

			aa = 0;

		if(bb ==0)

			bb = 6;

		

	}

	while(1);

}



void delay(uint z)

{

	uint x,y;

	for(x = z;x>0;x--)

		for(y=110;y>0;y--);



}

 作业2：用数码管的动态扫描制成一个秒表，让数码管后两位显示%1秒

//程序虽然写出来了，但是调试可是花了我大力气， 对于数码管而言   

//要记住，不要使用delay（）郭天祥那个延时函数，因为这里涉及的是微妙级别的数    

//display这个函数执行的时间不能大于10微秒，这就必须消影，

//高速度扫描时，不消影后果很严重，每次段选完以后要进行关灯   

//timu:有两个动态扫描的方法和定时器1在数码管的前三位显示表秒

//精确到%1，即后两位显示百分之一秒，一直循环下去 

//关于定时器装初值问题，这里有一个规律，就是，12Mhz的晶振，计数恋

//计数N毫秒，就装入N乘以1000 

#include<reg52.h>

#define uint unsigned int

#define uchar unsigned char



uint num;

uchar aa,bb,ge,shi,bai,tt;

sbit  wela =P2^6;

sbit dula = P2^7;

void init();

void delay(uint z);

void display(uchar bai,uchar shi,uchar ge);

uchar code table[]={0x3f,0x06,0x5b,0x4f,

					0x66,0x6d,0x7d,0x07,

					0x7f,0x6f,0x77,0x7c,

					0x39,0x5e,0x79,0x71};



main()

{

	init();

	while(1)

	{

		if(aa==10)

		{	aa = 0;

			num++;

			if(num == 1000)

			num=0;

		}

		ge = num%10;

		shi = num/10%10;

		bai = num/100;

		

		display(ge,shi,bai);

	}



}



void init()

{

	TMOD = 0x11;

	TH1 = (65535-1000)/256;

	TL1 = (65535-1000)%256;



	EA = 1;//开总中断

	ET1 = 1;//开定时器1中断

	TR1 = 1;//启动定时器1

	aa = 0;



}



void display(uchar bai,uchar shi,uchar ge)

{

	wela = 1;

	P0 = 0xfe;

	wela = 0;

	P0 = 0x0;//消影操作



	dula = 1;

	P0 = table[ge];

	dula = 0;

	tt = 25;

	while(tt--);

	dula = 1;//关灯操作，高速度扫描时必备 

	P0 = 0;

	dula = 0;



	wela = 1 ;

	P0 = 0xfd;

	wela = 0;

	P0= 0x0;//消影操作



	dula = 1;

	P0 = table[shi];

	dula = 0;

	tt = 25;

	while(tt--);

	dula = 1;//关灯操作，高速度扫描时必备 

	P0 = 0;

	dula = 0;





	wela = 1;

	P0 = 0xfb;

	wela = 0;

	P0 = 0x0;//消影操作



	dula = 1;

	P0 = table[bai];

	dula =0;

	tt=25;

	while(tt --);

	

	dula = 1;//关灯操作，高速度扫描时必备 

	P0 = 0;

	dula = 0;



}

 



void time1()interrupt 3

{

	TH1 = (65535-1000)/256;

	TL1 = (65535-1000)%256;

	aa ++;



}

 作业3：利用动态扫描，和定时器1，在数码管上显示从765432开始以十分之一秒的速度往下递减直至765398，并且保持住此数，与此同时利用定时器0以500ms的速度进行流水灯从上往下移动，但数码管减到停止时，实验板上流水灯也停止，然后开始闪烁，3秒后，（用T0）计时，数码管上显示HELLO，并且保持住

//这个程序也调试了我好久，这里主要是标志位的使用  

//当一个事件接二连三的发生时，这里要考虑设置标志位

//定时器的关闭后，再次启动，需要重新装入初值，还要重新启动   



#include<reg52.h>

#include<intrins.h>



#define uint unsigned int

#define uchar unsigned char



sbit dula = P2^2;

sbit wela = P2^3;



uchar aa,bb,cc,dd,ee,ff,bai,shi,ge,flag,flag1;

uint num,tt;



void init();//一个机器周期执行时间就是1um  

void display(uchar cc,uchar dd,uchar ee,uchar bai,uchar shi,uchar ge);



uchar code table[]={0x3f,0x06,0x5b,0x4f,

					0x66,0x6d,0x7d,0x07,

					0x7f,0x6f,0x77,0x7c,

					0x39,0x5e,0x79,0x71,

					0x76,0x79,0x38,0x3f,0};



void main()

