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what's the kernel parameter of "mem=76M@0x80000000 mem=384M@0x88000000" mean?

baolei zhang
Prodigy 230 points

now I have questions in the kernel parameters.

In uboot, there are Environment Variables as flows:

mem_size=mem=76M@0x80000000 mem=384M@0x88000000
mmcargs=setenv bootargs  ${mem_size} ......

what does it mean of   "mem=76M@0x80000000 mem=384M@0x88000000" ,  I have read www.kernel.org/doc/Documentation/kernel-parameters.txt , the format of "mem" kernel param is "mem=nn[KMG]" ,  does the Linux kernel in TI psp have extended the "mem" kernel parameter for it's special use?(for DSP use? )

  • 0
    Prodigy 230 points

    Hi friends!

    My RAM memory is 512M in total,  I guess "mem=76M@0x80000000 mem=384M@0x88000000" mean  the linux kernel use 76M memory started at 0x80000000, and dsp use 384 memory started at 0x88000000?

    Are there any detailed documents for this?

    Thanks for any help!

  • 0 in reply to baolei zhang
    TI__Intellectual 2785 points

    mem= parameter in bootarg implies that it is managed by kernel only, so as above 76M@0X80000000 & 384@0X88000000 both 480M (76M+384M) managed by kernel only.

    DSP memory partition is different, refer to your software delivery documents to know more about the memory partition in your release.

    In case if you use EZSDK for TI81XX devices then memory map document at link http://processors.wiki.ti.com/index.php/EZSDK_Memory_Map would help.

    Best Regards

    Velan

  • 0
    TI__Intellectual 2725 points

    Hi,

    mem=76M@0x80000000 mem=384M@0x88000000 means there are 2 memory partitions.  one 76MB starting at 0x80000000 & the other 384Mb at 0x88000000. so a kernel has to handle a total of 76M + 384M = 460MB.

    Regards

    Gururaja