How to use cut to get the last field of /proc/pid/maps?

Question

I need to do some processing on the list of .so loaded by a process. I tried cut with space as a delimiter but I didn't succeed. How to correctly pipe the output of cat /proc/PID/maps into cut to get the last field ?

Does this answer your question? How to print only last column? --- I think all you need to know is how to get a last field. — FelixJN, Dec 08 '21 at 11:46

score 2 · Accepted Answer · answered Dec 08 '21 at 11:42

Getting the last field often is a bit tricky. Options are

awk '{print $NF}' /proc/PID/maps

(note that awk returns the last field with an entry, this will return 0 instead of an empty field for inode=0 entries)

or double-reversing a line with selecting the first field in between:

rev /proc/PID/maps | cut -d' ' -f1 | rev

Use grep to match characters except for space, then match till the end of line:

grep -o '[^ ]*$' /proc/PID/maps

score 1 · Answer 2 · answered Dec 08 '21 at 06:09

You can rely on maps’ fixed padding:

cut -c74- /proc/.../maps

(on 64-bit platforms).

Extracting the last field in all cases, e.g. with awk '{ print $NF }', is misleading with maps since lines can omit the backing file or use (“[heap]” etc.) entirely.

score -1 · Answer 3 · answered Dec 08 '21 at 05:50

-1

Cut is a bit limited. If you have in file, file: 12345 abcde 12438 you can parse it with cat file | awk '{print $2}' to get abcde which you can pipe or redirect.

answered Dec 08 '21 at 05:50

Brian

158

How to use cut to get the last field of /proc/pid/maps?

3 Answers3