9.19键盘的应用

作业1:数码管前三位数显示一个跑表,从000到999,之间以百分之一秒的速度运行,当按下key1时跑表停止,再次按下时跑表开始

#include<reg52.h>
#define uint unsigned int
#define uchar unsigned char 

sbit dula =P2^6;
sbit wela = P2^7;
sbit key1 = P3^7;
uchar aa,flag,bai,shi,ge,tt;
uint num,bb;
void init();
void keyscan();
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();
	P3 = 0xff;
	while(1)
	{
		display(bai,shi,ge);
		if(key1==0)//检测是否按下
		{
			delay(10);
			if(key1==0)//确实按下了
			{
				TR0 = 0;//定时器停止
				bb ++;//计算按下的次数
				while(!key1)
						display(bai,shi,ge);
				if(bb %2==0)
				{
					TH0 =(65536-50000)/256;//装初值
					TL0 =(65536-50000)%256;	
					TR0 = 1;//重启定时器
				}
			}
		}
	}

}



void delay(uint z)
{
	uint x,y;
	for(x=z;x>0;x--)
		for(y=110;y>0;y--);
}

void init()
{
	TMOD = 0x01;
	TH0 =(65536-50000)/256;
	TL0 =(65536-50000)%256;
	EA = 1;
	ET0 = 1;
	TR0 = 1;

}

void time0()interrupt 1
{
	TH0 =(65536-50000)/256;
	TL0 =(65536-50000)%256;
	aa++;

		if(aa==2)
		{
			aa = 0;
			num ++;
			if(num==1000)
				num=0;
		}

	bai = num/100;
	shi = num/10%10;
	ge = num%10;

} 
void display(uchar bai,uchar shi,uchar ge)
{
    wela = 1;
    P0 = 0xfe;
    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;
 
    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[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;
 
}

 在上题目的基础上,用另外三个独立键盘实现按下第一个时计数停止,按下第二个时计数开始,按下第三个时计数清零重新开始

#include<reg52.h>
#define uint unsigned int
#define uchar unsigned char 

sbit dula =P2^6;
sbit wela = P2^7;
sbit key1 = P3^3;
sbit key2 = P3^4;
sbit key3 = P3^5;
sbit key4 = P3^6;
uchar aa,flag,bai,shi,ge,tt;
uint num,bb;
void init();
void keyscan();
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();
	P3 = 0xff;
	while(1)
	{
		display(bai,shi,ge);
		if(key1==0)//检测是否按下
		{
			delay(10);
			if(key1==0)//确实按下了
			{
				TR0 = 0;//定时器停止
				while(!key1)
						display(bai,shi,ge);
							TR0 = 1;//重启定时器
			
			}
		}
		if(key2==0)
		{
			delay(10);
			if(key2==0)
				{
					TR0 = 0;//定时器停止
					while(!key2)
					display(bai,shi,ge);
				}
		}
	
		if(key3==0)
		{
			delay(10);
			if(key3==0)
				{
					TH0=(65536-50000)/256;
					TL0 = (65536-50000)%256;
					TR0 = 1;//重启定时器
					while(!key3)
					display(bai,shi,ge);//不加入这句话有很大的bug  
				}
		}

		if(key4==0)
		{
			delay(10);
			if(key4==0)
				{
					num = 0;//重新开始计数
					while(!key4)
					display(bai,shi,ge);
				}
		}

	}

}



void delay(uint z)
{
	uint x,y;
	for(x=z;x>0;x--)
		for(y=110;y>0;y--);
}

void init()
{
	TMOD = 0x01;
	TH0 =(65536-50000)/256;
	TL0 =(65536-50000)%256;
	EA = 1;
	ET0 = 1;
	TR0 = 1;

}

void time0()interrupt 1
{
	TH0 =(65536-50000)/256;
	TL0 =(65536-50000)%256;
	aa++;

		if(aa==2)
		{
			aa = 0;
			num ++;
			if(num==1000)
				num=0;
		}

	bai = num/100;
	shi = num/10%10;
	ge = num%10;

} 
void display(uchar bai,uchar shi,uchar ge)
{
    wela = 1;
    P0 = 0xfe;
    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;
 
    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[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;
 
