reverse the order of characters in a string

Question

In string \"12345\", out string \"54321\". Preferably without third party tools and regex.

Answer 1

read word

reve=`echo "$word" | awk '{for(i=length($0); i>0;i--) printf (substr($0,i,1));}'`
echo  "$reve"

Answer 2

This reverses the string "in place":

a=12345
len=${#a}
for ((i=1;i<len;i++)); do a=$a${a: -i*2:1}; done; a=${a:len-1}
echo $a

or the third line could be:

for ((i=0;i<len;i++)); do a=${a:i*2:1}$a; done; a=${a:0:len}

or

for ((i=1;i<len;i++)); do a=${a:0:len-i-1}${a: -i:i+1}${a:len-i-1:1}; done

Answer 3

If var=12345:

bash for((i=0;i<${#var};i++)); do rev="$rev${var:~i:1}"; done

sh c=$var; while [ "$c" ]; do rev=$rev${c#"${c%?}"}; c=${c%?}; done

echo "var: $var, rev: $rev"

Run it:

$ rev
var: 12345, rev: 54321

Answer 4

Presume that a variable 'var' has the value '123'

var="123"

Reverse the string and store in a new variable 'rav':

rav=$(echo $var | rev)

You'll see the 'rav' has the value of '321' using echo.

echo $rav

Answer 5

For those without rev (recommended), there is the following simple awk solution that splits fields on the null string (every character is a separate field) and prints in reverse:

awk -F '' '{ for(i=NF; i; i--) printf("%c", $i); print "" }'

The above awk code is POSIX compliant. As a compliant awk implementation is guaranteed to be on every POSIX compliant OS, the solution should thus not be thought of as "3rd-party." This code will likely be more concise and understandable than a pure POSIX sh (or bash) solution.

(; I do not know if you consider the null string to -F a regex... ;)

Answer 6

This can of course be shortened, but it should be simple to understand: the final print adds the newline.

echo 12345 | awk '{for (i = length($0); i > 0; i--) {printf("%s", substr($0, i, 1));} print "";}'