求解答cy8cplc10 plc调制解调器的问题

读取AN52478本地地址似乎是特定节点的逻辑地址。
源地址是向主机发送消息的地址。
HTTP://www. CyPress？COM/？RID＝37956 AN52478
问候，Dana。



     以下为原文
  Reading AN52478 local address seems to be logical address of the specific node.
    Source address is the address from which a message was sent to master.
     
        
              
    http://www.cypress.com/?rID=37956     AN52478
     
    Regards, Dana.

嗨，Dana，
是的，你是对的……但是我知道。我只是想知道它们是否应该是一样的。显然，您发送数据的源地址应该是发送数据的节点的本地地址。那么，他们为什么给出了两个地址？第二个问题呢？我应该将数据发送到地址（本地）0x00到0xFF的节点，并等待答案，以检查它们是否存在？这要花很多时间。有最简单的方法吗？最好的问候…罗伯特



     以下为原文
  Hi Dana,
    Yes, you're right...but I knew about it. I'm just still wondering if they should be the same. It seems to be clear that the source address you send the data from should be the local address of the node that sent the data. So, for what reasons did they gave both addresses? What about the second questions? Should I send data to nodes from address (local) 0x00 to 0xFF and wait for the answer in order to check if they exist? It would take a lot of time. Is there a simplest way? Best regards...Robert

当你查看你的Windows操作系统时，你会发现它是这样工作的：有一个请求，并且在给定的时间限制内，所有的网络设备都必须回答。扫描所有可能的地址并等待答案需要一些时间，但是当您保存结果并仅在上电或用户请求时检查（用户通常知道网络什么时候变）按钮或您设计的任何内容。在上电之后，只需检查已经知道的设备，它们是否仍然存在，将加快三个初始化阶段。这些信息在断电后保存到EEPROM或程序到Flash中。
最后，当需要舒适性时，可以使用主机空闲时扫描连接的新设备的时间。这将需要一些时间，直到那些被认可，但这样你可以检测到网络的变化。
鲍勃



     以下为原文
  When you look at your windows OS, you will see that it works exactly that way: There is a request and within a given time-limit all network devices have to answer. Scanning all possible addresses and waiting for an answer takes some time but when you save the results and check only at power-up or at a user request (Users usually know when the net has changed) with a button or whatever you design. After a power-up just checking for the already known devices whether they are still there will speed up the3 initialization phase. To have informations survive a power-off store them into eeprom or program into flash.
    Lastly, when comfort is required, you may use times when your master is idle to scan for new devices connected. That will take some time until those get recognized but that way you can detect network changes.
     
    Bob

真的没有别的办法吗？我假设在PLCYM模式寄存器（0x05）中的位“杂乱掩码”应该被清除来做这件事吗？这意味着，如果目的地地址不匹配其本地地址，那么从属器会丢弃所有RX消息，并且不会响应主控器。因此，我应该等待设置StasuStxxNoApACK或StasuStxxNoYRESP位吗？它将持续大约256×0.5秒（没有ACK）或256×1.5秒（没有响应）。最好的问候…罗伯特



     以下为原文
          Isn't really no other way to do this? I assume that bit "Promiscuous_MASK" in the PLC_Mode Register (0x05)should be cleared to do this? This means that the Slave Drops all the RX Messages if Destination Address does not match its Local Address and will not respond the Master? So, this way I should wait for the Status_TX_NO_ACK or Status_TX_NO_RESP bits being set? It would last about 256*0.5 seconds (for No ACK) or 256*1.5 seconds (for No RESPONSE). Best regards...Robert