reverse the order of characters in a string

Question

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.

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

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

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

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.