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I have an array with some elements and in some function the following:

echo -n "Select the option: "
read option

echo "option: '${option}'"
echo "option: '$option'"

if [[ $option -ge 1 && option -le ${#array[@]} ]]; then
 echo "go go go"
 ...
else
 ...
fi

About the input - if a number higher than the array's length/size is written then the else block is executed as expected - same for a negative number - or even 0. Until here all is expected

But if a simple character as s, d is written then the if block is executed and it is not expected.

select the option: d
option: 'd'
option: 'd'
go go go

According with my understanding -ge and -le is applied for numbers - so it should fail if a character is used how a number comparison - it should go straight to the else block

How to fix this?

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4
  • 1
    Check that the option is a number. $option =~ ^(0|[1-9][0-9]*)$ Commented Feb 11, 2022 at 18:24
  • @choroba yes, I did do a research about that, but wondering about this behavior. Commented Feb 11, 2022 at 21:23
  • 1
    You left out $ after && in if [[ $option -ge 1 && option -le ${#array[@]} ]]; then. Intentionally? Why not if [[ $option -ge 1 && $option -le ${#array[@]} ]]; then? Commented Feb 16, 2022 at 16:04
  • @JohnSmith good catch - seems I was very tired I did not realise that. Furthermore interesting an error wasn't risen (variable not declared) Commented Feb 16, 2022 at 16:23

1 Answer 1

Reset to default
2

What's happened here is (for you) an unexpected evaluation. For reference here's your original expression

[[ $option -ge 1 && option -le ${#array[@]} ]]

First, $option is evaluated. In this case it's d. This is then substituted into the [[ … ]] expression to get this:

[[ d -ge 1 && option -le 12 ]]

This is evaluated. BUT because [[ … ]] uses bash's expression evaluation the d and option are treated as names of variables so there's another level of expansion (being aware that implicitly d=0):

[[ 0 -gt 1 && d -le 12 ]]

The first part fails, so the second (erroneous) expression is skipped.

There are a couple of solutions here that you can consider:

  1. Instead of using [[ … ]] use the older but acceptable [ … ]. This doesn't perform this double expansion. But that won't catch the "non-numeric" error when $option isn't numeric, so you'll still get an error

    [ "$option" -ge 1 ] && [ "$option" -le ${#array[@]} ]

  2. Protect the expression comparison by validating that the characters in the string are only numeric.

    [[ "$option" =~ ^(0|([1-9][0-9]*))$ && "$option" -ge 1 && "$option" -le "${#array[@]}" ]]
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  • if The first part fails, so the second (erroneous) expression is skipped. - it that part fails so why enter in the if statement? - in some way confuse that part ... pls consider to update it. Commented Feb 11, 2022 at 21:28
  • I thought that either "$option" or "${option}" was for the scenario with the value is consider a String Commented Feb 11, 2022 at 21:33
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    @ManuelJordan yes indeed thank you for that. (Did you know you could have edited it yourself, and you'd have been welcome to do so?) Commented Feb 11, 2022 at 21:54
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    @ManuelJordan The thing about those literal strings (d, option) getting processed further is particular to arithmetic contexts. It's the thing allows doing a=3; b=4; c=$((a+b)) to get the value 7 in c. Just that if the variable is unset or empty, it's taken as zero. Within [[ .. ]], the numeric operators evaluate their operands as arithmetic expressions, so that applies. But if you'd used [ .. ] instead, it would be different: foo=x; [ "$foo" -ge 1 ] would give an error (in Bash and most other shells, that is (but not in ksh)). Commented Feb 11, 2022 at 22:20
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    Double quoting variables in [[ "$x" -eq 1 ]] doesn't prevent recursive evaluation, e.g. a=1; x=a; [[ "$x" -eq 1 ]] && echo yes prints "yes". Commented Feb 11, 2022 at 22:38

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