• Locked Locked
  • Replies 2 replies
  • Subscribers 98 subscribers
  • Views 398 views
  • Users 0 members are here
Support feedback
Options
Options
Related

This thread has been locked.

If you have a related question, please click the "Ask a related question" button in the top right corner. The newly created question will be automatically linked to this question.

Where can I find d_can.c for linux AM3359?

Christopher Piggott
Intellectual 570 points
Other Parts Discussed in Thread: AM3359

Hi,

I'm trying to find d_can.c (the DCAN driver for AM3359).  It's referred to in the WIki:

http://processors.wiki.ti.com/index.php/Sitara_DCAN_Linux_Driver

but it doesn't seem to be in the linuxezsdk-bone package.  c_can is there, as is ti_hecc ... but not d_can.

Where can I find it?

 

--Chris

 

 

  • TI__Expert 5450 points

    Christopher,

    If you are pulling the 05.03.00.00 SDK release that release was done before the DCAN support was added.  However, you can see the latest kernel sources at http://arago-project.org/git/projects/?p=linux-am33x.git;a=summary and the d_can.c sources are available in the kernel.  The next release of the SDK will contain these sources as well.

    Chase Maupin

  • in reply to Chase Maupin
    Intellectual 570 points

    Oh, that explains it -- thank you!  I knew that with such a complete explanation of how it works on the wiki that code MUST exist.  I'll wait a little while for the next revision, if I can't wait any longer I'll just build that driver externally against the current kernel.

    I have taken a look at c_can because I was kind of curious how this all works.  I found the c_can_regs and can_if_regs structs for the C device (I understand it's not the AM3359 one, I'm just looking at it for educational purposes).  I also found:

     

     

            struct c_can_regs __iomem *regs;

    I'm wondering what exactly places that symbol "regs" at the correct place in the memory map?  It looks like it comes from something called platform_get_resource(), I'm not familiar with that.