单片机实验3提示
Posted zhangrelay
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键盘与LCD实验
对4×4矩阵式键盘电路的键值进行编码,编程实现在LCD液晶显示器上显示每个按键的ASCII码。
(⊙﹏⊙)呃这个和去年一样
专栏:
https://blog.csdn.net/zhangrelay/category_6638850.html
#include "8052.h"
#define uchar unsigned char
#define uint unsigned int
#define lcden P1_2
#define lcdrs P1_0
#define rw P1_1
/* IO引脚分配定义 */
#define KEY_IN_1 P2_4 //矩阵按键的扫描输入引脚1
#define KEY_IN_2 P2_5 //矩阵按键的扫描输入引脚2
#define KEY_IN_3 P2_6 //矩阵按键的扫描输入引脚3
#define KEY_IN_4 P2_7 //矩阵按键的扫描输入引脚4
#define KEY_OUT_1 P2_0 //矩阵按键的扫描输出引脚1
#define KEY_OUT_2 P2_1 //矩阵按键的扫描输出引脚2
#define KEY_OUT_3 P2_2 //矩阵按键的扫描输出引脚3
#define KEY_OUT_4 P2_3 //矩阵按键的扫描输出引脚4
#define LSA P1_5 //LED位选译码地址引脚A
#define LSB P1_6 //LED位选译码地址引脚B
#define LSC P1_7 //LED位选译码地址引脚C
unsigned char disBuf=0;
uchar table1[] = "Welcome To CSLG!";
uchar table2[] = "0123456789ABCDEF";
uchar table3[] = " ";
uchar num;
const unsigned char KeyCodeMap[4][4] = //矩阵按键到标准键码的映射表
'0', '1', '2', '3' , //
'4', '5', '6', '7' , //
'8', '9', 'a', 'b' , //
'c', 'd', 'e', 'f' //
;
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 ;
/* 函数声明 */
void KeyScan();
void KeyDriver();
void KeyAction(unsigned char keycode);
/* 按键驱动函数,检测按键动作,调度相应动作函数,需在主循环中调用 */
void KeyDriver()
unsigned char i, j;
static 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
;
for (i=0; i<4; i++) //循环检测4*4的矩阵按键
for (j=0; j<4; j++)
if (backup[i][j] != KeySta[i][j]) //检测按键动作
if (backup[i][j] != 0) //按键按下时执行动作
KeyAction(KeyCodeMap[i][j]); //调用按键动作函数
backup[i][j] = KeySta[i][j]; //刷新前一次的备份值
/* 按键扫描函数,需在定时中断中调用,推荐调用间隔1ms */
void KeyScan()
unsigned char i;
static unsigned char keyout = 0; //矩阵按键扫描输出索引
static unsigned char keybuf[4][4] = //矩阵按键扫描缓冲区
0xFF, 0xFF, 0xFF, 0xFF, 0xFF, 0xFF, 0xFF, 0xFF,
0xFF, 0xFF, 0xFF, 0xFF, 0xFF, 0xFF, 0xFF, 0xFF
;
//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][i] & 0x0F) == 0x00)
//连续4次扫描值为0,即4*4ms内都是按下状态时,可认为按键已稳定的按下
KeySta[keyout][i] = 0;
else if ((keybuf[keyout][i] & 0x0F) == 0x0F)
//连续4次扫描值为1,即4*4ms内都是弹起状态时,可认为按键已稳定的弹起
KeySta[keyout][i] = 1;
//执行下一次的扫描输出
keyout++; //输出索引递增
keyout &= 0x03; //索引值加到4即归零
switch (keyout) //根据索引值,释放当前输出引脚,拉低下次的输出引脚
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;
void KeyAction(unsigned char keycode)
disBuf = keycode;
void delay(uint z)
uint x, y;
for(x = z; x > 0; x--)
for(y = 110; y > 0; y--);
void write_com(uchar com)
lcdrs = 0;
P0 = com;
delay(5);
lcden = 1;
delay(5);
lcden = 0;
void write_data(uchar date)
lcdrs = 1;
P0 = date;
delay(5);
lcden = 1;
delay(5);
lcden = 0;
void init()
rw = 0;
lcden = 0;
write_com(0x38);
write_com(0x0e);
write_com(0x06);
write_com(0x01);
void main()
init();
write_com( 0x00 | 0x80 );
for(num = 0; num < 16; num++)
write_data(table1[num]);
delay(100);
write_com( 0x40 | 0x80 );
for(num = 0; num < 16; num++)
write_data(table2[num]);
delay(100);
write_com( 0x40 | 0x80 );
for(num = 0; num < 16; num++)
write_data(table3[num]);
delay(100);
while(1)
KeyScan();
KeyDriver();
write_com( 0x40 | 0x80 );
write_data(disBuf);
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