Serial Rx Buffer

hai,

I am Trying to Understand the arduino Serial Communication. The Hardware and hardware serial Both Works On the Interrupts.. Right. So Whenever it received the Data it is Stored into Serial Buffer WithOut Interrupting the Other Work If I Check the Serial Buffer Using Serial.avallable() >0(or some value) then i conform that i have some data in buffer and i can read it.

If i don't read the data insufficient intervals, Does data gets stuffed into buffer until it reached Overflow.

Suppose If I receive 6 bytes in a turn. with 9600 baud rate. it takes 62 milliSeconds. and if I dont read the Buffer For 3 turns say 2 milliSeconds, and when i tried to read the data then i get the from the First turn.

If i don't read the data insufficient intervals, Does data gets stuffed into buffer until it reached Overflow.

Not merely until it reaches overflow, but forever thereafter. There is no indication of overflow, data is silently overwritten.

Suppose If I receive 6 bytes in a turn. with 9600 baud rate. it takes 62 milliSeconds.

Suppose it take 6.2 milliseconds, because that's the minimum time it takes.

and if I dont read the Buffer For 3 turns say 2 milliSeconds, and when i tried to read the data then i get the from the First turn.

I don't understand that sentence

Quote and if I dont read the Buffer For 3 turns say 2 milliSeconds, and when i tried to read the data then i get the from the First turn. I don't understand that sentence

If I set a device to Continuously send the data to Software Serial.. if we assume that data Serially coming is [R0005/r,R0004/r,R0003/r,R0002/r,R001/r] first i read the data R0001/r

after sometime say... 3*6.2 milliseconds if i read the data What is the value of data I get? R0002/r? or R0005/r

If the transmitter is sending continuously, the receive buffer will fill up in 64 /1.04 = 61.5milliseconds.

Serial.read() always gives you the "oldest" byte in the input buffer - and it only gives you a single byte. If you need to get "R001/r" you need to do 5 reads.

This is not a correct representation [R0005/r,R0004/r,R0003/r,R0002/r,R001/r] it should be [..../r2000R/r100R]

I don't think it is correct to say that, when the buffer is full, the last byte is overwritten. My understanding is that the subsequent incoming bytes are just ignored and lost until space is made in the buffer by removing some of the data.

...R

This is not a correct representation [R0005/r,R0004/r,R0003/r,R0002/r,R001/r] it should be [..../r2000R/r100R]

More likely, it is meant to be 2000R\r100R\r etc

AWOL: Not merely until it reaches overflow, but forever thereafter. There is no indication of overflow, data is silently overwritten.

If I read this right, it just discards the extra:

inline void store_char(unsigned char c, ring_buffer *buffer)
{
  int i = (unsigned int)(buffer->head + 1) % SERIAL_BUFFER_SIZE;

  // if we should be storing the received character into the location
  // just before the tail (meaning that the head would advance to the
  // current location of the tail), we're about to overflow the buffer
  // and so we don't write the character or advance the head.
  if (i != buffer->tail) {
    buffer->buffer[buffer->head] = c;
    buffer->head = i;
  }
}

Serial.read() always gives you the “oldest” byte in the input buffer - and it only gives you a single byte. If you need to get “R001/r” you need to do 5 reads.

after sometime say… 3*6.2 milliseconds if i read the data
What is the value of data I get? R0002/r? or R0005/r

So I get R0002/r if I do 5 reads…

Is there anyway I can read newest byte received?
and to flush the Rx buffer Whenever I don’t want data
Serial.flush() is for Tx buffer… Right?

[quote author=Nick Gammon link=msg=1949114 date=1415177053] If I read this right, it just discards the extra:

inline void store_char(unsigned char c, ring_buffer *buffer)
{
  int i = (unsigned int)(buffer->head + 1) % SERIAL_BUFFER_SIZE;

  // if we should be storing the received character into the location
  // just before the tail (meaning that the head would advance to the
  // current location of the tail), we're about to overflow the buffer
  // and so we don't write the character or advance the head.
  if (i != buffer->tail) {
    buffer->buffer[buffer->head] = c;
    buffer->head = i;
  }
}

[/quote]

Oops, sorry, yes, discarded.