矩阵键盘扫描问题

无域思维 2010-3-7 20:40 楼主

PD2-PD5  和 PB5-PB7控制矩阵键盘
谁帮我看下，下面的程序有什么问题
#include
#include

#define uchar unsigned char
#define uint unsigned int

#pragma data:code
const uchar Duan[]={0x3f,0x06,0x5b,0x4f,0x66,0x6d,0x7d,0x07,0x7f,0x6f,0x77,0x7c,0x39,0x5e,0x79,0x71,0x00};
const uchar Numtk[]={10,11,12,3,6,9,2,5,8,1,4,7,16};
const uchar Act[]={0x06,0x05,0x03};

void delay_ms(uint n)
{
uint i,j;
for(i=0;i for(j=0;j<562;j++);
}

uchar key_press()                      //判断是否有键按下
{
uchar in;
PORTD=PORTD&0XC3;       //PD2-PD5输出0
PORTB=PORTB|0XE0;       //PB5-PB7上拉有效
in=PINB;
in=in>>5;
in=in&0x07;
if(in!=0x07)
{
delay_ms(10); //延时再判断
if(in!=0x07)
return 1;
}
else
return 0;
}

uchar key_scan()                //有键按下时，判断按下了那个键
{
uchar i,k,j;

for(i=0;i<4;i++)
{
     PORTD=PORTD|0X3C;    //先把PD2-PD5设置为1
PORTD=PORTD&(~BIT(i+2));//PD2-PD5分别置0
j=PINB;
j=j>>5;
j=j&0x07;
for(k=0;k<3;k++)       //判断PB5-PB7哪个为0
{
   if(j==Act[k])
      return (3*i+k);
}

}
return 12;
}
uchar key,num;
void main()
{
DDRD=DDRD|BIT(1);          //打开PD1引脚的指示灯
PORTD=PORTD&(~BIT(1));
DDRB=DDRB|BIT(4);    //打开第一位数码管
PORTB=PORTB|BIT(4);
DDRA=0xff; //PA0-PA8为输出0,关闭段选
PORTA=0x00;
DDRD=DDRD|0X3C;          //PD2-PD5设置为输出0
//PORTD=PORTD&0XC3;
DDRB=DDRB&0X1F;       //PB5-PB7设置为输入，上拉有效
//PORTB=PORTB|0Xe0;

while(1)
{
if(key_press())
{
num=key_scan();
key=Numtk[num];
PORTA=Duan[key];
delay_ms(1);
}
}
}