case construction for basic Arithmetics calculation

the scrip (q4.sh) should perform the following calcuation (+, -, / and *) it should be used like this:
q4.sh number1 operation number2

I wrote it already but the "*" does not work.

#!/bin/bash
#Date: 2010.10.19
# un script qui utilisera une instruction case pour effectuer des opérations arithméties :
#+ addition
#- subtraction
#x multiplication
#/ division
#Le nom du script doit etre 'q4' et il s'utilisera comme suit :
#$ ./q4 20 / 3


case $2 in "+" )
   echo "`expr $1 + $3`" ;;
"-" )
   echo "`expr $1 - $3`" ;;
"*" )
   echo "`expr $1 * $3`" ;;
"/" )
   echo "`expr $1 / $3`" ;;
esac

#END

Please give me your feedbacks how to improve it. and fix the "*". Only bash.
BR.

Have you tried calling the script like this

./q4 2 \* 2

Just tested but it returns an error

alioune@baccus ~/bashtuto/bashExercise $ ./q4.sh 2 \* 2
expr: syntax error

alioune@baccus ~/bashtuto/bashExercise $ 

Try:

echo "`expr $1 \* $3`" ;;

file modified

case $2 in "+" )
   echo "`expr $1 + $3`" ;;
"-" )
   echo "`expr $1 - $3`" ;;
"*" )
   echo "`expr $1 \* $3`" ;;
"/" )
   echo "`expr $1 / $3`" ;;
esac

test

alioune@baccus ~/bashtuto/bashExercise $ ./q4.sh 2 + 2
4
alioune@baccus ~/bashtuto/bashExercise $ ./q4.sh 2 - 2
0
alioune@baccus ~/bashtuto/bashExercise $ ./q4.sh 2 / 2
1
alioune@baccus ~/bashtuto/bashExercise $ ./q4.sh 2 * 2
alioune@baccus ~/bashtuto/bashExercise $ 
alioune@baccus ~/bashtuto/bashExercise $ 

Still no output for "*".

You need to call it like this:

./q4.sh 2 \* 2

---------- Post updated at 11:53 ---------- Previous update was at 11:45 ----------

You need to always escape the * on the command line to avoid expansion by the shell.

In the script, instead of using expr you can do this:

case $2 in
  +)  echo "$(( $1 + $3 ))" ;;
  -)  echo "$(( $1 - $3 ))" ;;
  \*) echo "$(( $1 * $3 ))" ;;
  /)  echo "$(( $1 / $3 ))" ;;
esac

then you do not need the inside the arithmetic expression and you are using a shell builtin, which is faster.

In this case you could even do this:

echo "$(($@))"

but that is probably not what you are after

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