Question

I have a problem with my first bash script. I fill an array in for loop and when I try to get an item from it I always get the first element.

for (( i = 0; i < ${#*}; i++ )); do
hash=$(md5 -q ${@:$i:1})
modifiedNames[$i]=${@:$i:1}$hash 
done

echo ${modifiedNames[1]}

for instance if I call my script like this: ./script.sh file1 file2 i get file1[file1hash]

Thanks in advance!

Answer 1

$@ isn’t a real array – it’s a shell “Special Parameter”, and you need to be a bit more careful with it than other arrays.

The reason for the behaviour you’re seeing is that the exact behaviour of the ${parameter:length:offset} syntax is special-cased when parameter is @, and the behaviour is not consistent with the behaviour you’d get if @ was a real array.

Here’s the relevant documentation (bold emphasis mine):

${parameter:offset:length}

… If parameter is @, the result is length positional parameters beginning at offset. If parameter is @, the result is length positional parameters beginning at offset. …

The positional parameters are $0, $1, $2, …, so with this syntax it’s behaving as if $@ contained the script name ($0) as well as the parameters to the script ($1, $2, …). This is inconsistent with "$@" expanding to "$1" "$2" ..., but that’s life.

You should be able to simplify things (and fix the script) by making a new array instead of using $@ directly, i.e.

new_array=("$@")
for (( i = 0; i < ${#new_array}; i++ )); do
    hash=$(md5 -q ${new_array[@]:$i:1})
    modifiedNames[$i]=${new_array[@]:$i:1}$hash 
done

echo ${modifiedNames[1]}