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 > what am i misunderstanding here? > > rday > -- This space was intended to be left blank. -- To unsubscribe from this list: send an email with "unsubscribe kernelnewbies" to ecartis@xxxxxxxxxxxx Please read the FAQ at http://kernelnewbies.org/FAQ
