CODEWITHSUNDEEP
bash
Ivan V.
Ivan V.

Reputation: 8081

Confused about bash arithmetic comparison

I've seen other answers about bash integer checks and comparisons, however the results I'm getting are very confusing to me. Suppose I have this script:

if [[ $1 -eq $1 ]] ;then
  echo "number"
else
  echo "not number"
fi

if (( $1 >= 0 )) ;then
  echo "number"
else
  echo "not number"
fi

If I pass a string for parameter one , I get back "number".

Upvotes: 0

Views: 101

Answers (1)

choroba
choroba

Reputation: 241908

That's because string is understood as a variable whose value is 0, and 0 >= 0 is true. Try with > (but it will report 0 as not number - but it already misclassifies all negative integers).

Cf:

a=1
b=a
x=b
(( x > 0 )) && echo 1
a=0
(( x > 0 )) || echo 0

or even

$ a=x
$ x=a
$ (( x > 0 ))
bash: ((: a: expression recursion level exceeded (error token is "a")

Bash tries hard to resolve the variable:

(
    for l in a{1..1023} ; do
        printf "$l\n$l="
    done
    echo 1
    echo '((a1>0))'
) | tail -n+2 | bash

Upvotes: 2

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