CODEWITHSUNDEEP
linuxshell
escargot agile
escargot agile

Reputation: 22379

Boolean return value from shell function behaving strangely

I've defined this function (with some debugging printouts):

listcontains() {
  for item in $1
  do
    echo $2 = $item ?
    if [ "$2" == "$item" ]; then
      echo YES!
      return 1;
    fi
  done
  echo NO!
  return 0
}

And I'm using it like this:

list="1 2"
if listcontains "$list" 3; then echo Y; else echo N; fi

I'm expecting N, but the result is Y, and the debug output makes it even more strange:

3 = 1 ?
3 = 2 ?
NO!
Y

What did I do wrong?

Upvotes: 0

Views: 45

Answers (1)

kamituel
kamituel

Reputation: 35950

In shell, you use 0 to denote true (and some positive number, i.e. 1, to denote false). It's the same notion as for scripts exit values - in unix world, 0 means success.

So to fix, just reverse return values:

listcontains() {
  for item in $1
  do
    echo $2 = $item ?
    if [ "$2" == "$item" ]; then
      echo YES!
      return 0;
    fi
  done
  echo NO!
  return 1
}

Example:

$ list="1 2"
$ if listcontains "$list" 3; then echo Y; else echo N; fi
3 = 1 ?
3 = 2 ?
NO!
N

$ list="1 3"
$ if listcontains "$list" 3; then echo Y; else echo N; fi
3 = 1 ?
3 = 3 ?
YES!
Y

Upvotes: 1

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