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How to use a relative path in xdcpath to specify platform location?

David Aldrich92784
Genius 3720 points

Hi

I am using CCS 5.5 and SYS/BIOS 6.35.  I am using a custom platform (called MY_PLATFORM here to obsure true name), which I specify in my xdcpath as follows:

xdcpath="C:/SVNProj/<my_proj>/trunk/RTSC_platform;C:/ti/edma3_lld_02_11_07_04/packages;C:/ti/ipc_3_00_00_20/packages;C:/ti/bios_6_35_04_50/packages;C:/ti/pdk_keystone2_1_00_00_11/packages;C:/ti/salld_keystone2_02_00_00_06/packages;C:/ti/ccsv5/ccs_base;" xdc.tools.configuro -o configPkg -t ti.targets.elf.C66 -p MY_PLATFORM -r release -c "C:/ti/ccsv5/tools/compiler/c6000_7.4.2"

The path of my platform is:

C:/SVNProj/<my_proj>/trunk/RTSC_platform\MY_PLATFORM

I want to specify the platform relative to the current project location, rather than use an absolute path.  This is possible for other project resources by using PROJECT_LOC, but the xdcpath does not appear to recognise that symbol.

How can I do this please?

Best regards

David

  • 0
    TI__Mastermind 37985 points

    David,
    I don't know where exactly you are using PROJECT_LOC, but this is what worked for me. In the RTSC tab, I clicked on Add... to add a repository and I just typed ${PROJECT_LOC} in the text box. This is the result:

    Then, when I build the project that path is corectly expanded:

    "C:\\ti\\ccsv5\\utils\\bin\\gmake" -k all
    'Building file: ../app.cfg'
    'Invoking: XDCtools'
    "C:/ccsproducts/xdctools_3_25_03_72/xs" --xdcpath="C:/ccsproducts/bios_6_35_04_50/packages;C:/ti/ccsv5/ccs_base;C:/Users/a0868339/workspace_v5_5/customPlatformTest;" xdc.tools.configuro -o configPkg -t ti.targets.C64P -p tisb.my64board -r release -b

    Is that what you tried to do?

  • 0 in reply to Sasha Slijepcevic
    Genius 3720 points

    Hi Sasha

    Your solution worked nicely.

    Thanks

    David