Numerical if statement in a Bash Script using $'s

Question

I have written a short Bash Script:

   for j in 0 1 2 3 4 5
   do
       for (( i=$j; i <= 5; i++ ));
       do
           if [ $(($(($i - $j)) > 1)) ]; then
               echo "True" 
           else
               echo  "False"
           fi
       done
   done

I expect this script to output a mixture of Trues and Falses however it only outputs Trues. I have checked and seen that $(($(($i - $j)) > 1)) is correctly producing 0s and 1s as it is supposed to but apparently the if statement is not registering these and always assumes "True".

Am I missing something in my code?

Thank you very much

James

$(($(($i - $j)) > 1)) wut? Try this (( i - j > 1))

user3442743
– user3442743

2014-06-23 14:59:08 +00:00
Commented Jun 23, 2014 at 14:59 — user3442743
– user3442743, Commented Jun 23, 2014 at 14:59
if (( i - j > 1 )); then

devnull
– devnull

2014-06-23 15:00:36 +00:00
Commented Jun 23, 2014 at 15:00 — devnull
– devnull, Commented Jun 23, 2014 at 15:00

dogbane · Accepted Answer · 2014-06-23 14:59:55Z

2

Your script is checking if [ 0 ] and if [ 1 ] which isn't what you think. It will always return true because it is checking that the length of the string 0 and 1 is not zero.

Change it to:

for j in {0..5}
do
    for (( i=j; i<=5; i++ ))
    do
        if (( i - j > 1 ))
        then
            echo True
        else
            echo False
        fi
    done
done

Note that the use of the $ prefix for variables within ((...)) is optional.

answered Jun 23, 2014 at 14:59

dogbane

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Comments

Paul Evans · Accepted Answer · 2014-06-23 15:09:20Z

0

You're using if incorrectly, you're testing for string length when you want an arithmetic comparison:

for j in {0..5}; do
    for i in $(seq $j 5); do
        if (( i - j > 1 )); then
            echo "True"
        else
            echo "False"
        fi
    done
done

answered Jun 23, 2014 at 15:09

Paul Evans

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