Can I issue an OPEN command to an already open Socket?

Question

Hi, I am a beginner with networking messages and I am just learning what I can do with a unit that consists of a PIC32MX795 host cpu and a W5500 network chip. I have opened a socket to transmit UDP messages and I have noticed that when I issue an OPEN command to this socket the TX_RD register is reset to ‘0’. This makes life a bit easier as it means that I don’t have to worry about the buffer wrapping around as it would if I “followed the rules” and added my message into the buffer at the offset where the TX_RD register was pointing. So: my question is - can I issue an OPEN command to a socket that has already been opened ? Will this have any bad effects on other aspects of the chip’s performance ? Thanks for any thoughts PhilipJ

Eugeny · Answer

You must follow the workflow documented in the datasheet, see W5100 datasheet section 5.2. This is the only guaranteed way for operation. Therefore you must close socket before opening it. And you always must respect TX_RD pointer (and other pointers) and not assume it will always be 0.

Eugeny · Answer

Yes, algorithms described in W5100 datasheet apply. You can’t, and should not reset pointers, they must be operated (read and some of them updated when needed) as described in that section.

PhilipJ · Answer

OK thanks PhilipJ

PhilipJ · Answer

Hi again, I hope I am not being a pain but I have another oddity happening. I set up my code as you suggested: read TX_FSR (this returns 0x0800 as I have 8 sockets) read TX_RD write data into the TX_BUF starting at this address add my message size (in this test case 250 bytes) to TX_RD and write this into TX_WR issue a SEND command This is working fine for the first 8 times but when I try to send it a ninth time the problem occurs: My code is working the buffer pointers as modul0 0x0800 and the data is written into the buffer so that it fills it up to address 0x07FF and then wraps around to write the rest from address 0x0000 The TX_WR pointer becomes 0x00CA. What actually happens is the W5500 sends 43 messages, each of 1472 bytes and then a 44th message of 442 bytes - total of 63,738 bytes. It’s almost as if the TX_RD and TX_WR pointers go through the full count of 0 to 65535 and don’t wrap around at 0x07ff. So my question is: What is the correct process when the TX_RD pointer is close to the end of the buffer? regards PhilipJ EDIT: I have read the W5500 datasheet about registers TX_RD and TX_WR and it does seem to say (on page 54) that these registers will go all the way to 0xFFFF and then wrap around to 0. So how does this work when the buffer is only 2048 bytes long? How do I know where to put my message data if both TX_RD and TX_WR and pointing at say 0x08ca ?

PhilipJ · Answer

Hi, but there is no MASK register in the W5500 chip. HOWEVER, I have worked it out, it seems the W5500 chip has a “virtual” 64k TX buffer (w5599_ds_v109e.pdf page 28) which you can write into BUT only up to the maximum of FREE BYTES (2048 or less_. The block you write is then “mapped” by the chip into the actual 2k bytes of memory allocated for TX messages!! So all my messing about with modulo 2048 maths was messing up the pointers causing the chip to send 0x10000 - 0x07d0 + 250 bytes in 44 seperate messages. regards PhilipJ