How to add a string to a number

I have been recently trying to create a script that can convert a binary number to a decimal. This is what I've got so far:

#!/bin/bash

echo "Specify no of digits"
read digits

if [[ $digits == 1 ]]; then
        echo "Enter 1 st digit"
        read 1
elif [[ $digits == 2 ]]; then
        echo "Enter 1 st digit"
        read 1
        echo "Enter 2 nd digit"
        read 2
elif [[ $digits == 3 ]]; then
        echo "Enter 1 st digit"
        read 1
        echo "Enter 2 nd digit"
        read 2
        echo "Enter 3 rd digit"
        read 3
elif [[ $digits > 3 ]]; then
        echo "Enter 1 st digit"
        read 1
        echo "Enter 2 nd digit"
        read 2
        echo "Enter 3 rd digit"
        read 3
        for digitno in {4..$digits};
        do
                echo "Enter $digitno th digit"
                read $digitno
                ($nodigits++)
        done
echo "$4"
else
        echo "Please enter a valid no of digits. Type './binary_decoder.sh'"
        exit 1
fi

It's a pretty long script, I know. But please try and take the time to examine this script.

If you look at any of the read lines inside the if conditional, you will see that the variables to which the read statements are assigning the numbers to are themselves numbers. With Bash syntax, that will not work. I want for the variables to be like n1, n2, n3, n4... and so on. But, if you see inside the elif [[ $digits > 3 ]]; then statement, you can see that there is a for loop that allows for an infinite no of digits to be decoded. Now, I do not know any way to add the string n to the number in the variable $digitno. But I was wondering that if any of you can maybe figure out how to add the string n to the $digitno variable.

Any help will be appreciated.

You can add a string to a number using simple concatenation:

$ i=3
$ echo n$i
n3

however that doesn't help much with your real goal here, which seems to be how to assign an indeterminate number of user inputs to indexed variables.

As you've already discovered, you can't use variables named 1, 2, 3 etc. in a read command. Aside from the fact that bash variable names must at least begin with an alphabetic character or an underscore, the expansions $1, $2, $3 and so on are reserved for the shell's positional parameters.

If you really want to use $1 ... $n in your script, you can actually do so using the set shell builtin. Note that whereas POSIX only requires support for parameters up to $9, bash supports an arbitrary number (although for indexes above 9 you will need to use braces to disambiguate between, for example, ${10} as the 10^th positional parameter and $10 as the concatenation of $1 with literal 0). For example:

#!/bin/bash

set --
while : ; do
  read -n1
  case $REPLY in
    [01]) set -- "$@" "$REPLY"
    ;;
    *) break
    ;;
  esac
done

for ((i=1; i<=$#; ++i)); do
  printf 'Digit #%d = %d\n' "$i"  "${!i}"
done

The user enters a sequence of 0 and 1 characters, terminating the sequence by hitting any other character (including newline):

$ ./bin2dec
1011010110
Digit #1 = 1
Digit #2 = 0
Digit #3 = 1
Digit #4 = 1
Digit #5 = 0
Digit #6 = 1
Digit #7 = 0
Digit #8 = 1
Digit #9 = 1
Digit #10 = 0

Alternatively, you could do essentially the same with a user-defined array:

#!/bin/bash

arr=()
while : ; do
  read -n1
  case $REPLY in
    [01]) arr+=("$REPLY")
    ;;
    *) break
    ;;
  esac
done

for ((i=0; i<${#arr[@]}; ++i)); do
  printf 'Digit #%d = %d\n' "$i"  "${arr[i]}"
done

Note the different indexing; although both arrays are zero-based, the zeroth element of $@ is reserved for the name of the script file.