点阵如何显示
时间:10-02
整理:3721RD
点击:
如图注释大部分地方理解了,只是对于数字字母是如何显示的比较困惑,或者说红色的二维数组是根据什么编出来的?求解,非常感谢。
#include<reg51.h>
unsigned char code tab[]={0x7f,0xbf,0xdf,0xef,0xf7,0xfb,0xfd,0xfe,}; //列选通控制
unsigned char code digittab[18][8]={
{0x00,0x00,0x3e,0x41,0x41,0x41,0x3e,0x00}, //0
{0x00,0x00,0x00,0x00,0x21,0x7f,0x01,0x00}, //1
{0x00,0x00,0x27,0x45,0x45,0x45,0x39,0x00}, //2
{0x00,0x00,0x22,0x49,0x49,0x49,0x36,0x00}, //3
{0x00,0x00,0x0c,0x14,0x24,0x7f,0x04,0x00}, //4
{0x00,0x00,0x72,0x51,0x51,0x51,0x4e,0x00}, //5
{0x00,0x00,0x3e,0x49,0x49,0x49,0x26,0x00}, //6
{0x00,0x00,0x40,0x40,0x40,0x4f,0x70,0x00}, //7
{0x00,0x00,0x36,0x49,0x49,0x49,0x36,0x00}, //8
{0x00,0x00,0x32,0x49,0x49,0x49,0x3e,0x00}, //9
{0x00,0x00,0x7F,0x48,0x48,0x30,0x00,0x00}, //P
{0x00,0x00,0x7F,0x48,0x4C,0x73,0x00,0x00}, //R
{0x00,0x00,0x7F,0x49,0x49,0x49,0x00,0x00}, //E
{0x00,0x00,0x3E,0x41,0x41,0x62,0x00,0x00}, //C
{0x00,0x00,0x7F,0x08,0x08,0x7F,0x00,0x00}, //H
{0x00,0x00,0x00,0xFF,0xFF,0x00,0x00,0x00}, //I
{0x00,0x7F,0x10,0x08,0x04,0x7F,0x00,0x00}, //N
{0x7C,0x48,0x48,0xFF,0x48,0x48,0x7C,0x00} //中
};
unsigned int timecount1 , timecount2; //定义的变量
unsigned char cntx , cnty ; //cntx列控制数的变量 cnty用来控制 字符与字符之间的切换
void main(void)
{
cnty=0;
cntx=0;
while(1)
{
if(cnty<18) //红色
{
P1=0xFF; //绿色选通都关闭掉
P2=tab[cntx]; // 列线
P0=digittab[cnty][cntx]; // 行线
}
else //绿色
{
P2=0xFF; //红色的列选通都关闭掉
P1=tab[cntx]; // 列线
P0=digittab[cnty-18][cntx]; // 行线
}
//用于控制动态扫描的速度
if(++timecount1>=50) //每50个扫描周期cntx才加一。
{
timecount1=0;
if(++cntx>=8) cntx=0; //要知道为什么是8,显示的每个数字用了八个十六进制数。
}
//用于控制动字符间的切换速度
if(++timecount2>=20000) //每20000个扫描周期cnty才加一
{
timecount2=0;
if(++cnty>=36)cnty=0; //红色显示16行绿色显示16行。所以是36
}
}
}
#include<reg51.h>
unsigned char code tab[]={0x7f,0xbf,0xdf,0xef,0xf7,0xfb,0xfd,0xfe,}; //列选通控制
unsigned char code digittab[18][8]={
{0x00,0x00,0x3e,0x41,0x41,0x41,0x3e,0x00}, //0
{0x00,0x00,0x00,0x00,0x21,0x7f,0x01,0x00}, //1
{0x00,0x00,0x27,0x45,0x45,0x45,0x39,0x00}, //2
{0x00,0x00,0x22,0x49,0x49,0x49,0x36,0x00}, //3
{0x00,0x00,0x0c,0x14,0x24,0x7f,0x04,0x00}, //4
{0x00,0x00,0x72,0x51,0x51,0x51,0x4e,0x00}, //5
{0x00,0x00,0x3e,0x49,0x49,0x49,0x26,0x00}, //6
{0x00,0x00,0x40,0x40,0x40,0x4f,0x70,0x00}, //7
{0x00,0x00,0x36,0x49,0x49,0x49,0x36,0x00}, //8
{0x00,0x00,0x32,0x49,0x49,0x49,0x3e,0x00}, //9
{0x00,0x00,0x7F,0x48,0x48,0x30,0x00,0x00}, //P
{0x00,0x00,0x7F,0x48,0x4C,0x73,0x00,0x00}, //R
{0x00,0x00,0x7F,0x49,0x49,0x49,0x00,0x00}, //E
{0x00,0x00,0x3E,0x41,0x41,0x62,0x00,0x00}, //C
{0x00,0x00,0x7F,0x08,0x08,0x7F,0x00,0x00}, //H
{0x00,0x00,0x00,0xFF,0xFF,0x00,0x00,0x00}, //I
{0x00,0x7F,0x10,0x08,0x04,0x7F,0x00,0x00}, //N
{0x7C,0x48,0x48,0xFF,0x48,0x48,0x7C,0x00} //中
};
unsigned int timecount1 , timecount2; //定义的变量
unsigned char cntx , cnty ; //cntx列控制数的变量 cnty用来控制 字符与字符之间的切换
void main(void)
{
cnty=0;
cntx=0;
while(1)
{
if(cnty<18) //红色
{
P1=0xFF; //绿色选通都关闭掉
P2=tab[cntx]; // 列线
P0=digittab[cnty][cntx]; // 行线
}
else //绿色
{
P2=0xFF; //红色的列选通都关闭掉
P1=tab[cntx]; // 列线
P0=digittab[cnty-18][cntx]; // 行线
}
//用于控制动态扫描的速度
if(++timecount1>=50) //每50个扫描周期cntx才加一。
{
timecount1=0;
if(++cntx>=8) cntx=0; //要知道为什么是8,显示的每个数字用了八个十六进制数。
}
//用于控制动字符间的切换速度
if(++timecount2>=20000) //每20000个扫描周期cnty才加一
{
timecount2=0;
if(++cnty>=36)cnty=0; //红色显示16行绿色显示16行。所以是36
}
}
}
再有就是里面的一些变量定义以后没有初始化,就用,会不会不太合适,他们的初始值就一定是零吗?