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I'm having trouble assigning values to a specific bash index, but apparently only when the index variable is set using a while read loop.

Taking this code as a test example:

#!/bin/bash

read -d '' TESTINPUT << 'EOF'
1,100
2,200
8,300
EOF

declare -A ARRAY
echo "$TESTINPUT"| while read _l; do
    i=$(cut -d, -f1 <<< $_l)
    j=$(expr $i + 0)
    value=$(cut -d, -f2 <<< $_l)
    ARRAY[$j]=$value
done

for i in {4..6}; do
    ARRAY[$i]=$i
done

for i in {1..10}; do
    echo "$i ${ARRAY[$i]}"
done

The output clearly shows that in the case of the while loop, the array variables don't get set, while with the for loop, using a range of {4..6} doesn't seem to have any issue.

$ ./test_array.sh
1 
2 
3 
4 4
5 5
6 6
7 
8 
9 
10

Notice I also tried to convert the index variable to an integer using

j=$(expr $i + 0)

But that doesn't seem to work either.

Any ideas?

steinocaro
  • 1
  • yeah, it's the pipe. see the linked Q's. `shopt -s lastpipe` or rearrange that to not use a pipe. `echo foo | blah` could be replaced with `blah <<< foo` – ilkkachu Dec 04 '20 at 10:45
  • Other things: you could use `IFS=, read -r a b` to split the fields without running `cut`. Also, Bash doesn't really have types for variables, so `$(expr $i + 0)` probably doesn't do anything (except it would remove leading zeroes). If you have just integer indexes, you could use a regular, non-associative array, `declare -a` instead of `-A`. Though note that leading zeroes would make Bash interpret the numbers as octal. – ilkkachu Dec 04 '20 at 10:50
  • Thanks. I finally noticed it. I solved the problem. – steinocaro Dec 04 '20 at 11:07

0 Answers0