How to disable special characters in bash? [duplicate] - bash

Question

This question already has an answer here:
Prevent globbing in a bash script
(1 answer)
Closed 6 years ago.
I've created a script which performs a calculation. For example:
count 1 + 3
1 + 3 = 4,
- and / also works, but if I type
count 1 * 3, I got
Should be number operand number
Here is a part of a script:
if [ "$#" -ne 3 ]; then
echo "Should be number operand number"
exit 1
fi
...
elif [ "$2" = "*" ]; then
result=`echo $1 \* $3 | bc`
echo "$1 * $3 = $result"
Yes, if I type \ before * in command line it would work, but I need it to run just with a *.
I've tried to use set -f inside the script it did not work (yes it disables special characters if I type in bash itself but it is not what I need). There is also a shopt command that controls shell behavior, so I've tried to enable some options like shopt -s globstar, nullglob, promptvars, etc. It did not work. I put the command before if statement maybe it is wrong.
I would appreciate if someone corrects me or tells the other way to disable interpretation of a special character from inside the script.

Answer 1

This exact problem is mentioned in the POSIX specification of expr, saying that the tool has a "rather difficult syntax" for the reason you describe.
This is definitely a known problem, in other words, and POSIX itself leaves it unsolved. You should really consider working with Unix and do the same, rather than trying to work against it.
Some options are:
Require the entire expression to one quoted argument, e.g. count "1 * 2"
This is a pragmatic, fail-fast way of ensuring that the argument is always quoted, so that any introduction of * still works. The bash builtin let does this.
read the expression yourself, e.g. just run count and then type in 1 * 2.
This avoids the shell touching your data, so that you can interpret it however you want. It's what bc and dc do.
Live with the fact that passing * will require careful quoting.
This is what expr does.
Use a magic alias hack that only works on certain shells under very specific circumstances.
This is what no one does, because it's fragile and unpredictable, even though it in some cases allows the syntax you want.

Answer 2

Always quote when you echo. You need to quote the variable reference as well:
me$ FOO="BAR * BAR"
me$ echo "$FOO"
BAR * BAR
If you don't use double quote, bash will see * and perform filename expansion. * expands to all files in your current directory.
However in saying this you would have to be using single quotes for the *:
#!/bin/bash
if [ "$2" = "*" ]; then
result=`echo $1 \* $3 | bc`
echo "$1 * $3 = $result"
fi
The following would work:
» count.sh 2 '*' 1
2 * 2 = 4
There is nothing you can do in your program, because the * expansion is done by the shell (in contrast to Windows, where it's done by the program).
Here's another code example using a menu-driven approach instead:
#!/bin/bash
while true; do
read -p "what's the first number? " n1
read -p "what's the second number? " n2
PS3="what's the operation? "
select ans in add subtract multiply divide; do
case $ans in
add) op='+' ; break ;;
subtract) op='-' ; break ;;
multiply) op='*' ; break ;;
divide) op='/' ; break ;;
*) echo "invalid response" ;;
esac
done
ans=$(echo "$n1 $op $n2" | bc -l)
printf "%s %s %s = %s\n\n" "$n1" "$op" "$n2" "$ans"
done
what's the first number? 5
what's the second number? 4
1) add
2) subtract
3) multiply
4) divide
what's the operation? /
invalid response
what's the operation? 4
5 / 4 = 1.25000000000000000000
In short, quote everything where you do not require the shell to perform token splitting and wildcard expansion.