reverse the order of characters in a string - bash

Question

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

Answer 1

I know you said "without third-party tools", but sometimes a tool is just too obviously the right one, plus it's installed on most Linux systems by default:
[madhatta#risby tmp]$ echo 12345 | rev
54321
See rev's man page for more.

Answer 2

Simple:
var="12345"
copy=${var}
len=${#copy}
for((i=$len-1;i>=0;i--)); do rev="$rev${copy:$i:1}"; done
echo "var: $var, rev: $rev"
Output:
$ bash rev
var: 12345, rev: 54321

Answer 3

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 4

rev | tail -r (BSD) or rev | tac (GNU) also reverse lines:
$ rev <<< $'12\n34' | tail -r
43
21
$ rev <<< $'12\n34' | gtac
43
21
If LC_CTYPE is C, rev reverses the bytes of multibyte characters:
$ LC_CTYPE=C rev <<< あの
��め�
$ export LC_ALL=C; LC_ALL=en_US.UTF-8 rev <<< あの
のあ

Answer 5

A bash solution improving over #osdyng answer (my edit was not accepted):
var="12345" rev=""
for(( i=0 ; i<${#var} ; i++ )); do rev="${var:i:1}$rev"; done
echo "var: $var, rev: $rev"
Or an even simpler (bash) loop:
var=$1 len="${#var}" i=0 rev=""
while (( i<len )); do rev="${var:i++:1}$rev"; done
echo "var: $var, rev: $rev"
A POSIX solution:
var="12345" rev="" i=1
while [ "$i" -le "${#var}" ]
do rev="$(echo "$var" | awk -v i="$i" '{print(substr($0,i,1))}')$rev"
: $(( i+=1 ))
done
echo "var: $var, rev: $rev"
Note: This works on multi byte strings. Cut solutions will work only in ASCII (1 byte) strings.

Answer 6

Some simple methods of reversing a string
echo '!!!esreveR si sihT' | grep -o . | tac | tr -d '\n' ; echo
echo '!!!esreveR si sihT' | fold -w 1 | tac | tr -d '\n' ; echo
Convert to hex values then reverse
echo '!!!esreveR si sihT' | xxd -p | grep -o .. | tac | xxd -r -p ; echo
echo '!!!esreveR si sihT' | xxd -p | fold -w 2 | tac | xxd -r -p ; echo

Answer 7

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 8

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 9

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 10

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 "";}'

Answer 11

Nobody appears to have posted a sed solution, so here's one that works in non-GNU sed (so I wouldn't consider it "3rd party"). It does capture single characters using the regex ., but that's the only regex.
In two stages:
$ echo 123456 | sed $'s/./&\\\n/g' | sed -ne $'x;H;${x;s/\\n//g;p;}'
654321
This uses bash format substitution to include newlines in the scripts (since the question is tagged bash). It works by first separating the input string into one line per character, and then by inserting each character into the beginning of the hold buffer.
x swaps the hold space and the pattern space, and
H H appends the (current) pattern space to the hold space.
So for every character, we place that character into the hold space, then append the old hold space to it, thus reversing the input. The final command removes the newlines in order to reconstruct the original string.
This should work for any single string, but it will concatenate multi-line input into a single output string.

Answer 12

Here is another simpler awk solution:
awk 'BEGIN{FS=""} {for (i=NF; i>0; i--) s=s $i; print s}' <<< '123456'
654321

Answer 13

Try Perl:
echo 12345 | perl -nle 'print scalar reverse $_'
Source: Perl one-liners

Answer 14

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