红外发射求助
#ifndef STC8_H#define STC8_H
#include <STC8G.H>
#endif
#include<intrins.h>
#include "Timer0.h"
//以12MHZ烧录
void Delay3000ms(void) //@11.0592MHz
{
unsigned char data i, j, k;
_nop_();
i = 169;
j = 80;
k = 87;
do
{
do
{
while (--k);
} while (--j);
} while (--i);
}
void main()
{
P5M0 |= 0x10; P5M1 &= ~0x10;
Timer0_Init();
P54 = 0;
while(1)
{
SendCode(0xfc,0xAf);//1111 1100 0110 1001
Delay3000ms();
}
}#ifndef STC8_H
#define STC8_H
#include <STC8G.H>
#endif
bit OP=0; //红外发射管的亮灭控制位
unsigned int count =0; //载波发射时间
unsigned int Lcount =0; //发射载波后,等待的时间
void Timer0_Init(void) //26微秒@12.000MHz
{
AUXR &= 0x7F; //定时器时钟12T模式
TMOD &= 0xF0; //设置定时器模式
TL0 = 0xE6; //设置定时初始值
TH0 = 0xFF; //设置定时初始值
TF0 = 0; //清除TF0标志
TR0 = 1; //定时器0开始计时
ET0 = 1;
EA = 1;
}
void SendSingleCode(unsigned char _code)
{
while (Lcount>0);//直到当前不在发射数据
switch (_code)
{
case 0://560/26=21.6,560/26=21.6
count = 22;
Lcount = 22;
break;
case 1://560/26=21.6,1690/26=65
count = 22;
Lcount = 65;
break;
case 2://引导码9000/26=346,4500/26=173
count = 346;
Lcount = 173;
break;
case 3://重复码 9000/26=346,2250/26=86.5
count = 346;
Lcount = 87;
break;
case 4://最后一位时,需要给个短暂的高电平,让接收端识别出最后一位的时间
count = 10;
break;
case 5://最后保持一定时间的低电平,用于重复码过短的问题
Lcount = 3800;
break;
}
}
void SendStart()
{
SendSingleCode(2);
}
void SendRepeat()
{
SendSingleCode(3);
SendSingleCode(4);//短暂的高电平
SendSingleCode(5);//长时间的低电平
}
<font color="#ff8c00">void test()
{
SendStart();
SendSingleCode(0);
SendSingleCode(0);
SendSingleCode(0);
SendSingleCode(0);
SendSingleCode(1);
SendSingleCode(1);
SendSingleCode(1);
SendSingleCode(1);
SendSingleCode(1);
SendSingleCode(1);
SendSingleCode(1);
SendSingleCode(1);
SendSingleCode(0);
SendSingleCode(0);
SendSingleCode(0);
SendSingleCode(0);
SendSingleCode(0);
SendSingleCode(0);
SendSingleCode(0);
SendSingleCode(0);
SendSingleCode(1);
SendSingleCode(1);
SendSingleCode(1);
SendSingleCode(1);
SendSingleCode(1);
SendSingleCode(1);
SendSingleCode(1);
SendSingleCode(1);
SendSingleCode(0);
SendSingleCode(0);
SendSingleCode(0);
SendSingleCode(0);
SendSingleCode(4);//短暂的高电平
SendSingleCode(5);//长时间的低电平
}</font>
<font color="#ff0000">void SendCode(unsigned char _address,unsigned char _command)//地址码,命令码
{
unsigned char temp ,i;
SendStart();
for (i = 0; i < 8; i++)
{
temp = _address>>i;
temp &= 0x01;
SendSingleCode(temp);
}
for (i = 0; i < 8; i++)
{
temp = _address>>i;
temp &= 0x01;
temp = ~temp;
SendSingleCode(temp);
}
for (i = 0; i < 8; i++)
{
temp = _command>>i;
temp &= 0x01;
SendSingleCode(temp);
}
for (i = 0; i < 8; i++)
{
temp = _command>>i;
temp &= 0x01;
temp = ~temp;
SendSingleCode(temp);
}
SendSingleCode(4);//短暂的高电平
SendSingleCode(5);//长时间的低电平
}</font>
void timeint(void) interrupt 1
{
if (count>0) //如果是待发送的有效数据flag=1，就在此产生载波（亮灭交变）
{
OP=~OP;
count--;
}
else
{
OP = 0; //发送低电平
if (Lcount>0)
{
Lcount--;
}
}
P54 = OP; //往发射模块的 IO口发送出去
}
问题挺奇怪的,看我代码标红的部分,我用了4个for循环分别发送地址码,地址反码,命令码,命令反码
发送的地址码和命令码分别是0xfc,0xAf
但是接收端只能收到16位数据,正常应该是32位啊,而且正好是反码没有
我又写了个test(),一位一位的发,不用for循环,又正常了
所以我想知道是不是什么导致for循环出问题了
试了下把SendCode里的四个for循环换成了:
i=0;
temp = _address>>i;
temp &= 0x01;
SendSingleCode(temp);
i++;
...
这样一行一行写,就又是正常了{:sweat:} 最终把代码改成这个了,虽然正常收发了,但是还是不知道原因是什么,难道是不能在for循环里取反?{:sweat:}
void SendCode(unsigned char _address,unsigned char _command)//地址码,命令码
{
unsigned char temp ,i;
SendStart();
for (i = 0; i < 8; i++)
{
temp = _address>>i;
temp &= 0x01;
if(temp)
{
SendSingleCode(1);
}
else
{
SendSingleCode(0);
}
}
for (i = 0; i < 8; i++)
{
temp = _address>>i;
temp &= 0x01;
if(!temp)
{
SendSingleCode(1);
}
else
{
SendSingleCode(0);
}
}
for (i = 0; i < 8; i++)
{
temp = _command>>i;
temp &= 0x01;
if(temp)
{
SendSingleCode(1);
}
else
{
SendSingleCode(0);
}
}
for (i = 0; i < 8; i++)
{
temp = _command>>i;
temp &= 0x01;
if(!temp)
{
SendSingleCode(1);
}
else
{
SendSingleCode(0);
}
}
SendSingleCode(4);//短暂的高电平
SendSingleCode(5);//长时间的低电平
} wnagming 发表于 2024-6-30 21:13
你用11.0592M，程序执行时间与定时器中断时间刚好契合，也就是程序中要发送下一位了，定时器刚好把上一位的 ...
你这应该涉及底层机器语言执行了吧,我对这不怎么熟
你的意思是定时器把for给打断了之后,当前的for被放弃了,然后执行下一个for?
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