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From: Paolo Bonzini <pbonzini@redhat.com>
To: 男6龙双A <5019096@qq.com>, qemu-devel <qemu-devel@nongnu.org>
Subject: Re: [Qemu-devel] why guest memory size not equal to my setting?
Date: Mon, 12 Jun 2017 14:46:56 +0200	[thread overview]
Message-ID: <cc7b29f5-b6d1-8ac9-187d-abe77b6b0beb@redhat.com> (raw)
In-Reply-To: <tencent_7AD3D98771C01E340454EDB8@qq.com>

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On 10/06/2017 15:54, ÄÐ6ÁúË«A wrote:
> Hello Qemu-devel,
> 
> Recently I'm trying to study vm memory allocation on qemu-kvm environment.
> I found some interesting here:
> 
> I have create a 8GB(8388608 k) memory guest using Centos 7. but when I using dmesg to show the init memory,
> it was 9437184 k,around 9216MB. I would like to know the gap?

> I know qemu will init two memory region:
> system_memory = g_malloc(sizeof(*system_memory));
> and system_io = g_malloc(sizeof(*system_io));
> 
> Does those gap point to the region of system_io ?

No, these are two different address spaces.  Guest RAM is allocated by
memory_region_allocate_system_memory as a single region of the size you
specified.

The guest memory map doesn't place all the memory contiguously.  For the
"pc" machine type, from 3GB to 4GB there is a hole for memory mapped
registers of PCI devices.  Linux reports this hole as "absent" memory:

[    0.000000] Memory: 7372140k/9437184k available (6244k kernel code,
1049100k absent, 1015944k reserved, 4178k data, 1604k init)

Note that Linux 3.11 or newer doesn't report absent pages anymore
(commit 46a841329a6c, "mm/x86: prepare for removing num_physpages and
simplify mem_init()", 2013-07-03).

Thanks,

Paolo

      reply	other threads:[~2017-06-12 12:47 UTC|newest]

Thread overview: 2+ messages / expand[flat|nested]  mbox.gz  Atom feed  top
2017-06-10 13:54 [Qemu-devel] why guest memory size not equal to my setting? =?gb18030?B?xNA2wfrLq0E=?=
2017-06-12 12:46 ` Paolo Bonzini [this message]

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