Print given number in reverse order [closed]

Question

clear
echo "Enter a number"
read n
sd=0
rev=0

while [ $n -gt 0 ]
do
    sd=$(( $n % 10 ))
    rev=$(( $rev *\ 10 + $sd ))
    n=$(( $n / 10 ))
done

echo "Reverse number of entered digit is $rev"

From the above code I am unable to get the required output. Instead, this error is displayed.

./Display: line 17: 0 *\ 10 + 4 : syntax error: operand expected (error token is "\ 10 + 4 ")
Reverse number of entered digit is 0

I don't know why this error is displayed, please help me to figure it out.

Answer 1

Numbers are text too. Text can be reversed with rev without any arythmetic.

#!/bin/bash
clear
read -p "Enter a number: " num
echo $num | rev

Answer 2

The error had already pointed out the problem clearly: 0 *\ 10 + 4 isn't a valid arithmetic expression.

I believe *\ is a typo for *. Fix this typo and the script will be OK.

Answer 3

*\ is not the valid expression. Instead, try * which removes the special meaning of the character *. The character * is a wildcard character, which is why the error appeared.

Answer 4

echo "Enter a number"
read n
sd=0
rev=0

while [ $n -gt 0 ]
do
    sd=$(( $n % 10 ))
    rev=`expr $rev \* 10 + $sd` 
    n=$(( $n / 10 ))
done

echo "Reverse number of entered digit is $rev"

Answer 5

* is a wildcard character. Hence make sure that you add a '\' before using it

Answer 6

echo "Enter a number"
read n
sd=0
rev=0

while [ $n -gt 0 ]
do
    sd=$(( $n % 10 ))
    rev=$(( $rev * 10 + $sd ))
    n=$(( $n / 10 ))
done

echo "Reverse number of entered digit is $rev"

This should work *\ is not the correct form just use * for this

Answer 7

echo "enter the number"
read n
rev=0
sd=0
while [$n -gt 0]
do
    sd=`expr $n%10`
    rev=`expr $rev\*10+$sd`
    n=`expr $n/10`
done
echo "the reverse number s: $rev"

Answer 8

for i in $*; do
  x=$i"  "$x
done;
echo "The Reverse order is $x"