Convert command line arguments into an array in Bash
Actually your command line arguments are practically like an array already. At least, you can treat the $@
variable much like an array. That said, you can convert it into an actual array like this:
myArray=( "$@" )
If you just want to type some arguments and feed them into the $@
value, use set
:
$ set -- apple banana "kiwi fruit"
$ echo "$#"
3
$ echo "$@"
apple banana kiwi fruit
Understanding how to use the argument structure is particularly useful in POSIX sh, which has nothing else like an array.
Maybe this can help:
myArray=("$@")
also you can iterate over arguments by omitting 'in':
for arg; do
echo "$arg"
done
will be equivalent
for arg in "${myArray[@]}"; do
echo "$arg"
done
Actually the list of parameters could be accessed with $1 $2 ...
etc.
Which is exactly equivalent to:
${!i}
So, the list of parameters could be changed with set,
and ${!i}
is the correct way to access them:
$ set -- aa bb cc dd 55 ff gg hh ii jjj kkk lll
$ for ((i=0;i<=$#;i++)); do echo "$#" "$i" "${!i}"; done
12 1 aa
12 2 bb
12 3 cc
12 4 dd
12 5 55
12 6 ff
12 7 gg
12 8 hh
12 9 ii
12 10 jjj
12 11 kkk
12 12 lll
For your specific case, this could be used (without the need for arrays), to set the list of arguments when none was given:
if [ "$#" -eq 0 ]; then
set -- defaultarg1 defaultarg2
fi
which translates to this even simpler expression:
[ "$#" == "0" ] && set -- defaultarg1 defaultarg2
Here is another usage :
#!/bin/bash
array=( "$@" )
arraylength=${#array[@]}
for (( i=0; i<${arraylength}; i++ ));
do
echo "${array[$i]}"
done