Using bash store a value to an array in a while loop failing

graysky · 2016-10-07 12:16:30

Goal: have the script store a value in a growing array.

Expected result: Create an array with 5 members, in this test case they are all the same value of 123 but they should be stored once per pass through the loop.

Actual result: An array with only 1 member is defined.

#!/bin/bash
x=0
limit=5
while [[ "$x" -lt "$limit" ]]; do
  diff=123
  Arr=["$x"]="$diff"
  echo "Index $x has value ${Arr[$x]}"
  x=$(( x + 1 ))
done

echo "${#Arr[@]}"

Result:

% ./arr
Index 0 has value [0]=123
Index 1 has value
Index 2 has value
Index 3 has value
Index 4 has value
1

Why am I not getting 5 members?

Last edited by graysky (2016-10-07 12:31:16)

lolilolicon · 2016-10-07 12:30:56

  Arr=["$x"]="$diff"

dude, remove the first =

graysky · 2016-10-07 12:32:28

lolilolicon wrote:

  Arr=["$x"]="$diff"
dude, remove the first =

<< HEADSMACK >> Thank you! Works as expected now.

lolilolicon · 2016-10-07 12:36:51

"Better" version:

#!/bin/bash

declare -a Arr
limit=5
for ((x=0; x<limit; ++x)); do
  diff=123
  Arr[x]=$diff
  echo "Index $x has value ${Arr[x]}"
done

Note the indexed array subscript [x], like variables in (()), they undergo arithmetic expansion, so no need to write it as [$x].

Last edited by lolilolicon (2016-10-07 12:42:21)

Arch Linux

#1 2016-10-07 12:16:30

Using bash store a value to an array in a while loop failing

#2 2016-10-07 12:30:56

Re: Using bash store a value to an array in a while loop failing

#3 2016-10-07 12:32:28

Re: Using bash store a value to an array in a while loop failing

#4 2016-10-07 12:36:51

Re: Using bash store a value to an array in a while loop failing

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