51单片机C语言程序4个独立按键实现对数码管数字显示的加减清零等

2024-12-02 23:56:37
推荐回答(3个)
回答1:

#include 
//P0是数码管。P1是LED.P2是按键
sbit KEY_OUT_1 = P2^3;
sbit KEY_OUT_2 = P2^2; 
sbit KEY_OUT_3 = P2^1;
sbit KEY_OUT_4 = P2^0;

#define uchar unsigned char
#define ulint unsigned long
#define Frequency 10//定时器中断时间 = f*T
#define Time 1//一个周期1ms
#define TubeNumber 6//数码管个数
#define KeyLine 4//矩阵按键行数 
#define KeyColumn 4//矩阵按键列数 
//数码管真值表
uchar code LED_Number[]={0x0C,0xF9,0xA4,0xB0,0x99,0x92,0x82,0xF8,0x80,0x90};
//uchar code LED_Alphabet[]={0x88,0x83,0xC6,0xA1,0x86,0x8E,0x89,0xC7,0x8C,0xC1,0x91,0x9C};
/*0~9
                              A~F   (b      、d为小写)        H    L    P    U   y    o  */
uchar  LED_Buff[TubeNumber]={0xff,0xff,0xff,0xff,0xff,0xff};
/*数码管显示缓冲区,0xff确保初始时都不亮.
不可写成uchar code LED_Buff[]。code定义变量写入room,不可修改*/
// 矩阵按键编号到标准盘码的映射表 
uchar code KeyCodeMap[4][4] = { 
( 0x31, 0x32, 0x33, 0x26 ), // 数字键 1、数字键 2、数字键 3、向上键
( 0x34, 0x35, 0x36, 0x25 ), // 数字键 4、数字键 5、数字键 6、向左键 
( 0x37, 0x38, 0x39, 0x28 ), // 数字键 7、数字键 8、数字键 9、向下键 
( 0x30, 0x1B, 0x0D, 0x27 )};  // 数字键 0、ESC 键  、回车键  、向右键
uchar StaFlag[KeyLine][KeyColumn] ={(1,1,1,1),(1,1,1,1),(1,1,1,1),(1,1,1,1)};//按键是否稳定标志 
void StartTime0();
void TubeDisplay(ulint sec);
ulint pow(x,y);
void TubeScan();
void KeyAction(uchar keycode);
void KeyDriver();
void KeyScan();
void main ()
{
P1 = 0x08;//使能U3,选择数码管。
StartTime0();
while(1)KeyDriver();
}
//定时器0启动函数
void StartTime0()
{
EA = 1;
ET0 = 1;
TMOD = 0x01;
TH0 = (65536 - Time * 100) / 256;
TL0 = (65536 - Time * 100) % 256;
PT0 = 1;
/*定时器0优先中断控制位。
IP 这个寄存器的每一位,表示对应中断的抢占优先级,每一位的复值都是0,当我们把某一位设置为1的时候,这一位优先级就比其它位的优先级高了。
比如我们设置了 PT0位为1后, 当单片机在主循环或其他中断程序执行时,一旦TO发生中断,作为更高优先级,程序马上执行T0.若在T0程序执行时,
其他中断程序发生中断,仍执行TO直到T0中断结束后再执行其他程序。 
*/  
}
//中断服务函数 
void To_time0()interrupt 1 using 0
{
static uchar cnt;//记录TO中断次数
// static ulint sec;//记录经过秒速
//判断是否溢出 
if (TF0 == 1)
{
TF0 = 0;
TH0 = (65536 - Time * 100) / 256;
TL0 = (65536 - Time * 100) % 256;
}