}

 作业3:按下16个矩阵键盘依次在数码管上显示1到16的平方

#include<reg52.h>
#define uint unsigned int
#define uchar unsigned char
sbit dula = P2^6;
sbit wela = P2^7;

uchar num1,num,temp,tt;
uint aa;
uchar keyscan();
void display(uchar bai,uchar shi,uchar ge);
void delay(uint z);
uchar code table[]={0x3f,0x06,0x5b,0x4f,
					0x66,0x6d,0x7d,0x07,
					0x7f,0x6f,0x77,0x7c,
					0x39,0x5e,0x79,0x71};
main()
{

	while(1)
	{
		display(aa/100,aa/10%10,aa%10);
		num1 = keyscan();
		aa = num1*num1;
	
	}
}

void delay(uint z)
{
	uint x,y;
	for(x=z;x>0;x--)
		for(y=110;y>0;y--);
}

void display(uchar bai,uchar shi,uchar ge)
{
    wela = 1;
    P0 = 0xfe;
    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;
 
    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[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;
 
}


uchar keyscan()
{
		P3 = 0xfe;//1111 1110
		temp = P3;
		temp = temp&0xf0;//11110000
		while(temp!=0xf0)//如果有键摁下
		{
			delay(5);//再次检测P3口
			temp = P3;
			temp = temp&0xf0;
			while(temp!=0xf0)
			{
				temp = P3;	
				switch(temp)
				{
					case 0x7e:num = 1;
						break;
					case 0xbe:num = 2;
						break;
					case 0xde:num = 3;
						break;
					case 0xee:num = 4;
						break;
					
				}
				while(temp!=0xf0)//松手检测
				{
					temp = P3;
					temp = temp&0xf0;
					display(aa/100,aa/10%10,aa%10);//这句话不加,有很大的bug  
				}
			
			}
		}

		
		P3 = 0xfd;//1111 1101
		temp = P3;
		temp = temp&0xf0;//11110000
		while(temp!=0xf0)//如果有键摁下
		{
			delay(5);//再次检测P3口
			temp = P3;
			temp = temp&0xf0;
			while(temp!=0xf0)
			{
				temp = P3;	
				switch(temp)
				{
					case 0x7d:num = 5;
						break;
					case 0xbd:num = 6;
						break;
					case 0xdd:num = 7;
						break;
					case 0xed:num = 8;
						break;
					
				}
				while(temp!=0xf0)//松手检测
				{
					temp = P3;
					temp = temp&0xf0;
					display(aa/100,aa/10%10,aa%10);
				}
		
			}
		}



		P3 = 0xfb;//1111 1011
		temp = P3;
		temp = temp&0xf0;//11110000
		while(temp!=0xf0)//如果有键摁下
		{
			delay(5);//再次检测P3口
			temp = P3;
			temp = temp&0xf0;
			while(temp!=0xf0)
			{
				temp = P3;	
				switch(temp)
				{
					case 0x7b:num = 9;
						break;
					case 0xbb:num = 10;
						break;
					case 0xdb:num = 11;
						break;
					case 0xeb:num = 12;
						break;
					
				}
				while(temp!=0xf0)//松手检测
				{
					temp = P3;
					temp = temp&0xf0;
					display(aa/100,aa/10%10,aa%10);
				}
	
			}
		}

		P3 = 0xf7;//1111 0111
		temp = P3;
		temp = temp&0xf0;//11110000
		while(temp!=0xf0)//如果有键摁下
		{
			delay(5);//再次检测P3口
			temp = P3;
			temp = temp&0xf0;
			while(temp!=0xf0)
			{
				temp = P3;	
				switch(temp)
				{
					case 0x77:num = 13;
						break;
					case 0xb7:num = 14;
						break;
					case 0xd7:num = 15;
						break;
					case 0xe7:num = 16;
						break;
					
				}
				while(temp!=0xf0)//松手检测
				{
					temp = P3;
					temp = temp&0xf0;
					display(aa/100,aa/10%10,aa%10);
				}
	
			}
		}
		
		return num;

}

 其实,单片机的功能模块有很多的示例代码可以参考,也正是由于c语言具有极好的移植性,所以大大加快了现在点子产品的开发周期

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posted @ 2013-09-19 16:08  跨七海的风  阅读(292)  评论(0编辑  收藏  举报