bash for loop running more than the length of array

Question

#!/bin/bash 
arrck=
arrrlt=

if true; then
arrck+=("1")
arrrlt+=("one")
else
    echo "something"
fi 

for i in ${!arrck[*]} 
do
    echo "${arrck[i]} is ${arrrlt[i]}"
done


output 1-- 

 is
1 is one

output 2--tracing on  

 ./Building_block.sh
+ arrck=
+ arrrlt=
+ true
+ arrck+=("1")
+ arrrlt+=("one")
+ for i in '${!arrck[*]}'
+ echo ' is '
 is
+ for i in '${!arrck[*]}'
+ echo '1 is one'
1 is one

Why is the loop running with no value assigned to i ? i can say it tried to execute echo "${arrck[]} is ${arrrlt[]}" and picking the blank space as value.

after commenting the declaration, it is running as expected just 1 time( length of the array) ``` arrck= arrrlt= ``` — Pradeep
– Pradeep, Commented Aug 29, 2019 at 20:46
There are 2 cells/elements in the array: echo ${!arrck[@]} ==> 0 1 — markp-fuso
– markp-fuso, Commented Aug 29, 2019 at 20:47
@chepner but they do create an initial cell with index 0 for the 2 arrays arrck and arrrlt — markp-fuso
– markp-fuso, Commented Aug 29, 2019 at 21:03

markp-fuso · Accepted Answer · 2019-08-29 22:26:27Z

2

When you assign a value to a variable you are actually assigning a value to cell/position 0 in an array, eg:

$ x=5

$ echo ${x}
5

$ echo ${!x[@]}
0

$ echo ${x[0]}
5

$ echo ${x[@]}
5

In your example the first set of commands assign the empty string to index position '0' of the 2 arrays arrck[] and arrrlt[]:

$ arrck=
$ arrrlt=

$ echo "${!arrck[@]} : ${!arrrlt[@]}"
0 : 0

The if/then/else block then appends a second set of values to your arrays with indexes 1 and values of 1 and one.

The for loop then loops through the 2 available indices for the arrck[] array, namely 0 and 1.

What you probably want to do is start out by dropping/deallocating your arrays (as opposed to creating cell 0), eg:

$ unset arrck
$ unset arrrlt
$ echo "${!arrck[@]:-undefined} : ${!arrrlt[@]:-undefined}"
undefined : undefined

edited Aug 29, 2019 at 22:26

answered Aug 29, 2019 at 21:01

markp-fuso

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