On Jun 15, 2007, at 6:41 AM, Christian Kandeler wrote:
sc->sc_txdatasz = q_to_b(&tp->t_outq, sc->sc_txbuf, sc->sc_txfifosz);
This copies at most sc->sc_txfifosz (the size of the UART's FIFO)
bytes from
tp->outq to sc->sc_txbuf. After that, UART_TRANSMIT is called and
outputs the
characters just copied to the serial device.
Yes.
Now obviously, the number of
characters in tp->t_outq can be greater than the UART's FIFO size,
but I
can't for the life of me find a loop anywhere in the call tree. So
what
happens to the rest of the outq buffer?
The loop is formed by the transmit interrupt of the UART.
When UART_TRANSMIT() is called, the FIFO of the UART is
written and transmission starts. Eventually the UART will
raise a "transmission done" interrupt, which will trigger
the next iteration.
And what if the UART has no FIFO at
all (e.g. 8250)? sc->sc_txfifosz is zero in that case, so how does
anything
ever get printed then?
If the UART has no FIFO (in the 16550-sense), then sc_txfifosz
is still 1. There's obviously always at least 1 8-bit register
in the UART to which data can to be written or read. The use
of FIFO in the uart(4) is not limited to its definition in PC
hardware.
FYI,
--
Marcel Moolenaar
[EMAIL PROTECTED]
_______________________________________________
[email protected] mailing list
http://lists.freebsd.org/mailman/listinfo/freebsd-hackers
To unsubscribe, send any mail to "[EMAIL PROTECTED]"
