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!
I think your loop is behaving funny because it should start with i = 1 and go to i = ${#*}. Expansion of ${@:0:1} is giving file1 and so is ${@:1:1}.
Try for (( i = 1; i <= ${#*}; i++ ))
$@ 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]}
You are iterating through the arguments array starting at 0, which is the command name. So, if you call your script like ./script.sh file1 file2, then:
${@} = array(
[0] = ./script,
[1] = file1,
[2] = file2
)
So you are getting:
modifiedNamed = array(
[0] = md5(./script),
[1] = md5(file1)
)
You should change your for loop indexes to:
for (( i = 1; i <= ${#*}; i++ )); do
Actually, you are getting the second element, but your loop puts the same value in [0] and in [1].
I'm not certain what you are trying to do but it would probably be easier to iterate over the arguments and then keep a separate counter for the array.
n=0
for i in "$@"; do
. . .
n=$(($n + 1))
done
