newprime.sh：第 14 行：[：0+1：integer 表达式预期

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.