{

	init();

	bai = num/100;

	shi = num/10%10;

	ge = num%10;

	while(1)

	{

		if(flag1 != 1)

		{

			display(7,6,5,bai,shi,ge);

		}

		else

		{

			display(16,17,18,18,19,20);

		}

	}

	



}



void init()

{

	TMOD = 0x11;

	TH1 = (65535-50000)/256;

	TL1 = (65535-50000)%256;

	

	TH0 = (65535-50000)/256;

	TL0 = (65535-50000)%256;





	EA = 1;

	ET1 = 1;

	TR1 = 1;

	

	

	ET0 = 1;

	TR0 = 1;

	bb = 0;

	aa = 0xfe;

	ff = 0;

	num=432;

	flag = 0;

	flag1 = 0;



	

}



void display(uchar cc,uchar dd,uchar ee,uchar bai,uchar shi,uchar ge)

{

	wela = 1;

	P0 = 0xfe;

	wela = 0;

	P0 = 0x0;



	dula = 1;

	P0 = table[cc];

	dula = 0;

	P0 = 0xff;

	tt=50;

	while(tt--);

	dula = 1;

	P0=0;

	dula =0;



	wela = 1;

	P0 = 0xfd;

	wela = 0;

	P0 = 0x0;



	dula = 1;

	P0 = table[dd];

	dula = 0;

	P0 = 0xff;

	tt=50;

	while(tt--);

	dula = 1;

	P0=0;

	dula =0;



	wela = 1;

	P0 = 0xfb;

	wela = 0;

	P0 = 0x0;



	dula = 1;

	P0 = table[ee];

	dula = 0;

	P0 = 0xff;

	tt=50;

	while(tt--);

	dula = 1;

	P0=0;

	dula =0;



	wela = 1;

	P0 = 0xf7;

	wela = 0;

	P0 = 0x0;



	dula = 1;

	P0 = table[bai];

	dula = 0;

	P0 = 0xff;

	tt=50;

	while(tt--);

	dula = 1;

	P0=0;

	dula =0;



	wela = 1;

	P0 = 0xef;

	wela = 0;

	P0 = 0x0;



	dula = 1;

	P0 = table[shi];

	dula = 0;

	P0 = 0xff;

	tt=50;

	while(tt--);

	dula = 1;

	P0=0;

	dula =0;



	wela = 1;

	P0 = 0xdf;

	wela = 0;

	P0 = 0x0;



	dula = 1;

	P0 = table[ge];

	dula = 0;

	P0 = 0xff;

	tt=50;

	while(tt--);

	dula = 1;

	P0=0;

	dula =0;



}



void time_0()interrupt 1

{



	TH0 = (65535-50000)/256;

	TL0 = (65535-50000)%256;

	bb++;

	if(flag!=1)

	{

		if(bb==10)

		{

			bb = 0;

			aa = _crol_(aa,1);

			P1 = aa;

		}

		

	}

	else

	{

		if(bb%2==0)

		{

			

			P1 = ~P1;

			if(bb==30)

			{

				TR0 = 0;

				P1 = 0xff;

				flag1 = 1;

			}

		}

	

	}





}



void time1()interrupt 3

{

	TH1 = (65535-50000)/256;

	TL1 = (65535-50000)%256;

	ff++;

	if(ff==2)

	{	

		ff = 0;

		num --;

		if(num == 357)

			{

			TR0 = 0;

			TR1 = 0;

			flag = 1;

			

			TH0 = (65536-50000)/256;//重新装入初值

			TL0 = (65536-50000)%256;

			TR0 = 1;

			

			}

		bai = num/100;

		shi = num/10%10;

		ge = num%10;

	}

}

关于中断函数中写入多少程序好的分析：中断函数假设为50毫秒，程序执行的时间不要超过50毫秒，一个机器周期大约是1微妙，单周期指令，双周期指令，那么撑死他有1000,行，才是1毫秒，一般的不要太复杂，所以，在中断函数中，做一些有利于观察，有利于计算的就行了

 

 

版权所有，转载请注明链接地址：http://www.cnblogs.com/fengdashen/p/3322170.html