Modify IFS in bash while building and array

Question

I'm trying to build an array from 4 different arrays in bash with a custom IFS, can you lend me a hand please.

#!/bin/bash   
arr1=(1 2 3 4)
arr2=(1 2 3 4)  
arr3=(1 2 3 4)  
arr4=(1 2 3 4)
arr5=()
oldIFS=$IFS
IFS=\;
for i in ${!arr1[@]}; do  
    arr5+=($(echo ${arr1[i]} ${arr2[i]} ${arr3[i]} ${arr4[i]}))
done
IFS=$oldIFS
echo ${arr5[@]}

i what the output to be:

1 1 1 1;2 2 2 2;3 3 3 3;4 4 4 4 4 4

But it doesn't work the output is with normal ' '.

1 1 1 1 2 2 2 2 3 3 3 3 4 4 4 4 4 4

Any ideeas?

I tried IFS in different places: 1) In the for loop 2) Before arr5()

I tested it in the for loop and after IFS does change to ";" but it doesn't take effect in the array creation.

What is your end goal here? It would be easy to build that string without touching IFS. — Tom Fenech
– Tom Fenech, Commented Jan 10, 2019 at 13:53
yeah i know it's easy just add a ";" after ${arr4[i]}. But i get alot of spaces there and i have to do multiple operation on the string after to split it back in parts where i needit — Grim Edge
– Grim Edge, Commented Jan 10, 2019 at 13:59
also the 4 arrays are variable, so if i get and extra space or some other character in there all the operations after will need to be modified .. Basically i need to trust that the array delimiter will not change. — Grim Edge
– Grim Edge, Commented Jan 10, 2019 at 14:02

chepner · Accepted Answer · 2019-01-10 14:42:41Z

1

IFS is used during the expansion of ${arr5[*]}, not while creating arr5.

arr1=(1 2 3 4)
arr2=(1 2 3 4)  
arr3=(1 2 3 4)  
arr4=(1 2 3 4)
arr5=()
for i in ${!arr1[@]}; do  
    arr5+=("${arr1[i]}" "${arr2[i]}" "${arr3[i]}" "${arr4[i]}")
done
(IFS=";"; echo "${arr5[*]}")

Where possible, it's simpler to just change IFS in a subshell rather than try to save and restore its value manually. (Your attempt fails in the rare but possible case that IFS was unset to begin with.)

That said, if you just want the ;-delimited string and arr5 was a way to get there, just build the string directly:

for i in ${!arr1[@]}; do
  s+="${arr1[i]} ${arr2[i]} ${arr3[i]} ${arr4[i]};"
done
s=${s%;}  # Remove the last extraneous semicolon

edited Jan 10, 2019 at 14:42

answered Jan 10, 2019 at 14:36

chepner

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1 Comment

Grim Edge Over a year ago

Yep i used ${arr5[@]} instead of ${arr5[*]}. Now works like a charm. The second way has an extra space after ';" and i had to change "; " to ";" . Well now i understand IFS better. Thank you !

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