关于51 单片机的扫描问题、

2019-03-24 19:37发布

#include<reg52.h>
#define uint unsigned int
#define uchar unsigned char uchar code shumaguan[] = {0xc0,0xf9,0xa4,0xb0,0x99,0x92,0x82,0xf8,0x80,0x90};
uchar code table[] = {0xfe,0xfd,0xfb,0xf7,0xef,0xdf,0xbf,0x7f}; uchar key = 0; void delays(int ms)
{
 int i;
 int j;
 for(i = 0;i<ms;i++)
  for(j = 0; j<124;j++);
} void matrixkeyscan()
{
 uchar temp;  P0 = 0xfe;
 temp = P0;
 temp = temp & 0xf0;
 if(temp != 0xf0)
 {
  delays(10);
  temp = P0;
  temp = temp & 0xf0;
  if(temp != 0xf0)
  {
   temp = P0;
   switch(temp)
   {
    case 0xee:
      key = 0;
     break;
    case 0xde:
     key = 1;
     break;
    case 0xbe:
     key = 2;
     break;
    case 0x7e:
     key = 3;
     break;
   } 
   while(temp != 0xf0)
   {
    temp = P0 ;
    temp = temp & 0xf0;
   }
   //display()  ;
  }
 }  P0 = 0xfd;
 temp = P0;
 temp = temp & 0xf0;
 if(temp != 0xf0)
 {
  delays(10);
  temp = P0;
  temp = temp & 0xf0;
  if(temp != 0xf0)
  {
   temp = P0;
   switch(temp)
   {
    case 0xed:
      key = 4;
     break;
    case 0xdd:
     key = 5;
     break;
    case 0xbd:
     key = 6;
     break;
    case 0x7d:
     key = 7;
     break;
   } 
   while(temp != 0xf0)
   {
    temp = P0 ;
    temp = temp & 0xf0;
   }
   //display()  ;
  }
 }  P0 = 0xfb;
 temp = P0;
 temp = temp & 0xf0;
 if(temp != 0xf0)
 {
  delays(10);
  temp = P0;
  temp = temp & 0xf0;
  if(temp != 0xf0)
  {
   temp = P0;
   switch(temp)
   {
    case 0xeb:
      key = 8;
     break;
    case 0xdb:
     key = 9;
     break;
    case 0xbb:
     key = 10;
     break;
    case 0x7b:
     key = 11;
     break;
   } 
   while(temp != 0xf0)
   {
    temp = P0 ;
    temp = temp & 0xf0;
   }
   //display()  ;
  }
 }  P0 = 0xf7;
 temp = P0;
 temp = temp & 0xf0;
 if(temp != 0xf0)
 {
  delays(10);
  temp = P0;
  temp = temp & 0xf0;
  if(temp != 0xf0)
  {
   temp = P0;
   switch(temp)
   {
    case 0xe7:
      key = 12;
     break;
    case 0xd7:
     key = 13;
     break;
    case 0xb7:
     key = 14;
     break;
    case 0x77:
     key = 15;
     break;
   } 
   while(temp != 0xf0)
   {
    temp = P0 ;
    temp = temp & 0xf0;
   }
   //display()  ;
  }
 }
}
void main()
{
 while(1)
 {
  matrixkeyscan();
  P2 = table[0];
  P1 = shumaguan[key];
 }
}   今天,新讲的扫描,没有搞懂。 知道如何控制行,但,对于列的控制,很是不明白、。 期待,详细解释、 此帖出自小平头技术问答
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2条回答
j1a2s3o4n
1楼-- · 2019-03-25 03:38
先把第一行导通,读出列,如果4列有被按下的,那就不是原来写进P口的那个值,&0xf0的目的是为了方便,确定有按键被按下以后,再通过导通的行和按下的列的位置确定按下按键的确切位置,最后等待抬起,在抬起期间反复读取P口的值,抬起以后退出。如果第一行没有列被按下,那就导通第二行,,,
youq2011
2楼-- · 2019-03-25 06:55
????

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