Wrong behavior of xargs

Question

I was trying to solve this issue using find + xargs but I stuck with another issue

I am try to increasing a count using ((a++)) but not working . I have tried couple of combination of counting a value eg. let a++ a=$[ $a + 1] and so on..

See below Output

rax@ubuntu:~# find ./test/ -mindepth 1 | xargs -I{} -n1 bash -xc "((a++)) ; echo $a {}"
+ (( a++ ))
+ echo 0 ./test/abc1
0 ./test/abc1
+ (( a++ ))
+ echo 0 ./test/abc1/abc
0 ./test/abc1/abc
+ (( a++ ))
+ echo 0 ./test/abc2
0 ./test/abc2
+ (( a++ ))
+ echo 0 ./test/abc3
0 ./test/abc3
+ (( a++ ))
+ echo 0 ./test/abcparent
0 ./test/abcparent

EDIT :- xargs did all in same shell

eg. with pid

rax@ubuntu:~# a=0
rax@ubuntu:~# find ./test/ -mindepth 1 | xargs -L2 -I{}  bash -xc "echo $a {} ;((a++)) ; echo $a $$"
+ echo 0 ./test/abc1
0 ./test/abc1
+ (( a++ ))
+ echo 0 1314
0 1314
+ echo 0 ./test/abc1/abc
0 ./test/abc1/abc
+ (( a++ ))
+ echo 0 1314
0 1314
+ echo 0 ./test/abc2
0 ./test/abc2
+ (( a++ ))
+ echo 0 1314
0 1314
+ echo 0 ./test/abc3
0 ./test/abc3
+ (( a++ ))
+ echo 0 1314
0 1314
+ echo 0 ./test/abcparent
0 ./test/abcparent
+ (( a++ ))
+ echo 0 1314
0 1314

Shell variables do not maintain their values across different invocations of the shell, so what you're doing will never work. Try piping find's output to wc -l to count the lines of output. — Kyle Jones, Apr 04 '13 at 04:28
I used that in past, it was working , but I thing I forget something — Rahul Patil, Apr 04 '13 at 04:36
@KyleJones I am doing count and print in same shell invocation — Rahul Patil, Apr 04 '13 at 04:37
@Rahul: Kyle is right, and no you're not doing everything in the same shell: xargs -n1 will invoke one bash shell for each line. — Mat, Apr 04 '13 at 04:47
@Rahul: -n1 makes all the difference. (Also note that even in the second case, if there is too much input, you'll get multiple shells.) — Mat, Apr 05 '13 at 05:30
Multiple shell means different pid , but it's showing same pid using echo $$ — Rahul Patil, Apr 05 '13 at 07:27
@Mat & Kyle Jones thanks , You are Correct , I was Wrong . Yes, it was single quotes mistake — Rahul Patil, Apr 05 '13 at 10:00

zorlem · Answer 1 · 2013-04-05T09:59:44.473

The problem is that your variables are being expanded by bash before find is executed. Because of this $a is expanded to 0, $$ is expanded to the PID of your (login) shell, so the command that is executed by bash -xc is actually:

echo 0 {} ;((a++)) ; echo 0 1314

Here {} is replaced by xargs with the result from find and the value of $a is postincremented, but never printed. Try escaping the variables (as in \$a), or even better, putting the bash -xc arguments in single quotes (') instead of double quotes (") to see that it works.

You will also notice that the other commenters are correct, xargs -n1 (or with -I{}) executes a new shell for each argument, therefore incrementing $a will not work - the incremented value will not be visible to the next invocation of the bash -xc command.

Yes, it was single quotes mistake – Rahul Patil Apr 05 '13 at 09:58 — Rahul Patil, Apr 05 '13 at 09:58

Wrong behavior of xargs

1 Answers1