实验程序1：按S2键，让P1.0口发光管亮
#include     头文件
sbit d1=P1^0;         位声明
sbit key1=P3^4;       声明键盘输入口
void main()           主函数开始
{
       P3=0xff;                   让P3口熄灭
       while(1)                    进入大循环
              {
                     if (key1==0)        如果P3.4口变为低电平
                            d1=0;          P1.0口灯亮
                     else                否则
                            d1=1;          灯灭
              }
}
实验程序2：  按S2键，让数码管第一位按一次按键加1，在0~9之间循环
#include              头文件
#define uint unsigned int   
#define uchar unsigned char       宏定义
sbit d1=P1^0;                   位声明
sbit key1=P3^4;                 按键声明
sbit dula=P2^6;                  段选声明
sbit wela=P2^7;                 位选声明
uchar code table[]={             数组
0x3f,0x06,0x5b,0x4f,
0x66,0x6d,0x7d,0x07,
0x7f,0x6f,0x77,0x7c,
0x39,0x5e,0x79,0x71,
};
uchar num;                     定义变量num
void delay(uint z)                延时子程序
{
       uint x,y;
       for(x=z;x>0;x--)
              for(y=110;y>0;y--);      
}
 
void main()                     主函数
{
       wela=1;                    位选开
       P0=0xfe;                   让六位数码管第一位亮
       wela=0;                    位选锁存
 
       P3=0xff;                   关闭P3口
       while(1)                     开始大循环
              {
                     if (key1==0)              如果按键按下
                            {
                                   delay(10);            延时消除按键抖动
                                   if(key1==0)           然后确认按键按下
                                   {
                                          d1=0;              发光管亮
                                          num++;             变量num加1
                                          if(num==10)         
                                                 num=0;         当num加到10时，num返回0
                                   }
                                   while(!key1);            当按键按下时，KEY1的值等于0，！KEY1=1  当按键按下时不退出循环，一旦松手KEY的值=1，！1的值等于0 跳出循环
                            }
                           
                     else                              否则
                            d1=1;                       发光管灯灭
                            dula=1;                      段选开
                            P0=table[num];                让第一位显示num变量对应的数组的数
                            dula=0;                       段锁存
              }
}
 


实验程序3：矩阵键盘按键在六位数码管上显示相应数字
#include                      头文件
#define uint unsigned int                 
#define uchar unsigned char              宏定义
 
sbit key1=P3^4;                       
sbit dula=P2^6;                         段选
sbit wela=P2^7;                         位选
uchar code table[]={                     数组
0x3f,0x06,0x5b,0x4f,
0x66,0x6d,0x7d,0x07,
0x7f,0x6f,0x77,0x7c,
0x39,0x5e,0x79,0x71,0
};
uchar num,temp;                     变量 num temp 声明
void delay(uint z)                     延时子程序
{
       uint x,y;
       for(x=z;x>0;x--)
              for(y=110;y>0;y--);      
}
 
void main()                         主函数
{
       dula=1;                       
       P0=0;
       dula=0;
       wela=1;
       P0=0xc0;
       wela=0;                       一开始让段选位选关闭
       while(1)                       开始矩阵键盘扫描
              {
                     P3=0xfe;                     首先给P3口送11111110
                     temp=P3;                     把11111110的值赋给temp
                     temp=temp&0xf0;              再让11111110“与”11110000   如果这个值等于11110000说明没有按键被按下  数码管上什么都不显示
                     while(temp!=0xf0)              如果与的值不等于11110000说明有按键被按下了 进入while循环
                            {
                                   delay(5);                进入循环后先延时一会消除抖动
                                   temp=P3;                这时把按下的P3值送给TEMP 确认按下。比如说按下的是第一个键， P3为11101110
                                   temp=temp&0xf0;         11101110与11110000 等于11100000
                                   while(temp!=0xf0)          因为temp的值不等于11110000了 所以进入while循环
                                   {
                                          temp=P3;              把11100000送给P3口
 
                     switch(temp)                       进入switch case语句
                            {
                                   case 0xee:num=1;           如果是11101110 说明是按键1
                                          break;
                                   case 0xde:num=2;           如果是11011110 说明是按键2
                                          break;
                                   case 0xbe:num=3;           如果是10111110 说明是按键3
                                          break;
                                   case 0x7e:num=4;           如果是01111110 说明是按键4
                                          break;
                            }
                            while(temp!=0xf0)               松手检测   
                            {
                                   temp=P3;
                                   temp=temp&0xf0;
                            }
 
                           
                            dula=1;
                            P0=table[num-1];
                            dula=0;                        显示
                            }
                            }
                     P3=0xfd;                           扫描第二行
                     temp=P3;
                     temp=temp&0xf0;
                     while(temp!=0xf0)
                            {
                                   delay(5);
                                   temp=P3;
                                   temp=temp&0xf0;
                                   while(temp!=0xf0)
                                   {
                                          temp=P3;
 
                     switch(temp)
                            {
                                   case 0xed:num=5;
                                          break;
                                   case 0xdd:num=6;
                                          break;
                                   case 0xbd:num=7;
                                          break;
                                   case 0x7d:num=8;
                                          break;
                            }
                            while(temp!=0xf0)
                            {
                                   temp=P3;
                                   temp=temp&0xf0;
                            }
                           
                            dula=1;
                            P0=table[num-1];
                            dula=0;
                            }
                            }
                     P3=0xfb;                         扫描第三行
                     temp=P3;
                     temp=temp&0xf0;
                     while(temp!=0xf0)
                            {
                                   delay(5);
                                   temp=P3;
                                   temp=temp&0xf0;
                                   while(temp!=0xf0)
                                   {
                                          temp=P3;
 
                     switch(temp)
                            {
                                   case 0xeb:num=9;
                                          break;
                                   case 0xdb:num=10;
                                          break;
                                   case 0xbb:num=11;
                                          break;
                                   case 0x7b:num=12;
                                          break;
                            }
                            while(temp!=0xf0)
                            {
                                   temp=P3;
                                   temp=temp&0xf0;
                            }
                           
                            dula=1;
                            P0=table[num-1];
                            dula=0;
                            }
                            }
                     P3=0xf7;                              扫描第四行
                     temp=P3;
                     temp=temp&0xf0;
                     while(temp!=0xf0)
                            {
                                   delay(5);
                                   temp=P3;
                                   temp=temp&0xf0;
                                   while(temp!=0xf0)
                                   {
                                          temp=P3;
 
                     switch(temp)
                            {
                                   case 0xe7:num=13;
                                          break;
                                   case 0xd7:num=14;
                                          break;
                                   case 0xb7:num=15;
                                          break;
                                   case 0x77:num=16;
                                          break;
                            }
                            while(temp!=0xf0)
                            {
                                   temp=P3;
                                   temp=temp&0xf0;
                            }
                           
                            dula=1;
                            P0=table[num-1];
                            dula=0;
                            }
                            }
             
              }
 
}
 
[图片摘自郭天翔视频教程]
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