单片机故障排除篇--如何判定是硬件问题还是软件问题？

zhangrelay

发布于 2021-12-02 14:30:58

5850

发布于 2021-12-02 14:30:58

文章被收录于专栏：机器人课程与技术机器人课程与技术

一段代码写完了，编译下载都ok。

但是没有看到想要的结果？怎么排错？？？

借一个别人正常的板子是一种方法，但是镜像提供了全仿真工具，试一下吧。

新建项目：

添加time.c到工程中！

添加复用数码管。

当然程序要做微微调整。

单片机课堂思考题-2000秒倒计时？

参考上例程序：

先分配管脚和端口

#include<8052.h>                         
       
unsigned char smgduan[]={0xc0,0xf9,0xa4,0xb0,0x99,0x92,0x82,0x0f8,0x80,0x90};
unsigned char T0RH = 0; 
unsigned char T0RL = 0; 
unsigned int cnt;
int sec;
int sec_all=6666;
int count=0;
void LedScan(unsigned char pos,unsigned char dispcode);
void EXINT0() __interrupt 0;
void Timer0() __interrupt 1;
void EXINT1() __interrupt 2;
void ConfigTimer0(unsigned int);
void main()
{
	cnt = 0;            
	sec = sec_all;      
	ConfigTimer0(2); 
	IT0=1; 
	EX0=1;
	IT1=1; 
	EX1=1;
	PX1=1;	  				
        TR0=1;          
	while(1)
	{
		LedScan(0,smgduan[sec%10]);    
		LedScan(1,smgduan[sec/10%10]);  
		LedScan(2,smgduan[sec/100%10]); 
		LedScan(3,smgduan[sec/1000]);   
		if(count%2==0)
		{
		  TR0 = 1;
		}
		else
		{
	          TR0 = 0;
		}	
	}
}

void EXINT0() __interrupt 0     	
{
	count++;				
}
void EXINT1() __interrupt 2
{
	sec_all=0;
}

void ConfigTimer0(unsigned int ms)
{
	unsigned long tmp; 	     
	tmp = 11059200 / 12; 	 
	tmp = (tmp * ms) / 1000; 
	tmp = 65536 - tmp; 	     
	tmp = tmp + 18; 	     
	T0RH = (unsigned char)(tmp>>8); 	
	T0RL = (unsigned char)tmp;
	TMOD &= 0xF0; 		
	TMOD |= 0x01; 		
	TH0 = T0RH; 		
	TL0 = T0RL;
	ET0 = 1; 		   
}

void Timer0() __interrupt 1
{
	TH0 = T0RH; 	
	TL0 = T0RL;
	cnt++; 		
	if (cnt >= 5)
	{
		cnt = 0;
		sec--;   
		if (sec < 0)   
		{
			sec = sec_all; 
		}
	}
}

void LedScan(unsigned char pos,unsigned char dispcode)
{
 unsigned char i;
 for(i=0;i<=pos;i++)
 {		
	switch(pos)	 	{
		case(0):
		P1_0=0;P1_1=1;P1_2=1;P1_3=1; break;
		case(1):
		P1_0=1;P1_1=0;P1_2=1;P1_3=1; break;
		case(2):
		P1_0=1;P1_1=1;P1_2=0;P1_3=1; break;
		case(3):
		P1_0=1;P1_1=1;P1_2=1;P1_3=0; break;
	}
	P0=dispcode;	
 }	
}

#include<8052.h>                         
       
unsigned char smgduan[]={0xc0,0xf9,0xa4,0xb0,0x99,0x92,0x82,0x0f8,0x80,0x90};
unsigned char T0RH = 0; 
unsigned char T0RL = 0; 
unsigned int cnt;
int sec;
int sec_all=6666;
int count=0;
void LedScan(unsigned char pos,unsigned char dispcode);
void EXINT0() __interrupt 0;
void Timer0() __interrupt 1;
void EXINT1() __interrupt 2;
void ConfigTimer0(unsigned int);
void main()
{
	cnt = 0;            
	sec = sec_all;      
	ConfigTimer0(1); 
	IT0=1; 
	EX0=1;
	IT1=1; 
	EX1=1;
	PX1=1;	  				
        TR0=1;          
	while(1)
	{
		LedScan(0,smgduan[sec%10]);    
		LedScan(1,smgduan[sec/10%10]);  
		LedScan(2,smgduan[sec/100%10]); 
		LedScan(3,smgduan[sec/1000]);   
		if(count%2==0)
		{
		  TR0 = 1;
		}
		else
		{
	          TR0 = 0;
		}	
	}
}

void EXINT0() __interrupt 0     	
{
	count++;				
}
void EXINT1() __interrupt 2
{
	sec_all=0;
}

void ConfigTimer0(unsigned int ms)
{
	unsigned long tmp; 	     
	tmp = 11059200 / 12; 	 
	tmp = (tmp * ms) / 1000; 
	tmp = 65536 - tmp; 	     
	tmp = tmp + 18; 	     
	T0RH = (unsigned char)(tmp>>8); 	
	T0RL = (unsigned char)tmp;
	TMOD &= 0xF0; 		
	TMOD |= 0x01; 		
	TH0 = T0RH; 		
	TL0 = T0RL;
	ET0 = 1; 		   
}

void Timer0() __interrupt 1
{
	TH0 = T0RH; 	
	TL0 = T0RL;
	cnt++; 		
	if (cnt >= 2)
	{
		cnt = 0;
		sec--;   
		if (sec < 0)   
		{
			sec = sec_all; 
		}
	}
}

void LedScan(unsigned char pos,unsigned char dispcode)
{
 unsigned char i;
 for(i=0;i<=pos;i++)
 {		
	switch(pos)	 	{
		case(0):
		P1_0=0;P1_1=1;P1_2=1;P1_3=1; break;
		case(1):
		P1_0=1;P1_1=0;P1_2=1;P1_3=1; break;
		case(2):
		P1_0=1;P1_1=1;P1_2=0;P1_3=1; break;
		case(3):
		P1_0=1;P1_1=1;P1_2=1;P1_3=0; break;
	}
	P0=dispcode;	
 }	
}

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原始发表：2021-11-03 ，如有侵权请联系 cloudcommunity@tencent.com 删除

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