关于单片机8个I/0口 实现36个按键功能
算是闲着无聊弄了弄，不弄不知道一弄发现原来那么简单
先上图
这是整体仿真图，用51单片机接，数码管显示每个按键的返回值；
按键以8个起始 递减的链接，7、6、5... 因此共36个按键
下面是检测按键的简单程序
#include<reg51.h>
#include<intrins.h>
#define uchar unsigned char
#define uint unsigned int
#define LED P0
#define Keycheck P1
sbit LED1=P2^0;
sbit LED2=P2^1;
sbit LED3=P2^2;
sbit LED4=P2^3;
void delay(uchar);
uchar code NUM[]={0xc0,0xf9,0xa4,0xb0,0x99,0x92,0x82,0xf8,0x80,0x90,0x88,0x83,0xc6,0xa1,0x86,0x8e};//公阳
//0x3f,0x06,0x5b,0x4f,0x66,0x6d,0x7d,0x07,0x7f,0x6f,0x77,0x7c,0x39,0x5e,0x79,0x71 公阴
uint DATA=0;
void LEDplay(void);
void Keych(void);
void delay(uchar);
int main(void)
{
DATA=0X00;
while(1)
{
Keych();
LEDplay();
}
}
void Keych(void)
{
uchar i=0,temp,Keycode;
uchar Keynum=0xfe;
Keycheck=0xff;
if(Keycheck!=0xff) //I/O口全为高电平扫描列8个按键
{ delay(5);
if(Keycheck!=0xff)
Keycode=Keycheck;
}
else
{
while((Keynum&0x80)!=0) //每次一个低电平口，循环移动扫描按键
{
Keycheck=Keynum;
if(Keycheck!=Keynum)
{
delay(5);
if(Keycheck!=Keynum)
Keycode=Keycheck;
break;
}
else
Keynum=_crol_(Keynum,1);
}
}
//else Keycode=0x00; //无按键按下的返回值，这里为保持所找按键值，估不设定返回
if(temp!=Keycode)
{
temp=Keycode;
DATA=Keycode;
}
}
void LEDplay(void)
{
P2=0x00;
LED=NUM[DATA&0X0F];LED4=1;delay(3);LED4=0;
LED=NUM[DATA>>4]; LED3=1;delay(3);LED3=0;
LED=0x89;LED2=1;delay(3);LED2=0; //0x89为共阳数码管H
LED=NUM[0];LED1=1;delay(3);LED1=0;
}
void delay(uchar x)
{
uchar k;
while(x--)
for(k=0;k<128;k++);
}
哎呀~ 为啥行距那么宽呢~？ 下回再弄
按照图上的那种接法按键返回值分别为：
0XFE
0XFC 0XFD
0XF9 0XFA 0XFB
0XF3 0XF5 0XF6 0XF7
0XE7 0XEB 0XED 0XEE 0XEF
0XCF 0XD7 0XDB 0XDD 0XDE 0XDF
0X9F 0XAF 0XB7 0XBB 0XBD 0XBE 0XBF
0x3F 0X5F 0X6F 0X77 0X7B 0X7D 0X7E 0X7F
若返回值有出入纯粹 链接模式差异或程序上的的差异。原理一样即可；
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