checking if a string is a palindrome

Question

I am trying to check if a string is a palindrome in bash. Here is what I came up with:

#!/bin/bash
read -p \"Enter a string: \" string
if [[ $string|rev ==         


        

Answer 1

A bash-only implementation:

is_palindrome () { 
    local word=$1
    local len=$((${#word} - 1))
    local i
    for ((i=0; i <= (len/2); i++)); do
        [[ ${word:i:1} == ${word:len-i:1} ]] || return 1
    done
    return 0
}

for word in hello kayak; do
    if is_palindrome $word; then
        echo $word is a palindrome
    else
        echo $word is NOT a palindrome
    fi
done

Inspired by gniourf_gniourf:

is_palindrome() {
  (( ${#1} <= 1 )) && return 0
  [[ ${1:0:1} != ${1: -1} ]] && return 1
  is_palindrome ${1:1: 1}
}

I bet the performance of this truly recursive call really sucks.

Answer 2

Another variation without echo and unnecessary quoting within [[ ... ]]:

#!/bin/bash
read -p "Enter a string: " string
if [[ $(rev <<< "$string") == "$string" ]]; then
    echo Palindrome
fi

Answer 3

Maybe it is not the best implementation, but if you need something with pure sh

#!/bin/sh

#get character <str> <num_of_char>. Please, remember that indexing is from 1
get_character() {
  echo "$1" | cut -c "$2"
}

for i in $(seq $((${#1} / 2))); do
  if [ "$(get_character "$1" "$i")" != "$(get_character "$1" $((${#1} - i + 1)))" ]; then
    echo "NO"
    exit 0
  fi
done

echo "YES"

and canonical way with bash as well

for i in $(seq 0 $((${#1} / 2 - 1))); do
  if [ "${1:$i:1}" != "${1:$((${#1} - i - 1)):1}" ]; then
    echo "NO"
    exit 0
  fi
done

echo "YES"

Answer 4

Use $(command substitution):

#!/bin/bash
read -p "Enter a string: " string
if [[ "$(echo "$string" | rev)" == "$string" ]]; then
    echo "Palindrome"
fi