newprime.sh: line 14: [: 0+1: integer expression expected - bash

Question

what's wrong with my code? I am trying to print prime numbers upto n digits
echo Enter Number
read num
for (( i=2; $i <= $num ; i++ ))
do
c=0
for (( j=2; $j <= $i ; j++))
do
mod=$(($i % $j))
if [ "$mod" -eq 0 ]
then
c=`expr $c+1`
fi
done
if [ "$c" -eq 1 ]
then
echo $c
fi
done
I don't have any idea what I'm doing wrong. If someone could tell me how to fix it I would be thankful
newprime.sh: line 14: [: 0+1: integer expression expected
newprime.sh: line 14: [: 0+1: integer expression expected
newprime.sh: line 14: [: 0+1+1: integer expression expected
newprime.sh: line 14: [: 0+1: integer expression expected
newprime.sh: line 14: [: 0+1+1+1: integer expression expected
newprime.sh: line 14: [: 0+1: integer expression expected

Answer 1

expr requires parameters to be passed as separate arguments. Quoting the POSIX standard for expr:
The application shall ensure that each of the expression operator symbols [...] and the symbols integer and string in the table are provided as separate arguments to expr.
The code here is appending all the operators into a single argument, hence your problem.
Thus:
c=$(expr "$c" + 1)
...NOT...
c=$(expr $c+1)
But don't do that at all. It's more efficient and more readable to write:
c=$(( c + 1 ))

Answer 2

Optimized with less iterations POSIX shell version:
#!/usr/bin/env sh
printf %s 'Enter Number: '
read -r num
i=1
while [ "$i" -le "$num" ]; do
c=0
j=2
# Stop checking division when divisor power 2 is greater than number
# or we identifed a divisor
while [ "$((j * j))" -le "$i" ] && [ "$c" -eq 0 ]; do
c=$((i % j == 0))
j=$((j + 1))
done
if [ "$c" -eq 0 ]; then
printf '%s\n' "$i"
fi
i=$((i + 2))
done
Or using a function:
#!/usr/bin/env sh
is_prime() {
j=2
# Check j is a divisor of argument number, while j^2 is less than number
while [ "$((j * j))" -le "$1" ]; do
# If j is a divisor of number before the end of the loop
# number is not prime, so return 1
[ "$(($1 % j))" -eq 0 ] && return 1
j=$((j + 1))
done
}
printf %s 'Enter Number: '
read -r num
i=1
while [ "$i" -le "$num" ]; do
if is_prime "$i"; then
printf '%s\n' "$i"
fi
i=$((i + 2))
done

Answer 3

Don't use expr. Put your mathematical expression inside (( )) (or echo $(( )) to print the result), and the shell will evaluate it.
See what the expr output looks like, vs regular shell arithmetic:
$ expr 0+1
0+1
$ echo "$((0+1))"
1
-eq with test, or single square brackets (eg [ 1 -eq 2 ]) prints an error if both operands aren't integers. That's what's causing your error.
Here's a fast and concise way to list primes in bash. You can put it in a function or script:
for ((i=2; i<="${1:?Maximum required}"; i++)); do
for ((j=2; j<i; j++)); do
((i%j)) || continue 2
done
echo "$i"
done
edit: Just to explain something, if (([expression])) evaluates to 0, it returns non-zero (1) (failure). If it evaluates to any other number (positive or negative), it returns zero (true). When a number divides evenly in to i, the modulo (%) (remainder) is zero. Hence the command fails, we know it's not prime, and we can continue the outer loop, for the next number.