Bash script printing expression instead of result

Question

I am trying to calculate multiplication of number of arguments and first argument provided by the customer. here the arguments are 10, 15 so total number of arguments is 2. Now i want the shell to perform 2*10(since 10 is the first argument). Instead of the answer I am getting the code as result.

the program I've written:

r=$(echo "$1 \* $#")
echo "Following are the numbers you entered $@ "
echo "first number : $1. and second number: $2."
echo "total number of entered numbers: $#"
echo "expected result: $r"

result:

root@LAPTOP-J:~# bash test.sh 10 15
Following are the numbers you entered 10 15
first number : 10. and second number: 15.
total number of entered numbers: 2
expected result: 10 \* 2
root@LAPTOP-J5JNFL7K:~#

Answer 1

Well, what echo does, is to print the strings its given as arguments. Like an echo sounds like the original sounds coming back.

echo foo bar prints foo bar; echo "$var" prints the contents of $var (after the shell expanded the value); and echo "$1 \* $#" prints 10 \* 2 if those are the values of $1 and $#. (The asterisk isn't special within double-quotes, so the backslash is not removed.)

You may have confused that with expr, which could do arithmetic.

But there's no need for an external command for arithmetic in the shell, just use the arithmetic expansion $(( .. )), e.g.:

r=$(( $1 * $# ))

but note that if $1 contains something else than numbers, the results may be odd (or even run arbitrary commands embedded within $1, at least in Bash). For serious work, you may want to sanity check the value there first.