8*8LED点阵扫描问题！！

只看该作者 · 2011-5-13 21:10

假设P3 = table[i+j]; i是行扫描时的数组变量，j是列扫描。为什么是i+j而不是直接P3=table[j] ???

只看该作者 · 2011-5-13 21:34

研究了两天led点阵，貌似懂了，坐骑实物来发现自己什么都不懂啊！！

1万 · 2011-5-13 21:36

一行程序，能懂者，仙人也。

只看该作者 · 2011-5-13 23:24

#include <reg52.h>
typedef  unsigned char uint8;
typedef  unsigned int uint16;
typedef  unsigned long uint32;

code uint8 graph[] = {0xFF,0xFF,0xFF,0xFF,0xFF,0xFF,0xFF,0xFF,
   0xFF,0xC3,0xE7,0xE7,0xE7,0xE7,0xC3,0xFF,
   0xFF,0x99,0x00,0x00,0x00,0x81,0xC3,0xE7,
   0xFF,0x99,0x99,0x99,0x99,0x81,0xC3,0xFF,
   0xFF,0xFF,0xFF,0xFF,0xFF,0xFF,0xFF,0xFF
   };
sbit ENLED = P1^4;
sbit ADDR0 = P1^0;
sbit ADDR1 = P1^1;
sbit ADDR2 = P1^2;
sbit ADDR3 = P1^3;

uint8 i= 0,counter = 0;

void timer1_init()
{
TMOD |= 0x10;
TMOD &= 0xdf;
TH1 = 0xFC;
TL1 = 0x67;
TR1 = 1;
}
void int_init()
{
ET1 = 1;
EA  = 1;
}
void refresh_led()
{
   static uint8 j = 0;
switch(j)
   {
   case 0: ADDR0 = 0;ADDR1 = 0;ADDR2 = 0;break;
case 1: ADDR0 = 1;ADDR1 = 0;ADDR2 = 0;break;
case 2: ADDR0 = 0;ADDR1 = 1;ADDR2 = 0;break;
case 3: ADDR0 = 1;ADDR1 = 1;ADDR2 = 0;break;
case 4: ADDR0 = 0;ADDR1 = 0;ADDR2 = 1;break;
case 5: ADDR0 = 1;ADDR1 = 0;ADDR2 = 1;break;
case 6: ADDR0 = 0;ADDR1 = 1;ADDR2 = 1;break;
case 7: ADDR0 = 1;ADDR1 = 1;ADDR2 = 1;break;
default: break;
}
P0 = graph[i + j];
if(7 == j++) j = 0;

}

main()
{
ENLED = 0; ADDR3 = 0;
timer1_init();
int_init();
while(1);

}
void interrupt_timer1() interrupt 3
{
   TH1 = 0xFC;
   TL1 = 0x67;
   counter++;
   if(counter == 200)
      {
     counter = 0;
i++;
if(i>=32){i=0;}
   }
   refresh_led();

}