Global variable is reset if loop send output to pipe

Question

According the bash(1) man pages, when I run the following:

set -e
x=2
echo Start $x
while [ $((x--)) -gt 0 ]; do echo Loop $x; done | cat
echo End $x

The output will be:

Start 2
Loop 1
Loop 0
End 2

After the loop (runs as a subshell) the variable x reset to 2. But if I remove the pipe the x will be updated:

Start 2
Loop 1
Loop 0
End -1

I need to change the x but, I need the pipe too. Any idea how to get around this problem?

@devnull, I think it's just there for illustration: that adding a pipe to anything will mean the while loop is run in a subshell, and any changes to variable $x are lost when the subshell ends. — glenn jackman
– glenn jackman, Commented Jun 4, 2013 at 15:09
Yes, The real loop is long. Inside the loop I have some lines like: ping -c1 Machine || STOP="YES" At the end of the loop I grep(1) out the lines of "icmp_seq=" — Udi
– Udi, Commented Jun 5, 2013 at 6:41

chepner · Accepted Answer · 2013-06-04 15:05:21Z

3

bash always (at least as of 4.2) runs all non-rightmost parts of a pipeline in a subshell. If the value of x needs to change in the calling shell, you must rewrite your code to avoid the pipeline.

One horrible-looking example:

# If you commit to one bash feature, may as well commit to them all:
#   Arithmetic compound: (( x-- > 0 ))
#   Process substitution: > >( cat )
while (( x-- > 0 )); do echo Loop $x; done > >( cat )

answered Jun 4, 2013 at 15:05

chepner

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2 Comments

glenn jackman Over a year ago

@udi, read about process substitution here

Udi Over a year ago

@chepner, Thank you very much! I had a little typo and now it works.

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