bash Command line substitution in for loop

Question

I have a test script test.sh where I am trying to print out command line arguments.to the script but the following does not work as expected

`#!/bin/bash
 for((i=1;i<"$#";i++)) do
        printf "Position %s of argumentArray has  %s \n", $i  $(($i))
  done`

( my idea was that the (()) will do the mathematical evaluation resulting in $1 $2 etc.) Neither does

for((i=1;i<"$#";i++)) do
printf "Position %s of argumentArray has  %s \n", $i  $"$( eval echo $i )"
done

both gives as out put when run as follows

./test.sh  first second third 

Position 1 of argumentArray has  1 
Position 1 of argumentArray has  2
Position 1 of argumentArray has  3

instead of

Position 1 of argumentArray has  first 
Position 1 of argumentArray has  second
Position 1 of argumentArray has  third

I face the same problem with

for((i=1;i<="$#";i++)) 
  do
      case "$($i)" in
       .......

case evaluates to 1 ,2 3 etc insted of the actual parameter passed in.

Please help me in understanding where I am going wrong.

Gordon Davisson · Accepted Answer · 2011-10-22 15:24:49Z

1

You can use indirect expansion to do this fairly easily:

for((i=1;i<=$#;i++)) do
    printf "Position %s of argumentArray has  %s \n" $i  "${!i}"
done

I also fixed some minor problems in the above: the loop end condition should be i<=$# to print the last arg, there shouldn't be a comma after the format string for printf, and there should be double-quotes around the argument ("${!i}") in case it has any spaces or other funny characters.

answered Oct 22, 2011 at 15:24

Gordon Davisson

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Ray Toal · Accepted Answer · 2011-10-22 05:27:32Z

0

The commandline arguments can be accessed directly but if you want them by position you can do this:

arguments=($@)
for ((i = 0; i < ${#arguments[@]}; i++)); do
    echo "arguments[$i] = ${arguments[$i]}"
done

A run of the script:

$ ./args.sh first second third
arguments[0] = first
arguments[1] = second
arguments[2] = third

ADDENDUM

Information on Bash arrays:

answered Oct 22, 2011 at 5:27

Ray Toal

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1 Comment

Sibi JV Over a year ago

Thanks it works. Forgot about arays altogether I went on to do. ` for((i=1;i<"$#";i++)) do str=$( echo $( eval echo \$$( echo $i ))) echo $str printf "Position %s of batchProcessArray has %s \n", $i "$str" done ` but your answer is the best solution. thanks once again.

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