bash - Parameter as reference? -

i trying make function swap() this

swap(){ #swap $1 , $2 here } 

what want swap array let have

array[0]=12 array[1]=45 array[2]=99  swap $array[0] $array[1] 

so want swap array[0] becomes 45 , array[1] becomes 12 after swap function. thinking of doing swapping referencing $array[0] (such pointer in c) , $array[1] changed. found command eval , upvar seriously, don't understand thing. pretty new shell scripting , of documentation out there confused me lot.

bash arrays little tricky. clearest thing can write function take three arguments, array name , 2 indices swap. requires treat array global variable; cannot pass entire array single object in bash.

(this require bash 4 or later, introduced -g flag declare.)

swap () {     local name=$1     local a=$2     local b=$3      local aname=$name[$a]     local bname=$name[$b]     local tmp=${!aname}    # e.g. tmp=${array[$a]}     declare -g "${aname}=${!bname}"   # e.g. array[$a]=${array[$b]}     declare -g "${bname}=$tmp"        # e.g. array[$b]=$tmp } 

bash 4.3 introduces namerefs (declare -n) simplify letting declare local array acts alias global array.

swap () {   declare -n arr=$1   local a=$2 b=$3   local tmp=${arr[$a]}   arr[$a]=${arr[$b]}   arr[$b]=$tmp 

}