51单片机独立按键的使用 -

标题: 51单片机独立按键的使用 [打印本页]

作者: 107415890 时间: 2018-5-16 15:49
标题: 51单片机独立按键的使用
51单片机独立按键的使用，主函数的c文件如下，完整工程文件在附件
#include<reg52.h>
#define uchar unsigned char
#define uint unsigned int
sbit we = P2^7; //数码管位选
sbit du = P2^6; //数码管段选

sbit key_s2 = P3^0; //S2按键未定义

uchar code leddata[]={

            0x3F,  //"0"
            0x06,  //"1"
            0x5B,  //"2"
            0x4F,  //"3"
            0x66,  //"4"
            0x6D,  //"5"
            0x7D,  //"6"
            0x07,  //"7"
            0x7F,  //"8"
            0x6F,  //"9"
            0x77,  //"A"
            0x7C,  //"B"
            0x39,  //"C"
            0x5E,  //"D"
            0x79,  //"E"
            0x71,  //"F"
            0x76,  //"H"
            0x38,  //"L"
            0x37,  //"n"
            0x3E,  //"u"
            0x73,  //"P"
            0x5C,  //"o"
            0x40,  //"-"
            0x00,  //熄灭
            0x00  //自定义

                     };

void delay(uint z)
{
uint x,y;
for(x = z; x > 0; x--)
for(y = 114; y > 0 ; y--);
}

void main()
{
   uchar num = 0;
   we = 1;    //数码管位选
   P0 = 127;  //选通第一个数码管,该数码管接低电平，其余为高电平
   we = 0;
   while(1)
   {
      du = 1;
P0 = leddata[num];
du =0;
if(key_s2 == 0)
{
delay(5); //软件延时
if(key_s2 == 0)
{
num++;
}
while(!key_s2);//松手检测
}
if(num == 10)
{
num=0;
}

   }

}

1独立按键.zip

13.43 KB, 下载次数: 4, 下载积分: 黑币 -5

作者: xiaoyu. 时间: 2018-5-18 15:16
独立按键控制数码管时钟
#include<reg52.h>
#define uint unsigned int
#define uchar unsigned char
sbit dula=P2^6;
sbit wela=P2^7;
sbit K1=P3^4;
sbit K2=P3^5;
sbit K3=P3^6;
sbit K4=P3^7;
uchar code table[]={
0x3f,0x06,0x5b,0x4f,
0x66,0x6d,0x7d,0x07,
0x7f,0x6f,0x77,0x7c,
0x39,0x5e,0x79,0x71};
uint aa;
uchar num1,num2,shi1,ge1,shi2,ge2;

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

void display(uchar num1,uchar num2)
{
shi1=num1/10;
ge1=num1%10;
shi2=num2/10;
ge2=num2%10;

dula=1;
P0=table[shi1];
dula=0;
P0=0xff;
wela=1;
P0=0xfe;
wela=0;
delay(1);

dula=1;
P0=table[ge1];
dula=0;
P0=0xff;
wela=1;
P0=0xfd;
wela=0;
delay(1);

dula=1;
P0=table[shi2];
dula=0;
P0=0xff;
wela=1;
P0=0xfb;
wela=0;
delay(1);

dula=1;
P0=table[ge2];
dula=0;
P0=0xff;
wela=1;
P0=0xf7;
wela=0;
delay(1);
}

void key()
{
if(K1==0)
{
delay(5);
if(K1==0)
{
TR0=0;
while(!K1);
}
}

if(K2==0)
{
delay(5);
if(K2==0)
{
if(num1==0)
{
num1=60;
}
num1--;
while(!K2);
}
}

if(K3==0)
{
delay(5);
if(K3==0)
{
num1++;
if(num1>=60)
{
num1=num1-60;
}
while(!K3);
}
}

if(K4==0)
{
delay(5);
if(K4==0)
{
TR0=1;
while(!K4);
}
}
}

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

void main()
{
inint();
while(1)
{
key();
display(num1,num2);
}
}

void zhongduan()interrupt 1
{
TH0=(65536-50000)/256;
TL0=(65536-50000)%256;
aa++;
if(aa==20)
{
aa=0;
num2++;
if(num2==60)
{
num2=0;
num1++;
if(num1==60)
num1=0;
}
}
}

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