按键扫描程序分析
#include <reg52.h>
sbit KEY_IN_1 = P2^4; //矩阵按键的扫描输入引脚1
sbit KEY_IN_2 = P2^5; //矩阵按键的扫描输入引脚2
sbit KEY_IN_3 = P2^6; //矩阵按键的扫描输入引脚3
sbit KEY_IN_4 = P2^7; //矩阵按键的扫描输入引脚4
sbit KEY_OUT_1 = P2^3; //矩阵按键的扫描输出引脚1
sbit KEY_OUT_2 = P2^2; //矩阵按键的扫描输出引脚2
sbit KEY_OUT_3 = P2^1; //矩阵按键的扫描输出引脚3
sbit KEY_OUT_4 = P2^0; //矩阵按键的扫描输出引脚4
sbit ADDR0 = P1^0;
sbit ADDR1 = P1^1;
sbit ADDR2 = P1^2;
sbit ADDR3 = P1^3;
sbit ENLED = P1^4;
unsigned char code LedChar[] = {
0xC0,0xF9,0xA4,0xB0,0x99,0x92,0x82,0xF8,
0x80,0x90,0x88,0x83,0xC6,0xA1,0x86,0x8e
}; //数码管真值表
const unsigned char code KeyCodeMap[4][4] = { //矩阵按键编号到PC 标准键盘键码的映射表
{ '1', '2', '3', 0x26 }, //数字键1、数字键2、数字键3、向上键
{ '4', '5', '6', 0x25 }, //数字键4、数字键5、数字键6、向左键
{ '7', '8', '9', 0x28 }, //数字键7、数字键8、数字键9、向下键
{ '0', 0x1B, 0x0D, 0x27 } //数字键0、ESC 键、回车键、向右键 };
unsigned char KeySta[4][4] = { //全部矩阵按键的当前状态
{1, 1, 1, 1},
{1, 1, 1, 1},
{1, 1, 1, 1},
{1, 1, 1, 1}
}; //由于数组不能定义成bit 型，这里定义成unsigned char 型
unsigned char LedBuf[6] = { //数码管动态扫描显示缓冲区
0xC0, 0xFF, 0xFF, 0xFF, 0xFF, 0xFF
}; //// 要这个显示缓冲区干嘛?
void DisplayNum(unsigned long num);
void KeyAction(unsigned char keycode);
void main(void)
0{
unsigned char i, j;
unsigned char backup[4][4] = { //按键值备份，保存前一次的值
{1, 1, 1, 1},
{1, 1, 1, 1},
{1, 1, 1, 1},
{1, 1, 1, 1} };
//选择数码管进行显示
P0 = 0xFF;
ADDR3 = 1;
ENLED = 0;
//配置T0 工作在模式1，定时1ms
TMOD = 0x01;
TH0 = 0xFC;
TL0 = 0x67;
TR0 = 1;
ET0 = 1;
EA = 1;
while(1)
1{ //检索按键状态的变化
for (i=0; i<4; i++)
2{for (j=0; j<4; j++)
3{if (backup[j] ！= KeySta[j])
4{if (backup[j] == 0) //按键弹起时执行动作
5{KeyAction(KeyCodeMap[j]);}5
backup[j] = KeySta[j];}4
}3 }2 }1 }0
void KeyAction(unsigned char keycode)
{
static unsigned long result = 0; //用于保存运算结果
static unsigned long addend = 0; //用于保存输入的加数
if ((keycode>='0') && (keycode<='9')) //输入0-9 的数字
{addend = (addend*10) + (keycode-'0'); //原数据扩大10 倍，由新输入的
字填充其个位
DisplayNum(addend); //运算结果显示到数码管 }
else if (keycode == 0x26) //向上键用作加号，执行加法或连加运算
{result += addend; //进行加法运算
addend = 0;
DisplayNum(result); //运算结果显示到数码管}
else if (keycode == 0x0D) //回车键，执行加法运算（实际效果与加号并无区
别）{ result += addend; //进行加法运算
addend = 0;
DisplayNum(result); //运算结果显示到数码管 }
else if (keycode == 0x1B) //Esc 键，清零结果
{addend = 0;
result = 0;
DisplayNum(addend); //清零后的加数显示到数码管}
}
void DisplayNum(unsigned long num)
{
signed char i;
unsigned char buf[6];
for (i=0; i<6; i++) //把长整型数转换为6 位十进制的数组
{buf = num % 10; num /= 10; }
for (i=5; i>=1; i--) //从最高位起，遇到0 即转换为空格，遇到非0 即退出
{if (buf == 0)
{LedBuf = 0xFF;}
else
{break;}
}
for ( ; i>=0; i--) //剩余低位都如实转换为数字
{LedBuf = LedChar[buf];}
}
void InterruptTimer0() interrupt 1
{
unsigned char i;
static unsigned char ledcnt = 0; //数码管扫描计数器
static unsigned char keyout = 0; //矩阵按键扫描输出计数器
static unsigned char keybuf[4][4] = { //按键扫描缓冲区，保存一段时间内的扫描值
{0xFF, 0xFF, 0xFF, 0xFF},
{0xFF, 0xFF, 0xFF, 0xFF},
{0xFF, 0xFF, 0xFF, 0xFF},
{0xFF, 0xFF, 0xFF, 0xFF}};
TH0 = 0xFC; //溢出后进入中断重新赋值
TL0 = 0x67;
//将一行的4 个按键值移入缓冲区
keybuf[keyout][0] = (keybuf[keyout][0] << 1) | KEY_IN_1;
keybuf[keyout][1] = (keybuf[keyout][1] << 1) | KEY_IN_2;
keybuf[keyout][2] = (keybuf[keyout][2] << 1) | KEY_IN_3;
keybuf[keyout][3] = (keybuf[keyout][3] << 1) | KEY_IN_4;
//消抖后更新按键状态
for (i=0; i<4; i++) //每行4 个按键，所以循环4 次
{ if ((keybuf[keyout] & 0x0F) == 0x00)
{ //连续4 次扫描值为0，即16ms(4*4ms)内都只检测到按下状态时，可认为
按键已按下KeySta[keyout] = 0;}
else if ((keybuf[keyout] & 0x0F) == 0x0F)
{ //连续4 次扫描值为1，即16ms(4*4ms)内都只检测到弹起状态时，可认为
按键已弹起KeySta[keyout] = 1;}
}
//执行下一次的扫描输出
keyout++;
keyout &= 0x03;
switch (keyout)
{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; }
//执行数码管动态扫描显示
P0 = 0xFF;
switch (ledcnt)
{
case 0: ADDR0=0; ADDR1=0; ADDR2=0; break;
case 1: ADDR0=1; ADDR1=0; ADDR2=0; break;
case 2: ADDR0=0; ADDR1=1; ADDR2=0; break;
case 3: ADDR0=1; ADDR1=1; ADDR2=0; break;
case 4: ADDR0=0; ADDR1=0; ADDR2=1; break;
case 5: ADDR0=1; ADDR1=0; ADDR2=1; break;
default: break; }
P0 = LedBuf[ledcnt];
ledcnt++;
if (ledcnt >= 6)
{ ledcnt = 0; } }
大家仔细看一下程序，分析一下其功能是什么? 能实现吗?
可以实现一个简单一次运算的计算器