7 years, 11 months ago.

CAN transmit buffer status - how to read it properly?

In my program I need to send several CAN messages one after the other, and I know that there are only 3 transmitting buffers. I have been looking into the CAN registers, especially the global status register, but can't get it to successfully send more than 3 messages without using a wait(...) command.

I'm trying to access bit 2 of CAN2GSR using the following code, but it is not working:

CAN can(p30,p29);
unsigned int _gsr = 0;
char data8[8] = {0x10,0x27,0x28,0x3A,0x10,0x27,0x28,0x8A};
cmsg[1] = CANMessage(2,data8,8,CANData,CANStandard);

for(_stepper = 0; _stepper < 10; _stepper++)
{
    while(1)
    {
        _gsr = LPC_CAN2->GSR;
        if ((_gsr >> 2) & 0x1)
        {
            break;
        } 
    }
    cmsg[1].id = _stepper;
    can.write(cmsg[1]);
}

This code runs, but on the other side only 3 messages are received.

Am I accessing the right register in the right way?

Any help would be appreciated.

Thanks,

Adam.

1 Answer

7 years, 11 months ago.

It looks like can.write() returns 0 if the call fails and 1 if it succeeds. Could you try something like:

CAN can(p30,p29);
char data8[8] = {0x10,0x27,0x28,0x3A,0x10,0x27,0x28,0x8A};
cmsg[1] = CANMessage(2,data8,8,CANData,CANStandard);
 
for(_stepper = 0; _stepper < 10; _stepper++)
{
    while (0 == can.write(cmsg[1]));
}

Accepted Answer

Yh that actually works perfectly, thank you for making it so simple.

posted by Adam Osborn 08 Oct 2013