CODEWITHSUNDEEP
bashloopsshellfor-loopstep-through
uglyorgans
uglyorgans

Reputation: 15

How to print every 17th number, 0-200, skipping even numbers?

I need to write a Bash/Shell code that prints every 17th number from 0 to 200, skipping all even numbers.

This is the code I have right now (I know it doesn't work as intended):

for i in {0..200..17}
do
if [ $(($i % 2)) ]
then
  echo -n "${i}, "
fi
done;

I need to print an output that looks like this:

17, 51, 85, 119, 153, 187

I have the part where it steps through every 17th number, but I can't seem to figure out how to get it to skip every even number and every guide I've read through online doesn't help. Appreciation in advance!

Upvotes: 0

Views: 55

Answers (2)

Ed Morton
Ed Morton

Reputation: 204608

Change [ $(($i % 2)) ] into (( i % 2 )) so you're simply doing an arithmetic operation and testing the result (success=0/fail=non-zero exit status) rather than an arithmetic operation and then testing the output of that operation.

Output:

$ echo $((27 % 2))
1

$ echo $((26 % 2))
0

Exit status:

$ ((27 % 2)); echo $?
0

$ ((26 % 2)); echo $?
1

Upvotes: 2

Gilles Quénot
Gilles Quénot

Reputation: 185810

What I would do:

for i in {0..200..17}; do ((i % 2)) && echo "$i"; done


17
51
85
119
153
18

((...)) and $((...)) are arithmetic commands, which returns an exit status of 0 if the expression is nonzero, or 1 if the expression is zero. Also used as a synonym for "let", if side effects (assignments) are needed.

See http://mywiki.wooledge.org/ArithmeticExpression

Upvotes: 2

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