取字模时是8*40的点阵，为什么我将index改成40运行到最后会出现一行乱码
#include <reg52.h>
sbit ADDR0 = P1^0;
sbit ADDR1 = P1^1;
sbit ADDR2 = P1^2;
sbit ADDR3 = P1^3;
sbit ENLED = P1^4;
unsigned char code image[] = { //图片的字模表
0xFF,0xFF,0xFF,0xFF,0xFF,0xFF,0xFF,0xFF,
0x99,0x99,0x99,0x99,0x99,0x81,0xC3,0xFF,
0x99,0x00,0x00,0x00,0x81,0xC3,0xE7,0xFF,
0xC3,0xE7,0xE7,0xE7,0xE7,0xE7,0xC3,0xFF,
0xFF,0xFF,0xFF,0xFF,0xFF,0xFF,0xFF,0xFF
};
void main()
{
EA = 1; //使能总中断
ENLED = 0; //使能U4，选择LED点阵
ADDR3 = 0;
TMOD = 0x01; //设置T0为模式1
TH0 = 0xFC; //为T0赋初值0xFC67，定时1ms
TL0 = 0x67;
ET0 = 1; //使能T0中断
TR0 = 1; //启动T0
while (1);
}
/* 定时器0中断服务函数 */
void InterruptTimer0() interrupt 1
{
static unsigned char i = 0; //动态扫描的索引
static unsigned char tmr = 0; //250ms软件定时器
static unsigned char index = 39;
TH0 = 0xFC; //重新加载初值
TL0 = 0x67;
//以下代码完成LED点阵动态扫描刷新
P0 = 0xFF; //显示消隐
switch (i)
{
case 0: ADDR2=0; ADDR1=0; ADDR0=0; i++; P0=image[index-7]; break;
case 1: ADDR2=0; ADDR1=0; ADDR0=1; i++; P0=image[index-6]; break;
case 2: ADDR2=0; ADDR1=1; ADDR0=0; i++; P0=image[index-5]; break;
case 3: ADDR2=0; ADDR1=1; ADDR0=1; i++; P0=image[index-4]; break;
case 4: ADDR2=1; ADDR1=0; ADDR0=0; i++; P0=image[index-3]; break;
case 5: ADDR2=1; ADDR1=0; ADDR0=1; i++; P0=image[index-2]; break;
case 6: ADDR2=1; ADDR1=1; ADDR0=0; i++; P0=image[index-1]; break;
case 7: ADDR2=1; ADDR1=1; ADDR0=1; i=0; P0=image[index-0]; break;
default: break;
}
//以下代码完成每250ms改变一帧图像
tmr++;
if (tmr >= 250) //达到250ms时改变一次图片索引
{
tmr = 0;
if (index <= 8)
index = 39;
else
index--;
}
}
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