loop in bash does not perform integer comparison properly

Question

I have below code

i=0
for s in / - \\ \| 
do 
  printf "\rWaiting for application start to finish $i $s"
  sleep 1
  ((i++))
  if [[ $i -gt 30 ]]
  then 
    break 
  fi
done

The loop always ends after 3 iterations. Any reason as why?

It loops up to 3 (4 iterations) because the arguments in the for loop are 4: '/', '-', '\' and '|', isn't it? — Joao Donasolo
– Joao Donasolo, Commented Dec 8, 2021 at 13:19
Yeah, I think that' it. I've added more arguments to the array and it loops more times. Not sure how to solve it though — Joao Donasolo
– Joao Donasolo, Commented Dec 8, 2021 at 13:20

M. Nejat Aydin · Accepted Answer · 2021-12-08 16:42:48Z

1

This might be what you are looking for:

#!/bin/bash

a=('/' '-' '\' '|')
for ((i = 0; i < 30; ++i)); do
    printf '\rWaiting for application start to finish %d %s' \
            "$i" "${a[i%4]}"
    sleep 1
done
echo

answered Dec 8, 2021 at 16:42

M. Nejat Aydin

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Comments

BTR Naidu · Accepted Answer · 2021-12-08 13:24:27Z

0

My mistake. The for loop which has three arguments is exiting instead of the if condition failing.

answered Dec 8, 2021 at 13:24

BTR Naidu

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3 Comments

ashish_k Over a year ago

Why this has been posted as an answer?

chepner Over a year ago

@User123 Because it's the answer to the question?

BTR Naidu Over a year ago

The behavior was right and should not have been a question at all. Therefore the answer.

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