Delete positional parameters in Bash?

Question

You can skip positional parameters with shift but can you delete positional parameters by passing the position?

x(){ CODE; echo \"$@\"; }; x 1 2         


        

Answer 1

You can call set and reset the positional paramaters at any time for example

function q {
echo ${@}
set $2 $3 $4
echo ${@}
set $4
echo ${@}
}

q 1 2 3 4

then slice out what you dont want from the array, the below code does that... not sure if its the best way to do it though, was on stack looking for a better way ; )

#!/bin/bash


q=( one two three four five )

echo -e "
  (remove) { [:range:] } <- [:list:]
                | [:range:] => return list with range removed range is in the form of [:digit:]-[:digit:]
"

function remove {
  if [[ $1 =~ ([[:digit:]])(-([[:digit:]]))?   ]]; then
    from=${BASH_REMATCH[1]}
    to=${BASH_REMATCH[3]}
  else
    echo bad range
  fi;shift
  array=( ${@} )
  local start=${array[@]::${from}}
  local rest
  [ -n "$to" ] && rest=${array[@]:((${to}+1))}  || rest=${array[@]:((${from}+1))}
  echo ${start[@]} ${rest[@]}
}

q=( `remove 1 ${q[*]}` )
echo ${q[@]}