标题:
4x4键盘、数码管显示、数组应用等练习
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作者:
mabaoguang
时间:
2015-3-28 15:38
标题:
4x4键盘、数码管显示、数组应用等练习
//程序名称:简单加法
//用于练习4x4键盘、数码管显示、数组应用等
//4x4软键盘布局
// 1 2 3 A
// 4 5 6 B
// 7 8 9 C
// * 0 # D
// 程序中A加号,#回车,*ESC清零
// 数码管接P0、P2^4-P2^7
// 键盘接P1
//改编自
http://www.51hei.com/bbs/dpj-20218-1.html
// 1.3.6 包含综合应用的实用程序
#include <reg52.h>
sbit KEY_IN_1 = P1^4; //矩阵按键的扫描输入引脚1
sbit KEY_IN_2 = P1^5; //矩阵按键的扫描输入引脚2
sbit KEY_IN_3 = P1^6; //矩阵按键的扫描输入引脚3
sbit KEY_IN_4 = P1^7; //矩阵按键的扫描输入引脚4
sbit KEY_OUT_1 = P1^3; //矩阵按键的扫描输出引脚1
sbit KEY_OUT_2 = P1^2; //矩阵按键的扫描输出引脚2
sbit KEY_OUT_3 = P1^1; //矩阵按键的扫描输出引脚3
sbit KEY_OUT_4 = P1^0; //矩阵按键的扫描输出引脚4
sbit ADDR0 = P2^4;
sbit ADDR1 = P2^5;
sbit ADDR2 = P2^6;
sbit ADDR3 = P2^7;
sbit ENLED = P0;
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
{ 0x1B, '0', 0x0D, 0x27 }, //数字键0、ESC键、 回车键、 向右键
{ '7', '8', '9', 0x28 }, //数字键7、数字键8、数字键9、向下键
{ '4', '5', '6', 0x25 }, //数字键4、数字键5、数字键6、向左键
{ '1', '2', '3', 0x26 } //数字键1、数字键2、数字键3、向上键
};
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
};
void DisplayNum(unsigned long num);
void KeyAction(unsigned char keycode);
void main(void)
{
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)
{
//检索按键状态的变化
for (i=0; i<4; i++)
{
for (j=0; j<4; j++)
{
if (backup
[j] != KeySta
[j])
{
if (backup
[j] == 0) //按键弹起时执行动作
{
KeyAction(KeyCodeMap
[j]);
}
backup
[j] = KeySta
[j];
}
}
}
}
}
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,由新输入的数字填充其个位
DisplayNum(addend); //运算结果显示到数码管
}
else if (keycode == 0x26) //向上键A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[4];
for (i=0; i<4; i++) //把长整型数转换为4位十进制的数组
{
buf
= num % 10;
num /= 10;
}
for (i=3; 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=1; ADDR2=1; ADDR3=1; break;
case 1: ADDR0=1; ADDR1=0; ADDR2=1; ADDR3=1; break;
case 2: ADDR0=1; ADDR1=1; ADDR2=0; ADDR3=1; break;
case 3: ADDR0=1; ADDR1=1; ADDR2=1; ADDR3=0; break;
case 4: ADDR0=1; ADDR1=1; ADDR2=1; ADDR3=1; break;
default: break;
}
P0 = LedBuf[ledcnt];
ledcnt++;
if (ledcnt >= 5)
{
ledcnt = 0;
}
}
作者:
yangrui
时间:
2015-6-15 19:14
不错,值得借鉴一下
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