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点阵seo（点阵色沉图片）

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2022-12-19 17:17:41
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今天给各位分享点阵seo的知识，其中也会对点阵色沉图片进行解释，如果能碰巧解决你现在面临的问题，别忘了关注本站，现在开始吧！

本文目录一览：

1、单片机编程
2、LED点阵 CD-1088A/B 为8*8单色二极管点阵，共有16个引脚，如何连到51单片机上显示汉字呢？
3、求en或者em开头的英文单词
4、计算机软硬件系统的组成及主要技术指标

单片机编程

用取模软件提取点阵汉字信息，按照下面的程序，修改以下就可以了

下面是静止显示“农林大学”，你用点阵信息代码以下就显示单片机了。

#include REG51.h

#define uchar unsigned char

#define uint unsigned int

uchar disrow;

uchar code HZ[];

uchar BUFF[9];

void loadonelineandsend(void);

/********************************************/

/*******************************************************/

void main(void)

{

while(1)

{

for(disrow=0;disrow16;disrow++)

{

loadonelineandsend();

P1=0x10+disrow;

P1=0x20+disrow;

}

/******************************************************/

void loadonelineandsend(void)

{

uchar s;

int q;

q=0;

for(s=0;s4;s++)

{

BUFF[q]=HZ[32*s+disrow*2];

BUFF[q+1]=HZ[32*s+disrow*2+1];

SBUF=255-BUFF[q+1];

while(!TI);TI=0;

SBUF=255-BUFF[q];

while(!TI);TI=0;

}

/******************************************************/

/***********农林大学********************************************/

uchar code HZ[]=

{

0x01,0x00,0x01,0x00,0x7F,0xFE,0x41,0x02,0x81,0x04,0x02,0x00,0x02,0x08,0x05,0x18,

0x09,0x20,0x18,0xC0,0x28,0x80,0x48,0x40,0x88,0x30,0x0A,0x0E,0x0C,0x04,0x08,0x00,

0x08,0x20,0x08,0x20,0x08,0x20,0x08,0x24,0xFE,0xFE,0x08,0x20,0x18,0x60,0x1C,0x70,

0x2A,0xA8,0x28,0xAE,0x49,0x24,0x8A,0x20,0x08,0x20,0x08,0x20,0x08,0x20,0x08,0x20,

0x01,0x00,0x01,0x00,0x01,0x00,0x01,0x00,0x01,0x04,0xFF,0xFE,0x01,0x00,0x02,0x80,

0x02,0x80,0x02,0x40,0x04,0x40,0x04,0x20,0x08,0x10,0x10,0x0E,0x60,0x04,0x00,0x00,

0x22,0x08,0x11,0x08,0x11,0x10,0x00,0x20,0x7F,0xFE,0x40,0x02,0x80,0x04,0x1F,0xE0,

0x00,0x40,0x01,0x84,0xFF,0xFE,0x01,0x00,0x01,0x00,0x01,0x00,0x05,0x00,0x02,0x00,

};

上移程序：

#include REG51.h

#define uchar unsigned char

#define uint unsigned int

#define SPEED 3

uchar col,disrow;

uint word,hangword;

uchar code HZ[];

uchar BUFF[9];

void loadonelineandsend(void);

/********************************************/

/*******************************************************/

void main(void)

{

uchar i;

col=0;word=0;hangword=0;

while(1)

{

while(col16)

{

for(i=0;iSPEED;i++)

{

for(disrow=0;disrow16;disrow++)

{

loadonelineandsend();

P1=0x10+disrow;

P1=0x20+disrow;

}

col++;

}

col=0;

hangword=8*word++;

if(word=32)word=0;

}

/******************************************************/

void loadonelineandsend(void)

{

uchar s;

int q;

q=0;

for(s=0;s4;s++)

{

BUFF[q]=HZ[hangword+32*s+disrow*2];

BUFF[q+1]=HZ[hangword+32*s+disrow*2+1];

SBUF=255-BUFF[q+1];

while(!TI);TI=0;

SBUF=255-BUFF[q];

while(!TI);TI=0;

}

/******************************************************/

/***********农林大学********************************************/

uchar code HZ[]=

{

0x00,0x00,0x00,0x00,0x00,0x00,0x00,0x00,0x00,0x00,0x00,0x00,0x00,0x00,0x00,0x00,