For loop range in Bash

Question

Im trying a code in bash to generate prime nos as follows:

#!/bin/bash


echo "Enter till where u wish to generate"
read num
echo "Generating prime numbers from 2 to $num"

flag="prime"

for i in {2..$num}
do

for j in {2..$((${num}-1))}
 do

 [ $((${i}%${j})) -eq 0 ] && flag="nprime" || flag="prime"
 break
 done

 [ "$flag" == "prime" ] && echo "$i"

done

Upon execution, it throws an error because the for loop takes the sequence mentioned in the curly braces as it is not as a sequence. Could you guide me as to where am i going wrong ?

score 2 · Accepted Answer · answered Nov 11 '15 at 21:26

2

man bash in my version says:

A sequence expression takes the form {x..y[..incr]}, where x and y are either integers or single characters, and incr, an optional increment, is an integer.

You can't use variables in ranges. Try seq instead:

for i in $(seq 2 $num) ; do

Note that incr for seq goes between x and y.

answered Nov 11 '15 at 21:26

choroba

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Thank you! seq seems to be a better approach! – abhishek nair Nov 11 '15 at 21:33
The reason is the order of [shell expansions](https://www.gnu.org/software/bash/manual/bashref.html#Shell-Expansions) -- brace expansion occurs before parameter expansion. – glenn jackman Nov 11 '15 at 21:33

score 2 · Answer 2 · answered Nov 11 '15 at 21:27

2

Use:

for ((i=2; i<=$num; i++))

answered Nov 11 '15 at 21:27

Cyrus

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For loop range in Bash

2 Answers2