UNIX for Dummies Questions & Answers

Hey all,

How do I make a script print its arguments in reverse order?

Thanks

Code:

#!/bin/ksh

n=$(($#+1))
while ((n=n-1))
do
  echo $n
done

A minor fix ..... eval is needed.

Code:

#!/bin/ksh

n=$(($#+1))
while ((--n))
do
   eval echo \$${n}
done

Sorry this isn't working, it runs the script infinitely

Hi,

I am trying to write a script to sum the series of integers passed as command line arguments to the script.

Code:

#!/bin/bash

while ((n=n+$1)); do
    shift
done
echo $n

the n gives the correct value, ie sum of all the passed positional parameters, but with an error message n=n+: syntax error: operand expected (error token is "+"). I guess, this error message is due to the NULL value after shifting all the positional arguments. Now, my question is, how to make this script work proper? Also, I want to keep this script as compact as it is now.

Code:

while (($1)); do
   ((n+=$1))
   shift
done
echo $n