Because L1 cache is 8k deep. On Wed, Jun 29, 2016 at 10:31 PM, Shiva Shankar <[email protected]> wrote:
> Hi > I was going through arm vmlinux script file. and i found below statements > > /* > * first, the init task union, aligned > * to an 8192 byte boundary. > */ > > INIT_TASK_DATA(THREAD_SIZE) > > #define INIT_TASK_DATA(align) \ > . = ALIGN(align); \ > *(.data..init_task) > > As i know we align data structures on a n byte boundary to save access time > and for speed execution but i did not understand why do we need to align > this particular data structure on a 8192 byte boundary. > > sometimes it will be on 1024 or 2048 byte boundary. > > Any inputs would be appreciated!!! > > Regards > > > _______________________________________________ > Kernelnewbies mailing list > [email protected] > http://lists.kernelnewbies.org/mailman/listinfo/kernelnewbies > >
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