write shell script to count occurence of a perticular digit in inputted number.

#!/bin/bash

#shell script to count occurence of a perticular digit in inputted number.

echo -n "enter any number:-"

read number

echo -n "which digit number do u want to count:-"

read digit

len=`echo -n $number |wc -c`

echo "the length of number is:-$len"

count=0

while test $len -gt 0

do

flag=`echo -n $number |cut -c $len`

if test $flag -eq $digit

then

count=`expr $count + 1`

fi

len=`expr $len - 1`

done

echo "|$digit| occured |$count| times in number ($number)"

HackerRank Bash Challenges - Arithmetic Operations Solution

Task

We provide you with expressions containing +,-,*,^, / and parenthesis. None of the numbers in the expression involved will exceed 999.
Your task is to evaluate the expression and display the output correct to 3 decimal places.

Sample Input 1

5+50*3/20 + (19*2)/7

Sample Output 1

17.929

Sample Input 2

-105+50*3/20 + (19^2)/7

Sample Output 2

-45.929

Sample Input 3

(-105.5*7+50*3)/20 + (19^2)/7

Sample Output 3

 22.146

Solution

read line;

printf "%.3f" $(echo "scale = 4; $line" | bc);

Let's get started with some simple numerical computations in Bash

As can be observed from the examples below, there are several ways of making simple numerical calculations in Bash. Just trying to echo an expression wrapped in quotation marks will not work. Wrapping the expression in double parenthesis $((..)) evaluates it, but this is confined to integer computations. To evaluate expressions involving decimal places (floating points) "bc -l" is very useful.

~$ echo "5+5"

5+5

~$ echo "5+5"| bc

10

~$ echo "5+5"| bc -l

10

~$ echo "5+5.2"| bc -l

10.2

~$ echo "5+5.2"| bc

10.2

~$ echo "3/4"| bc

0

~$ echo "3/4"| bc -l

.75000000000000000000

~$ echo $((3+3))

6 

To display the final result by rounding it to a certain number of decimal places, "printf" with a format specified can accomplish the task by specifying the "scale" (number of decimal points). Note that the ordering of the numbers matters in this case, as demonstrated below.

~$ echo "scale = 2; 10 * 100 / 30" | bc

33.33

~$ echo "scale = 2; 10 / 30 * 100" | bc

33.00

~$ echo "scale = 2; (10 / 30) * 100" | bc

33.00

'Expr' is another way to accomplish such tasks.

~$ echo $(expr 5 + 5)

10

~$ echo $(expr 5 - 5 + 2 )

2

~$ echo $(expr 5 - 5 + 2 / 3 )

0

~$ echo $(expr 5 - 5 + 2 / 1 )

2

Be careful with spacing in such expressions! Bash is very sensitive to them.