Re: [PATCH] dynamic cache allocation

[phcoder <phcoder@gmail.com>]

Robert Millan wrote:
> On Sat, Mar 07, 2009 at 11:54:18PM +0100, phcoder wrote:
>> +      grub_disk_cache_table = (struct grub_disk_cache *)
>> +	grub_malloc (grub_disk_cache_num * sizeof (struct grub_disk_cache));
>> [...]
>> +   By default use up to one quarter of memory available for allocation  */
>> +#define GRUB_DISK_CACHE_DIVIDE	16536
> 
> I might be missing something, but I think this implies our cache will stop
> working if the memory map doesn't include a contigous block that is as big
> as one quarter of total memory.
> 

It uses the indirection. The space that is allocated by this code is 
only for the cache index. Every index entry is 20 bytes and describes a 
cache entry of 4096 bytes. So for cache to work we need just a block of 
size of 1/800 of free allocatable memory (not the total memory but only 
the memory which was announced to mm.c as free). On the BIOS system with 
3GB only 1/4 of available memory is announced as free (it's probably 
different with Vesa's new allocator). It means that cache index will 
take about 1MiB (about 4MiB if Vesa's allocator uses memory better)

> I think we should be more conservative and define an upper bound for how
> big our cache is supposed to be in a normal use case.  In most cases, we'll
> have as much memory available as we need for this purpose, and it's only a
> few situations (i.e. machines with very low memory) that might be a problem.
> 
> Thoughts?
> 


-- 

Regards
Vladimir 'phcoder' Serbinenko