Hi!

quoted

If this is an Intel x86 machine, it is impossible

quoted

for pages
to get fragmented in the first place. The hardware
allows any
page, from anywhere in memory, to be concatenated
into linear
virtual address space. Even the kernel address space
is virtual.
The only time you need physically-adjacent pages is
if you
are doing DMA that is more than a page-length at a
time. The
kernel keeps a bunch of those pages around for just
that
purpose.

So, if you are making a "memory defragmenter", it is
a CPU time-sink.
That's all.

What if the external fragmentation increases so much
that it is not possible to find a large sized block?
Then, is it not better to defragment rather than swap
or fail?

-Alok

All "blocks" are the same size, i.e., PAGE_SIZE. When RAM
is tight the content of a page is written to the swap-file
according to a least-recently-used protocol. This frees
a page. Pages are allocated to a process only one page at
a time. This prevents some hog from grabbing all the memory
in the machine. Memory allocation and physical page allocation
are two different things, I can malloc() a gigabyte of RAM on
a machine. It only gets allocated when an attempt is made
to access a page.

Alok is right. kernel needs to do kmalloc(8K) from time to time. And
notice that kernel uses 4M tables.
								Pavel
-- 
When do you have a heart between your knees?
[Johanka's followup: and *two* hearts?]
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