Print shell arguments in reverse order

Question

I am a bit stuck. My task is to print the arguments to my script in reverse order except the third and fourth.

What I have is this code:

#!/bin/bash

i=$#
for arg in "$@"
do
    case $i
    in
        3) ;;
        4) ;;
        *) eval echo "$i. Parameter: \$$i";;
    esac
    i=`expr $i - 1`
done

As I hate eval (greetings to PHP), I am looking for a solution without it but I am not able to find one.

How can I define the position of the argument dynamically?

PS: No its not a homework, I am learning shell for an exam so I try to solve old exams.

Answer 1

eval is the only portable way. Your script would be clearer if you explicitly looped on the index rather than the values (which you aren't using). Note that you don't need expr unless you want your script to run in antique Bourne shells; $((…)) arithmetic is in POSIX. Limit the use of eval to the smallest possible fragment; for example, don't use eval echo, assign the value to a temporary variable.

i=$#
while [ $i -gt 0 ]; do
  if [ $i -ne 3 ] && [ $i -ne 2 ]; then
    eval value=\$$i
    echo "Parameter $i is $value"
  fi
  i=$((i-1))
done

In bash, you can use ${!i} to mean the value of the parameter whose name is $i. This works when $i is either a named parameter or a number (denoting a positional parameter). While you're at it, you can make use of other bash convenience features.

for ((i=$#; i>0; i--)); do
  if ((i != 3 && i != 4)); then
    echo "Parameter $i is ${!i}"
  fi
done

Answer 2

This is a correct and non-dangerous use of eval. You fully control the content that you are evaling.

If it still gives you bad feelings, then if you don't care about portability, you can use Bash's ${!i} indirection syntax.

Answer 3

I keep a script reverse on my path that does this:

#!/bin/sh

if [ $# -gt 0 ]; then
    arg=$1
    shift
    reverse $@
    echo $arg
fi

Example usage:

$ reverse a b c
c
b
a