Re: how to interpret the output of "cat /proc/1/maps"?

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On Tue, 27 Mar 2007, Daniel Cheng wrote:

> Robert P. J. Day wrote:
> >   as i understand it, on my x86 system, the full 32-bit address space
> > is divided in a 3G/1G split -- the first 3G is for user space, while
> > the last 1G is the kernel address space, which consists mainly of
> > "logical" addresses that map, via a fixed offset, down to the
> > corresponding physical memory.
> >
> >   so how do i interpret the following (first part of) the output from
> > the command:
> >
> > # cat /proc/1/maps
> > 001d4000-001d5000 r-xp 001d4000 00:00 0          [vdso]
> > 00940000-00959000 r-xp 00000000 fd:00 166390     /lib/ld-2.5.so
> > 00959000-0095a000 r-xp 00018000 fd:00 166390     /lib/ld-2.5.so
> > 0095a000-0095b000 rwxp 00019000 fd:00 166390     /lib/ld-2.5.so
> > ...
> >
> > those start-end addresses are described as "virtual" addresses, but
> > they don't fall in the normal kernel address space, which i always
> > took to be in the interval [3G-4G).
> >
>
> /lib/ld-2.5.so is a userspace library, this is in the user address space

and only *seconds* after i posted this did i realize how silly my
question was.  duh.

rday
-- 
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Robert P. J. Day
Linux Consulting, Training and Annoying Kernel Pedantry
Waterloo, Ontario, CANADA

http://fsdev.net/wiki/index.php?title=Main_Page
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