Hi,<br><br>On further investigation I noticed that the size of /proc/kcore in the guest, is initially the size of memory<br>specified while booting the guest (256MB in this case).<br><br>After executing a program which I have written to read the contents of /proc/kcore using the open() system call, <br>
the size of the file increases to 1GB! and I am able to read upto 1GB of data!<br>(which is the size of the actual RAM on my machine)<br><br>Could someone please explain this? Does the read here also give the contents of the host memory?<br>
<br>Thank you so much in advance,<br>Sujit<br><br><div class="gmail_quote">On Thu, Apr 10, 2008 at 9:51 PM, Sujit Sanjeev <<a href="mailto:sujit771@gmail.com">sujit771@gmail.com</a>> wrote:<br><blockquote class="gmail_quote" style="border-left: 1px solid rgb(204, 204, 204); margin: 0pt 0pt 0pt 0.8ex; padding-left: 1ex;">
Hi,<br><br>On a machine with 1GB RAM, I booted an lguest guest with 256MB memory as follows:<br>./lguest --block='blockfile' --initrd=/boot/initrd.img-2.6.23-rc9 256m /boot/vmlinuz-2.6.23-rc9 root=/dev/lgba rw single<br>
<br>The size of /proc/kcore file in the host obtained using 'ls -la' is 1073422336, <br>while the same command executed within the guest shows the size as 1069228032. <br><br>Since the size of /proc/kcore represents the size of the physical memory (RAM), and I boot the guest with just 256MB <br>
of memory, shouldn't the size of the kcore file under guest be 256MB instead of around 1GB?<br><br>Will the contents of /proc/kcore in the guest also reflect the contents of the host memory?<br><br>Please let me know.<br>
<br>Thanks!<br><font color="#888888">Sujit<br><br><br>
</font></blockquote></div><br>