Bash shell Decimal to Binary base 2 conversion

Question

I'm looking for an easy way in Bash to convert a decimal number into a binary number. I have variables that need to be converted:

$ip1 $ip2 $ip3 $ip4

Is there a simple method to do this without looking at every individual number?

I would prefer not to have to write a lot of code.

Answer 1

Convert decimal to binary with bash builtin commands (range 0 to 255):

D2B=({0..1}{0..1}{0..1}{0..1}{0..1}{0..1}{0..1}{0..1})

echo ${D2B[7]}

00000111

echo ${D2B[85]}

01010101

echo ${D2B[127]}

01111111

---

To remove leading zeros, e.g. from ${D2B[7]}:

echo $((10#${D2B[7]}))

111

---

This creates an array with 00000000 00000001 00000010 ... 11111101 11111110 11111111 with bash‘s brace expansion. The position in array D2B represents its decimal value.

See also: Understanding code ({0..1}{0..1}{0..1}{0..1}{0..1}{0..1}{0..1}{0..1})

Answer 2

Decimal to binary conversion in Bash:

I'm using Ubuntu 14.04 to do this.

Convert the decimals 1 through 5 to binary.

el@apollo:~$ bc <<< "obase=2;1"
1
el@apollo:~$ bc <<< "obase=2;2"
10
el@apollo:~$ bc <<< "obase=2;3"
11
el@apollo:~$ bc <<< "obase=2;4"
100
el@apollo:~$ bc <<< "obase=2;5"
101

Bonus example:

el@apollo:~$ bc <<< "obase=2;1024"
10000000000

el@apollo:~$ bc <<< "obase=2;2^128"
100000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000

Answer 3

General method for converting an integer number into another representation with another base (but base<=10 because of using digits 0..9 for representation, only):

function convertIntvalToBase () # (Val Base)
{
   val=$1
   base=$2
   result=""
   while [ $val -ne 0 ] ; do
        result=$(( $val % $base ))$result #residual is next digit
        val=$(( $val / $base ))
   done
   echo -n $result
}

e.g.

convertIntvalToBase $ip1 2     # converts $ip1 into binary representation

Answer 4

Defined as a function in bash:

# to Binary:
toBinary(){
    local n bit
    for (( n=$1 ; n>0 ; n >>= 1 )); do  bit="$(( n&1 ))$bit"; done
    printf "%s\n" "$bit"
}

Answer 5

Decimal to Binary using only Bash

Any integer number can be converted ti binary using it::

touch dec2bin.bash && chmod +x "$_" && vim "$_"

And, then copy paste the following:

#!/bin/bash
num=$1;
dec2bin()
{
    op=2; ## Since we're converting to binary
    quo=$(( $num/ $op)); ## quotient
    rem=$(( $num% $op)); ## remainder
    array=(); ## array for putting remainder inside array
    array+=("$rem"); ## array expansion
        until [[ $quo -eq 0 ]]; do
            num=$quo; ## looping to get all remainder, untill the remainder is 0
            quo=$(( $num / $op));
            rem=$(( $num % $op));
            array+="$rem"; ## array expansion
        done
    binary=$(echo "${array[@]}" | rev); ## reversing array
    printf "$binary\n"; ## print array
}
main()
{
[[ -n ${num//[0-9]/} ]] &&
    { printf "$num is not an integer bruv!\n"; return 1;
    } || { dec2bin $num; }
}
main;

For example:

./dec2bin.bash $var
110100100

Integer must be added!!

./dec2bin.bash 420.py
420.py is not an integer bruv!

Also, another way using python: Much slower

python -c "print(bin(420))"
0b110100100

Hexadecimal to Binary using only Bash

Similarly, hexadecimal to binary, as follows using only bash:

#!/usr/local/bin/bash ## For Darwin :( higher bash :)
#!/bin/bash ## Linux :)
hex=$1;
hex2bin()
{
    op=2; num=$((16#$hex));
    quo=$(( $num/ $op));
    rem=$(( $num% $op));
    array=();
    array+=("$rem");
        until [[ $quo -eq 0 ]]; do
            num=$quo;
            quo=$(( $num / $op));
            rem=$(( $num % $op));
            array+="$rem";
        done
    binary=$(echo "${array[@]}" | rev);
    printf "Binary of $1 is: $binary\n";
}
main()
{
[[ -n ${hex//[0-9,A-F,a-f]/} ]] &&
    { printf "$hex is not a hexa decimal number bruv!\n"; return 1;
    } || { hex2bin $hex; }
}
main;

For example:

./hex2bin.bash 1aF
Binary of 1aF is: 110101111

Hex must be passed:

./hex2bin.bash XyZ
XyZ is not a hexa decimal number bruv!

Answer 6

To make @codaddict's answer a little more pretty, use this to prefix the output with 0b for "binary":

printf "0b%s\n" "$(echo "obase=2; $((num1 + num2))" | bc)"

Example:

num1=2#1111  # binary  1111 (decimal 15)
num2=2#11111 # binary 11111 (decimal 31)
printf "0b%s\n" "$(echo "obase=2; $((num1 + num2))" | bc)"

Output:

0b101110

This is decimal 46.

For details on the input base-2 formatting in bash, such as 2#1111 above, see the very end of my answer here: How to use all bash operators, and arithmetic expansion, in bash.

To have at least 8 digits in the output, use:

printf "0b%08d\n" $(echo "obase=2; $((num1 + num2))" | bc)

Source: David Rankin, in an answer to my question here.

Answer 7

I like to use dc for this. It's very concise:

$ n=50; dc -e "$n 2op"
110010

The commands here are as follows:

1. Push the number, n, on the stack, via shell expansion.
2. Push 2 on the stack, then use o to pop the stack and use 2 as the output radix.
3. p to print the top of the stack (which is just n), using the output radix set in step 2 (so print in binary).

If you want padding:

$ n=50; pad_size=16; printf "%0${pad_size}d\n" $(dc -e "$n 2op")
0000000000110010