Shell Script output odd

Question

I am new to Shell Script programming. I am attempting to make a program that outputs only gives back the odd arguments. For example, "I am a robot" should give "I a" only.

I thought I could use the shift command to through the arguments along with while and if statements, but I am unsure of how to convey a division by 2 with a remainder of 1 in the square brackets in the if statement ([])

#!/bin/sh
X=$1
shift
while [ $# -gt 0 ]; do
  if [ "HELP HERE" ]; then
     X=$1
     echo $X
  fi
  shift
 done
echo $X

nneonneo · Accepted Answer · 2012-10-05 07:00:47Z

3

How about just shifting twice?

while [ $# -gt 0 ]; do
    X=$1
    echo $X
    shift
    shift
done

answered Oct 5, 2012 at 7:00

nneonneo

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4 Comments

David Cosman Over a year ago

Thanks a lot, that is a much better idea

Charles Duffy Over a year ago

This would be better written as shift 2.

nneonneo Over a year ago

@CharlesDuffy: That does not work. shift 2 won't shift anything if $# < 2, and therefore if an odd number of arguments is provided it will loop forever.

Jonathan Leffler Over a year ago

But by that token, won't you get an error from the second shift? The ksh I use on Linux gives an error message (ksh: nothing to shift) when I do shift on an empty argument list (or shift 2 on a one entry argument list). ... Answer: bash is more sensible and doesn't generate a message (but doesn't shift anything when you try to shift more items than there are to shift). So, differences between shells — what fun. I didn't expect the error from ksh; I expected shift 2 to shift 1 when there aren't two left. (/bin/sh on Solaris gives an error but does do the shift.) Weird!

Taizo Ito · Accepted Answer · 2012-10-05 15:41:04Z

0

nneonneo's idea must be better than the following, but I'm going to present a way that checks if the argument is odd just in case.

#!/bin/sh
cnt=1
while [ $# -gt 0 ]; do
  X=$1

  # output the current argument when the counter is odd.
  if [ $(($cnt%2)) -eq 1 ]; then
    echo $X
  fi

  cnt=$(($cnt+1))
  shift
done

answered Oct 5, 2012 at 15:41

Taizo Ito

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Comments

Ярослав Рахматуллин · Accepted Answer · 2012-10-05 22:41:02Z

0

#!/bin/bash

declare -a ODD
declare -a EVEN
ODDIX=0
EVENIX=0

while [ $# -ne 0 ]; do
    if (( $#%2 == 1 ));then
        ODD[$ODDIX]=$1
        ODDIX=$((ODDIX+1))
    else        
        EVEN[$EVENIX]=$1
        EVENIX=$((EVENIX+1))
    fi
    shift
done

echo ODD: ${ODD[@]}
echo EVEN: ${EVEN[@]}*

answered Oct 5, 2012 at 22:41

Ярослав Рахматуллин

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