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嗨，我有一个PIC16F628A，我可以创建一个传入字节的简单回声，它工作得很好。但是我需要调整它，以便它能够处理8位数据和1个奇偶校验位。我试过几件事，但没用。我尝试改变方法接受9位，我尝试添加计时器忽略第一位，以及…但是这种情况并没有发生。希望得到任何帮助。这是我为发送和接收方法提供的代码（我在网上找到了，这是由于Charles）无符号char._._byte（void）{unsigned char RxChr；//保存收到的无符号char i的串行字节；//位索引i=8；//8位数据位以接收TMR0=(256-(SER_BAUD))；//加载TMR1值以偏移RxBit的中心，而(TMR0&1<<7)；//等待波特，而(i)//接收8位串行位，LSB优先{RxChr=(RxChr>>1)；//如果(RxPin==1)//保存数据位RxChr=RxCR SeriBIT & lt；lt；7；否则RXCHR＝RXCHCHR！SER_BIT<7;i-//Next Bit TMR0-=SER_BAUD；//load校正了波特值，而(TMR0&1<<7)；//waitthebaud}TMR0-=SER_BAUD；//等待停止位，确保串行端口是空闲的，而(TMR0&1<<7)；/返回RxChr;}，或者如果任何人有任何代码，则返回RxChr。KS
以上来自于百度翻译 以下为原文 Hi, I have a PIC16F628A, I can create a simple echo of an incoming byte and it works fine. However I need to adjust it so it can handle 8 bits of data and 1 parity bit. I've tried a few things and its not working. Ive tried changing the method to accept 9 bits, I've tried adding a timer to ignore the first bit as well... but it just ain't happening. Any help would be appreciated. This is the code I have for the send and receive methods (which i found online thanks to Charles) unsigned char receive_serial_byte(void) { unsigned char RxChr; // holds the serial byte that was received unsigned char i; // Bit Index i = 8; // 8 data bits to receive TMR0 =(256 - (SER_BAUD)); // load TMR1 value to offset ~center of RxBit while(TMR0 & 1<<7); // wait for baud while(i) // receive 8 serial bits, LSB first { RxChr = (RxChr>>1); // rotate right to store each bit if( RxPin == 1) // save data bit RxChr = RxChr | SER_BIT<<7; else RxChr = RxChr |!SER_BIT<<7; i--; // Next Bit TMR0 -= SER_BAUD; // load corrected baud value while(TMR0 & 1<<7); // wait for baud } TMR0 -= SER_BAUD; // wait for stop bit, ensure serial port is free while(TMR0 & 1<<7); return RxChr; } Or if anyone has any code that would be great. Thanks 
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19个回答
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为什么不在16F628 A中使用内置的UART模块呢？
以上来自于百度翻译 以下为原文 Why don't you use the built-in USART module in the 16F628A? |
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您好，有人可能会批评您的编码风格，但是您的主要问题是RX变量是无符号字符——当奇偶校验打开时只需要8位。（不知道当前的值，但我敢肯定它需要乘以2.BTW：标准UART移位LSBIT第一。因此，用0x80 Re.0x100处理RX缓冲器，当下一个比特到达时右移是相当有效的。
以上来自于百度翻译 以下为原文 Hello, one might criticize your coding style, but yor main problem is that your RX variable is unsigned char - holding only 8 bits where 9 are required when parity is on. (Don't know about the current value, but I'm quite sure it will need a multiplication by 2. BTW: standard UARTs shift LSBit first. Thus ORing the RX buffer with 0x80 resp.0x100 and shifting right when the next bit arrived is quite efficient. Regards |
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HI @ DU000 000 01 -所以基本上我只是想忽略传入奇偶校验，所以8位应该是好的，因此，无符号字符。我会想到第一个输入位的延迟，然后处理其他8位应该按照预期工作。一个简单的回音在没有输入奇偶校验位的情况下工作得很好，所以我认为我的方法和右移等方法可以吗？@1和0-我不太熟悉这些模块，因为我对这个还不熟悉，但是内置的模块是否接受传入的奇偶校验位（我猜是第9位）？任何例子都是很棒的。在接收到的8位数据+ 1位奇偶校验时，我找不到任何东西。
以上来自于百度翻译 以下为原文 Hi @du00000001 - so basically i just want to ignore the incoming parity so 8 bits should be ok hence the unsigned char. I would've thought a delay of the first incoming bit and then handling the other 8 bits should work as expected. A simple echo works fine without an incoming parity bit - so i think the method i have and shifting right etc is ok ? @1and0 - im not too familiar with the modules as I'm new to this , but does the built in module accept incoming parity bits (a 9th bit i guess) ? Any examples of this would be great. I can't find anything on receiving incoming 8 bits of data + 1 bit of parity. |
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有一个鲜为人知的文件，您可以阅读有关这一点。在74页上，您将看到USART接收机（RCSTA）的控制寄存器包含一个名为“RX9:9位接收启用位”的位和另一个名为“RX9D:9位接收数据（可以是奇偶校验位）”的位，因此您完全能够启用9位模式。，但是忽略接收到的奇偶校验位的值。
以上来自于百度翻译 以下为原文 There is a little known document where you can read about this. It is known as the PIC16F628A datasheet On page 74 you will observe that the control register for the USART receiver (RCSTA) contains a bit called "RX9: 9-bit Receive Enable bit" and another one called "RX9D: 9th bit of received data (Can be parity bit)" So you are perfectly able to enable 9-bit mode, but ignore the value of the received parity bit. |
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谢谢，不太熟悉USAT，所以我会四处看看。使用9位的任何明显的例子将是令人敬畏的。谢谢）
以上来自于百度翻译 以下为原文 @qhb thanks, not too familiar with the USART so ill have a look around. Any clear examples of this working with 9 bits would be awesome. Thanks :) |
