bash shell script, issues with a 32 bit decimal to binary converter - bash

Question

Having an issue with this
when I run it, the console repeats this over and over;
14: shellscript.sh: 2: not found
15: shellscript.sh: 32: not found
18: shellscript.sh: =1: not found
19: shellscript.sh: 0: not found
It seems like it's something to do with how bash handles redefining variables through arithmetic?
#!/bin/bash
echo "This script converts a user's number into an IP address."
echo -n "Input your number: "
read user
if [ $user -lt 4294967296 ]
then
exp=$((32))
num=$((0))
ipb=""
while [ $exp -gt 0 ]
do
bit=expr 2 ** $exp
exp=expr $exp - 1
if [ $bit+$num -le $user ]
then
$ipb="${ipb}1"
num=expr $num + $bit
else
$ipb="${ipb}0"
fi
done
echo $ipb
echo "done"
fi
Same as above but with comments to explain it.
#!/bin/bash
echo "This script converts a user's number into an IP address."
echo -n "Input your number: "
read user
#check if number is larger than 32bits
if [ $user < 4294967296 ]
then
#var exp is exponent that will be used to redefine var bit each loop cycle
#var num is var used to rebuild the user number with corresponding bits added to -
#var ipb is IP binary (not yet converted to 4 octet integers)
exp=$((32))
num=$((0))
ipb=""
#while the exponent is greater than 0 (exponent is 1 less as per binary order)
while [ $exp > 0 ]
do
#(Re)define bit var for line 23
bit=expr 2**$exp
#go to next lowest exponent
exp=expr $exp - 1
#If the current bit is added to our num var,will it be
#less than or equal to the user number?
if [ $bit + $num -le $user ]
then
#If so, redefine the ipb string var with a 1 on the end
#and redefine the num integer var added with the current
#iteration of the bit integer var's value
$ipb="${ipb}1"
num=expr $num + $bit
else
#if not, redefine the ipb string var with a 0 on the end
$ipb="${ipb}0"
fi
done
#output the IP binary
echo $ipb
echo "done"
fi
EDIT:
After some googling and help from shellcheck I got it to work. for some reason with my version of linux mint, the let command was the only thing that correctly took 2**31 as an exponent operation. Here's the code for anyone curious.
echo "This script converts a user's number into the 32 bit equivalent."
echo -n "Input a number below 4294967296: "
read user
echo ""
if [ $user -lt 4294967296 ]
then
exp=$((31))
num=$((0))
ipb=""
while [ $exp -ge 0 ]
do
let bit=2**$exp
let exp-=1
if (( $bit + $num <= $user ))
then
ipb="${ipb}1"
num=$(($num + $bit))
else
ipb="${ipb}0"
fi
done
fi
echo $ipb
Be sure to use bash instead of sh or ./ when running the script in terminal.

Answer 1

Your script has several issues. This one does not throw any errors.
#! /bin/bash
echo "This script converts a user's number into an IP address."
echo -n "Input your number: "
read user
if [ $user -lt 4294967296 ]
then
exp=$((32))
num=$((0))
ipb=""
while [ $exp -gt 0 ]
do
bit=$((2 ** $exp))
exp=$(expr $exp - 1)
if (( bit+num < user ))
then
ipb="${ipb}1"
num=$(expr $num + $bit)
else
ipb="${ipb}0"
fi
done
echo $ipb
echo "done"
fi
One major issue is that expr cannot calculate powers. You also do not use command substitution when you are trying to assign the result of expr to a variable.
Another major issue is that you cannot do math inside single [ ] brackets as you try to do in an if branch. Use (( )) instead.
There were a few other issue, e.g., $ipb="${ipb}1" expanding the variable that you actually want to assign.