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). what am i misunderstanding here? rday -- To unsubscribe from this list: send an email with "unsubscribe kernelnewbies" to ecartis@xxxxxxxxxxxx Please read the FAQ at http://kernelnewbies.org/FAQ
