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  1. Practice
  2. Linux Shell
  3. Bash
  4. The World of Numbers
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The World of Numbers

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votes

  • x0r19x91 + 1 comment

    My Solution

    #!/bin/bash
read X
read Y
printf "%s\n" $X{+,-,*,/}"($Y)" | bc
    • T_abhi541 + 1 comment

      "%s\n" what does this represent can you please tell me as I am new to this.

      • x0r19x91 + 1 comment

        print each string separated by a newline

        • kbcrum + 1 comment

          what is meant by piping it to bc

          • kbcrum + 1 comment

            oh its like dc but not reverse polish notation! thats a cool solution. better golf but less so for readibility. still slick

            • x0r19x91 + 0 comments

              Thanks

  • hpedrorodrigues + 2 comments

    This worked fine

    read x
read y

echo $((x + y))
echo $((x - y))
echo $((x * y))
echo $((x / y))
    • NandiChaurasiya + 0 comments
      [deleted]
    • ogkmasum + 0 comments

      Great. Simple and fine

  • superbenk + 2 comments

    The test cases conflict with the requirements. Test #4 & #5 expect things to work with x,y being greater than 100 or less than -100 which goes against what the rules state. So in the end my validation to fit the rules broke my submissions.

    • rablanken13Asked to answer + 0 comments

      I typically ignore the input restrictions.

    • metta_crawler + 0 comments

      The instructions are ambigous if you or the system will be applying the constraints to the input. Thanks for pointing out that the instructions are false too.

  • roszel + 1 comment

    I thought so

    signs=(+-*/)
for ((i=0;i<4;i++));do echo $((5"${signs:$i:1}"2)); done
7
3
10
2
    • codemonkey2 + 0 comments

      kind of art. it is working. thanks

  • CAP_hadd0ck + 1 comment

    Four hours or so and i've got this.

    #!/bin/bash
read x
read y

if [ "$x" -ge "-100" ] && [ "$y" -ge "-100" ] && [ "$x" -le "100" ] && [ "$y" -le "100" ] && [ "$y" -ne "0" ]; then
	echo "$x + $y" | bc
	echo "$x - $y" | bc
	echo "$x * $y" | bc
	echo "$x / $y" | bc
else
	echo "ERROR"

fi

    works like a charm. probably messy. i'm extremely new to bash-scripts.

    • [deleted] + 0 comments

      This is what I want. Thank you, it's really helpful.

  • shubhamitankr + 0 comments
    read x 
read y
echo `expr $x + $y`
echo `expr $x - $y`
echo `expr $x \* $y`
echo `expr $x / $y`

  • Arindam31 + 1 comment

    The word 'difference' is confusing. It mostly suggests a mod of both values.

    • quetonix + 0 comments

      For difference

      read x
read y
echo $(($x+$y))
if [ "$x" -eq "$y" ];then
  echo 0;
fi
if [ "$x" -gt "$y" ];then
  echo $(($x-$y))
fi
if [ "$x" -lt "$y" ];then
  echo $(($y-$x))
fi
echo $(($x*$y))
echo $(($x/$y))

  • cj_cawley + 0 comments

    Why does this not clear when it works fine on the os ?

  • nigelagoddard + 0 comments

    what did I miss? when I run I get the correct output. But "wrong answer" to solve the problem?

    x=5 y=2 num1="x+((y))" num3="x*((y))" echo "num2" echo "num4"

    Was it becasue I didn't lead with #!/bin/bash...?

  • __Z__ + 0 comments
    read x
read y
echo -e "$((x+y))\n$((x-y))\n$((x*y))\n$((x/y))"

    This worked for me!

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