if (cnt >= Frequency)
{
cnt = 0;
//sec++;
// Tube_Display(sec);
TubeScan();
KeyScan();
}
}
//数码管显示函数
void TubeDisplay(ulint nom)
{
uchar m = 2;//小数部分位数
uchar i;//传输索引
//秒速达到上限清零
if (nom > pow(10,TubeNumber - m))nom = 0;
//分别传输小数部分和整数部分 
for(i=0; i LED_Buff[i] = LED_Number[nom/pow(10,i)%10];
for(i=0; i<(TubeNumber - m); i++)
LED_Buff[i+m] = LED_Number[nom/pow(10,i)%10];
//点亮小数点
LED_Buff[m] &= 0x7f;
}
//平方运算函数 
ulint pow(x,y)//x为底,为幂
{
ulint p,i = 1;
//平方运算 
for(i=1; i<=y; i++)
p*=x;
//输出结果 
return p;
}
//数码管动态函数
void TubeScan()
{
static uchar i = 0;//动态扫描索引
//关闭所有段选位,数码管消隐
P0 = 0xff;
//for (i=0; i < Tube_number; i++)
P1 = (P1 & 0xf8) | i;//位选索引赋值到P1口低3位
P0 = LED_Buff[i];//缓冲区中的索引位置数据传输到P0口
if (++i >= TubeNumber)i=0;//索引递增循环,遍历整个缓冲区

//矩阵按键动作函数
void KeyAction(uchar keycode)
{
static ulint result;
ulint nom = 0;
//输入数字0~9 
if ((keycode >= 0x30) && (keycode <= 39))
{
nom = (nom*10) + (keycode - 0x30);//十进制整体左移,新数进入各位
TubeDisplay(nom); 
}
//输入方向键 
if ((keycode >= 0x25) && (keycode <= 28))
switch (keycode)
{
case 0x26:result += nom; nom = 0; TubeDisplay(result);
case 0x28:result -= nom; nom = 0; TubeDisplay(result);
case 0x25:result = 1; result *= nom; nom = 0; TubeDisplay(result);
case 0x27:result = 1; result /= nom; nom = 0; TubeDisplay(result);
}
else if (keycode == 0x0d)TubeDisplay(result);//输入回车键,输出最终结果 
else if (keycode == 0x1b)//输入ESC键,清零 
{
nom = result = 0;
TubeDisplay(nom);
}
}
//矩阵按键驱动函数 
void KeyDriver()
{
uchar l,c; 
static uchar backup[KeyLine][KeyColumn] = {(1,1,1,1),(1,1,1,1),(1,1,1,1),(1,1,1,1)};//按键值备份,保存前一次值 
for(l=0; l {
for(c=0; c {
if (backup[l][c] != StaFlag[l][c])
{//检测按键动作 
if (backup[l][c] == 1)//按键按下时执行
KeyAction(KeyCodeMap[l][c]);//调用动作函数 
backup[l][c] = StaFlag[l][c];//刷新前一次备份值
}
}
}

//矩阵按键扫描函数
void KeyScan()
{
uchar l = 0;//矩阵按键扫描输出索引
uchar c = 0;//矩阵按键扫描列索引
uchar keybuff [KeyLine][KeyColumn] = {(0xff,0xff,0xff,0xff),(0xff,0xff,0xff,0xff),
      (0xff,0xff,0xff,0xff),(0xff,0xff,0xff,0xff)};//矩阵按键扫描缓冲区
  //将一行的四个按键移入缓冲区 
  for(l=0; l  keybuff [l][c] = ((0xfe | (P2 >> (4 + l)) & 0x01));
//按键消抖 
for(l=0; l  {
if((keybuff [l][c] & 0x0f) == 0x00)//连续4次扫描都为0,判断4*4ms内都是按下状态,可认为按键已稳定按下
StaFlag[l][c] = 0;
else if ((keybuff [l][c] & 0x0f) == 0x0f)//连续4次扫描都为1,判断4*4ms内都是弹起状态,可认为按键已稳定弹起
  StaFlag[l][c] = 1;
}
for(c=0; c  {
switch (c) // 根据索引,释放当前输出脚 拉低下次的根据索引

case 0: KEY_OUT_4 = 1; KEY_OUT_1 = 0; break; 
case 1: KEY_OUT_1 = 1; KEY_OUT_2 = 0; break; 
case 2: KEY_OUT_2 = 1; KEY_OUT_3 = 0; break; 
case 3: KEY_OUT_3 = 1; KEY_OUT_4 = 0; break; 
default: break;
 }
}
}

回答2:

图片失真,看不清楚

回答3:

需要带做吗

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