51单片机实现点阵的编程C语言

道亦无名

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2015-01-29 09:28:42

实现的效果是

急求！！！

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道亦无名 2015-02-02

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貌似这个问题已经解决了

dontBeHappy 2015-02-02

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图我都看不懂我是大学生也是学单片机的就是过来看看多多照顾

玉怀一捧雪 2015-02-01

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这个跟数码管一样的道理啊！

worldy 2015-01-29

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引用 2 楼 u011046042 的回复:

这个肯定能写出来的，看来楼上的不是搞技术的是搞经济的

买个现成的肯定比自己做一个划算

道亦无名 2015-01-29

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#include<reg51.h>
#include<intrins.h>
#define NOP() _nop_()
sbit MOSIO=P3^0;
sbit R_CLK=P3^2;
sbit S_CLK=P3^1;
void HC595Send(unsigned int SendVal);
unsigned int Val;
unsigned char codetab[]={0xfe,0xfd,0xfb,0xf7,0xef,0xdf,0xbf,0x7f};
unsigned char codedigittab[24][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},
{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}
};
unsigned int timecount;
unsigned char cnta;
unsigned char cntb;

void HC595Send(unsigned int SendVal)
{
	unsigned char i;
	for(i = 0; i < 16; i++)
	{ 
		if((SendVal << i)&0x8000)
		{
			MOSIO=1;    //setdatalinehigh0X8000
		}
			
		else
		{
			MOSIO=0;   //MOSIO=1
		}
		S_CLK = 0;
		NOP();
		NOP();
		S_CLK = 1;
	}
	R_CLK = 0;         //setdatalinelow
	NOP();
	NOP();
	R_CLK = 1;
}
void main(void)
{
	TMOD=0x01;
	TH0 = (65536 - 3000) / 256;
	TL0 = (65536 - 3000) % 256;
	TR0 = 1;
	ET0 = 1;
	EA = 1;
	cntb = 0;
	while(1)
	{ 
		
	}
}
void timer0() interrupt 1 using 0
{
	TH0=(65536-3000) / 256;
	TL0=(65536-3000) % 256;
	if(cntb < 16)
	{
		P0 = ~codedigittab[cntb][cnta];
		Val = codetab[cnta]&0x00ff;
		Val <<= 8;
		Val = Val + 0x00ff;
		HC595Send(Val);
	}
	else
	{
		P0 = ~codedigittab[cntb - 24][cnta];
		Val = codetab[cnta];
		Val = Val + 0xFF00;
		HC595Send(Val);
	}
	if(++cnta >= 8)
		cnta = 0;
	if(++timecount >= 333)
	{
		timecount = 0;
		if(++cntb >= 36)
			cntb = 0;
	}
}

道亦无名 2015-01-29