Bash for loop printing the statement

Question

I was trying to print out some numbers from 1 to 20 with an increment of 2

#!/bin/bash

for i in {0..20..2} 
do
    echo $i
done

and this is what it has printed out

{0..20..2}

what I am doing wrong?

What version of bash are you using? Setting the step value is only after v4.0... — StefanR
– StefanR, Commented Oct 12, 2022 at 7:47
You may consider using a C-style for loop: for ((i=0; i<=20; i+=2)); do echo $i; done — M. Nejat Aydin
– M. Nejat Aydin, Commented Oct 12, 2022 at 8:07
@Jon, ...suboptimal. seq isn't part of bash and isn't specified by POSIX, so you pay external-executable startup costs, and whether it works (and how it works) is entirely at the grace of your operating system vendor. — Charles Duffy
– Charles Duffy, Commented Oct 12, 2022 at 12:59

ramsay · Accepted Answer · 2022-10-12 13:06:19Z

1

Using the built-in for loops syntax:

#!/bin/bash
for (( i=0; c<=20; c+=2 ))
do
    echo $i
done

answered Oct 12, 2022 at 12:56

ramsay

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1 Comment

Yes, using the built-in for loop probably is a better idea.