Create array from a list generated by grep

Question

I have a list of threads stored in a file. I can retrieve the threads name with a grep:

$ grep "#" stack.out
"MSC service thread 1-8" #20 prio=5 os_prio=0 tid=0x00007f473c045800 nid=0x7f8 waiting on condition [0x00007f4795216000]
"MSC service thread 1-7" #19 prio=5 os_prio=0 tid=0x00007f4740001000 nid=0x7f7 waiting on condition [0x00007f479531b000]
"MSC service thread 1-6" #18 prio=5 os_prio=0 tid=0x00007f4738001000 nid=0x7f4 waiting on condition [0x00007f479541c000]
. . .

As I'd need to manipulate the output of this list, I'd need to store these lines in an Array. I've found some examples suggesting this approach:

$ export my_array=( $(grep "#" stack.out) )

However if I browse through the Array I don't get the same result from my earlier grep:

$ printf '%s\n' "${my_array[@]}"

"MSC
service
thread
1-8"
#20
prio=5
os_prio=0
tid=0x00007f473c045800
nid=0x7f8
waiting
on
condition
[0x00007f4795216000]
"MSC
service
thread
1-7"
#19
prio=5
os_prio=0
tid=0x00007f4740001000
nid=0x7f7
waiting
on
condition
[0x00007f479531b000]

It seems that carriage returns are messing with my array assignment. Any help how to fix it ? Thanks !

gniourf_gniourf · Accepted Answer · 2016-12-28 15:15:47Z

5

This is an antipattern to populate an array! moreover, the export keyword is very likely wrong. Use a loop or mapfile instead:

With a loop:

my_array=()
while IFS= read -r line; do
    my_array+=( "$line" )
done < <(grep "#" stack.out)

or with mapfile (with Bash≥4):

mapfile -t my_array < <(grep "#" stack.out)

answered Dec 28, 2016 at 15:15

gniourf_gniourf

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1 Comment

Carla Over a year ago

Thanks for your help. Both replies were correct: I've chosen this one as you provided an additional way to do it.

Tom Fenech · Accepted Answer · 2016-12-28 15:16:26Z

5

The problem is that the output of grep is being split on all whitespace before the array is created, so every word becomes a separate element in the array.

Use mapfile to create your array instead:

mapfile -t my_array < <(grep "#" stack.out)

answered Dec 28, 2016 at 15:16

Tom Fenech

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1 Comment

Carla Over a year ago

Thank you very much for the explanation

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