On Fri, Aug 6, 2010 at 3:56 PM, Wolfgang Grandegger <[email protected]> wrote:
> To be clear, out-of-order messages are also not OK.
Yes I was looking at this with some interleave analysis. I think the
problem lies in the following race:
// suppose we come here with RXB0 filled with packet X
u8 intf = mcp251x_read_reg(spi, CANINTF);
u8 eflag;
int can_id = 0, data1 = 0;
// packet X+1 arrives and lands in RXB1
if (intf & CANINTF_RX0IF) {
mcp251x_hw_rx(spi, 0);
/* Free one buffer ASAP */
mcp251x_write_bits(spi, CANINTF, intf & CANINTF_RX0IF,
0x00);
}
// packet X+2 arrives and lands in RXB0
// unfortunately variable intf is old and doesn't show that we have
something in RXB1
// (*1)
if (intf & CANINTF_RX1IF)
mcp251x_hw_rx(spi, 1);
now on the next turn we will have packet X+2 in RXB0 and X+1 in RXB1
so we read them out-of-order. The easiest way to solve the problem is
to take out the line:
/* Free one buffer ASAP */
mcp251x_write_bits(spi, CANINTF, intf & CANINTF_RX0IF,
0x00);
but we risk to lose more packets. Perhaps also rereading intf in (*1)
could work but we must be careful when we reset CANINTF register to
not lose an interrupt for RXB0. I will do some tests but after next
week because I don't have the access to the hardware until then.
--
Christian Pellegrin, see http://www.evolware.org/chri/
"Real Programmers don't play tennis, or any other sport which requires
you to change clothes. Mountain climbing is OK, and Real Programmers
wear their climbing boots to work in case a mountain should suddenly
spring up in the middle of the computer room."
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