I want to generate System.map from vmlinuz,cause most of machines don't have the file System.map.In fact,vmlinuz are compressed to vmlinuz or bzImage.
It's any tool or script can do this?
I tried:
dd if=/boot/vmlinuz skip=`grep -a -b -o -m 1 -e $'\x1f\x8b\x08\x00' /boot/vmlinuz | cut -d: -f 1` bs=1 | zcat > /tmp/vmlinux
It was failed:
zcat: stdin: not in gzip format
32769+0 records in
32768+0 records out
To extract the uncompressed kernel from the kernel image, you can use the extract-vmlinux
script from the scripts
directory in the kernel tree (available at least in kernel version 3.5) (if you get an error like
mktemp: Cannot create temp file /tmp/vmlinux-XXX: Invalid argument
you need to replace $(mktemp /tmp/vmlinux-XXX)
by $(mktemp /tmp/vmlinux-XXXXXX)
in the script). The command is /path/to/kernel/tree/scripts/extract-vmlinux <kernel image> >vmlinux
.
If the extracted kernel binary contains symbol information, you should¹ be able to create the System.map
file using the mksysmap
script from the same subdirectory. The command here is NM=nm /path/to/kernel/tree/scripts/mksysmap vmlinux System.map
.
¹ The kernel images shipped with my distribution seem to be stripped, so the script was not able to get the symbols.
As Abrixas2 wrote, you will need a kernel image with symbol information in order to create System.map files and a packed vmlinuz image is not likely to have symbols in it. I can, however, verify that the script in your original post works with '-e' replaced with '-P' and '$' dropped, i.e.,
$ dd if=vmlinuz-3.8.0-19-generic skip=`grep -a -b -o -m 1 -P '\x1f\x8b\x08\x00' vmlinuz-3.8.0-19-generic | cut -d: -f 1` bs=1 | zcat > /tmp/vmlinux
gzip: stdin: decompression OK, trailing garbage ignored
I'm on ubuntu linux.
you can change $'\037\213\010\000'
to "$(echo '\037\213\010\000')"
in sh
bash$ N=$(grep -abo -m1 $'\037\213\010\000' vmlinuz-4.13.0-37-generic | awk -F: '{print $1+1}') &&
tail -c +$N vmlinuz-4.13.0-37-generic | gzip -d > /tmp/vmlinuz