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您可以忽略输入的奇偶校验位。虽然我觉得使用16位缓冲器应该更容易实现。
以上来自于百度翻译 以下为原文 You could ignore the incoming parity bit. Though I feel that using a 16-bit buffer should be easier to implement. Regards |
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我忽略了第一个比特——但没有运气……在LOL中做我的头
以上来自于百度翻译 以下为原文 @du00000001 im ignoring the first bit - but no luck...doing my head in lol |
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忽略第一位？TU-UART传输序列ISSTART BIT BIT 0位1…位7-奇偶位位，启动位用适当的“Seribit”很容易移出。BTW：您是否将CunrxxChr改为“无符号短”？
以上来自于百度翻译 以下为原文 Ignoring the first bit ? Tue UART transmit sequence is Start bit Bit 0 Bit 1 ... Bit 7 Parity bit Stop bit The start bit is easily shifted out with the appropriate 'SER_BIT'. BTW: did you change RxChr to 'unsigned short' ? |
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作为一个TopyAdvID到PoxSeriSerialyByter（）在POST 1中，只需等待奇偶校验位，因为你想忽略它：
以上来自于百度翻译 以下为原文 As a bandaid to your receive_serial_byte() in Post #1, just wait out the parity bit since you wanted to ignore it: // ... TMR0 -= 2*SER_BAUD; // wait for parity bit and stop bit while(TMR0 & 1<<7); return RxChr; } |
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@ 1和0没有工作，嗯…@ DU000 0 000 01意识到我有奇偶校验位错误…所以，8位的解决方案，我应该只是工作，因为奇偶性将被忽略？也许进来的数据有点太快了
以上来自于百度翻译 以下为原文 @1and0 that didn't work hmm... @du00000001 realised I've got the parity bit wrong... so the 8 bit solution i have should just work right as the parity would just be ignored ? maybe the incoming data is coming it a bit too quickly hmm |
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关于上面的代码）：在最后添加第二个“等待周期”（等待奇偶校验位，然后是停止位）。您确定发送器“在另一端”工作正常吗？
以上来自于百度翻译 以下为原文 Regarding your code on top) again: Add a 2nd 'wait cycle' at the end (awaiting the parity bit, then the stop bit). Are you sure the transmitter 'at the other end' is working correctly? |
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这取决于塞尔波德值对TMR0位7的影响。所以试试这个：
以上来自于百度翻译 以下为原文 That depends on the value of SER_BAUD affecting bit 7 of TMR0 or not. So try this: // ... TMR0 -= SER_BAUD; // wait for parity bit while(TMR0 & 1<<7); TMR0 -= SER_BAUD; // wait for stop bit while(TMR0 & 1<<7); return RxChr; } |
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@1以及0也不走运，传出数据仍然与传入数据不匹配。@du00000001yep，trasmitter绝对在发送字节。我可以在逻辑分析仪上看到。波特率是一个定制的，大约12100kbp.h。我还能试试什么？
以上来自于百度翻译 以下为原文 @1and0 no luck with that either, the outgoing data still isn't matching the incoming data... @du00000001 yep the trasmitter is definitely transmitting the bytes. i can see it on a logic analyser. The baud rate is a custom one at about 12100kbps. hmm what else can I try |
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12100 KBD？-将转换为12.1 MbD。如何回应通过IO引脚接收的比特，并观察RX与IO PIN的范围？
以上来自于百度翻译 以下为原文 12100 kBd? - would translate to 12.1 MBd. What about echoing the bits received via an IO pin and watch the RX vs. the IO pin with the scope ? |
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对不起，不是KBD，只是BPS，所以12100BPS。我去试试。
以上来自于百度翻译 以下为原文 @du00000001 sorry not kbd, just bps, so 12100bps. Ill give that a go. |
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TMR0=(256-(SER_BAUD))；//加载TMR1值以偏移RxBit的中心，而(TMR0&1<<7)；//等待波特，这不能确保在波特时间的中心开始采样，我也看不到等待开始
以上来自于百度翻译 以下为原文 TMR0 =(256 - (SER_BAUD)); // load TMR1 value to offset ~center of RxBit while(TMR0 & 1<<7); // wait for baud this does not ensure the start of the sampling at the center of the baud time also I do not see the wait for the start |
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@ I1CTG-我在while循环中等待开始位。我应该用什么来确保波特时代的中心？如果（RxPIN＝＝）！SeriBIT）/ /在这里等待串行字节起始位{sEndEySerialSyb字节（收件人序列化字节（））；//回音}
以上来自于百度翻译 以下为原文 @i1ctg - i have wait for start bit in the while loop. what should i use to ensure the centre of the baud time ? if(RxPin == !SER_BIT) // wait here for serial byte start bit { send_serial_byte(receive_serial_byte()); // Echo back } |
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现在我们正在接近这个问题的基础！增加约1/2位的延迟时间，并开始采样与开始位。嘿！
以上来自于百度翻译 以下为原文 Now we're approaching the basis of the problem! Add about 1/2 bit time of delay and start sampling with the start bit. Hey presto! |
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在开始后等待一半时间，验证开始位，等待全位时间，开始采样8/9位
以上来自于百度翻译 以下为原文 wait exactly half bit time after start, verify start bit, wait full bit time, start sampling 8/9 bits |
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