Re: Regarding memory fragmentation using malloc....
From: Pintu Agarwal
Date: Thu Apr 14 2011 - 08:25:04 EST
Hello Mr. Michal,
Thanks for your comments.
Sorry. There was a small typo in my last sentence (mitigating not *migitating* memory fragmentation)
That means how can I measure the memory fragmentation either from user space or from kernel space.
Is there a way to measure the amount of memory fragmentation in linux?
Can you provide me some references for that?
--- On Thu, 4/14/11, Michal Nazarewicz <mina86@xxxxxxxxxx> wrote:
> From: Michal Nazarewicz <mina86@xxxxxxxxxx>
> Subject: Re: Regarding memory fragmentation using malloc....
> To: "Américo Wang" <xiyou.wangcong@xxxxxxxxx>, "Pintu Agarwal" <pintu_agarwal@xxxxxxxxx>
> Cc: "Andrew Morton" <akpm@xxxxxxxxxxxxxxxxxxxx>, "Eric Dumazet" <eric.dumazet@xxxxxxxxx>, "Changli Gao" <xiaosuo@xxxxxxxxx>, "Jiri Slaby" <jslaby@xxxxxxx>, "azurIt" <azurit@xxxxxxxx>, linux-kernel@xxxxxxxxxxxxxxx, linux-mm@xxxxxxxxx, linux-fsdevel@xxxxxxxxxxxxxxx, "Jiri Slaby" <jirislaby@xxxxxxxxx>
> Date: Thursday, April 14, 2011, 5:47 AM
> On Thu, 14 Apr 2011 08:44:50 +0200,
> Pintu Agarwal <pintu_agarwal@xxxxxxxxx>
> > As I can understand from your comments that, malloc
> from user space will not have much impact on memory
> It has an impact, just like any kind of allocation, it just
> don't care about
> fragmentation of physical memory. You can have only
> 0-order pages and
> successfully allocate megabytes of memory with malloc().
> > Will the memory fragmentation be visible if I do
> kmalloc from
> > the kernel module????
> It will be more visible in the sense that if you allocate 8
> KiB, kernel will
> have to find 8 KiB contiguous physical memory (ie. 1-order
> >> No. When you call malloc() only virtual
> address space is allocated.
> >> The actual allocation of physical space occurs
> when user space accesses
> >> the memory (either reads or writes) and it happens
> page at a time.
> > Here, if I do memset then I am accessing the
> memory...right? That I am doing already in my sample
> Yes. But note that even though it's a single memset()
> call, you are
> accessing page at a time and kernel is allocating page at a
> On some architectures (not ARM) you could access two pages
> with a single
> instructions but I think that would result in two page
> faults anyway. I
> might be wrong though, the details are not important
> >> what really happens is that kernel allocates the
> >> pages and when
> >> it runs out of those, splits a 1-order page into
> >> 0-order pages and
> >> takes one of those.
> > Actually, if I understand buddy allocator, it
> allocates pages from top to bottom.
> No. If you want to allocate a single 0-order page,
> buddy looks for a
> a free 0-order page. If one is not found, it will
> look for 1-order page
> and split it. This goes up till buddy reaches
> > Is the memory fragmentation is always a cause of the
> kernel space program and not user space at all?
> Well, no. If you allocate memory in user space,
> kernel will have to
> allocate physical memory and *every* allocation may
> contribute to
> fragmentation. The point is, that all allocations
> from user-space are
> single-page allocations even if you malloc() MiBs of
> > Can you provide me with some references for migitating
> memory fragmentation in linux?
> I'm not sure what you mean by that.
> --Best regards,
> .o. | Liege of Serenely Enlightened Majesty of
> o' \,=./ `o
> ..o | Computer Science, Michal "mina86"
> Nazarewicz (o o)
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