From: Russell King on
On Wed, Mar 10, 2010 at 01:24:26PM -0800, Yinghai Lu wrote:
> +/* How much should we pad RAM ending depending on where it is? */
> +static unsigned long __init ram_alignment(resource_size_t pos)
> +{
> + unsigned long mb = pos >> 20;
> +
> + /* To 64kB in the first megabyte */
> + if (!mb)
> + return 64*1024;
> +
> + /* To 1MB in the first 16MB */
> + if (mb < 16)
> + return 1024*1024;
> +
> + /* To 64MB for anything above that */
> + return 64*1024*1024;
> +}

This doesn't make sense for generic code.

1. All architectures do not have RAM starting at physical address 0.
2. What about architectures which have relatively little memory (maybe
16MB total) split into four chunks of 4MB spaced at 512MB ?

Other comments:

1. It doesn't support mem=size(a)base, which is used extensively on ARM.
2. How does memory get allocated for creating things like page tables?

Currently, bootmem supports ARM very well with support for flatmem,
sparsemem and discontigmem models (the latter being deprecated). Can
this code support all three models?

Where are patches 1 to 4?

Lastly, why exactly is bootmem being eliminated? Bootmem offers more
flexible functionality than this e820 code appears at first read-through
seems to.

--
Russell King
Linux kernel 2.6 ARM Linux - http://www.arm.linux.org.uk/
maintainer of:
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