Unexpected output from Bash for loop using curly brackets

Question

I expected the following loop to print numbers:

#!/bin/bash
# Basic range in for loop
for value in {1..5}
do
    echo $value
done

but I get the following output in a single line: {1..5}

What am I doing wrong?

most likely you are running the script wrongly, for ex: `sh script.sh`.. — Sundeep, Sep 19 '16 at 03:50
[different ways to run bash script](http://stackoverflow.com/documentation/bash/300/hello-world/1136/hello-world#t=201609190352481927018) — Sundeep, Sep 19 '16 at 03:53
yes, your script is written in `bash` but you are running it as `sh` script which is a different scripting language... see the link I posted for details... — Sundeep, Sep 19 '16 at 04:10

score 1 · Answer 1 · answered Sep 19 '16 at 03:57

1

{1..5} won't necessarily work like you expect it to in every circumstance. A better way to do this (to my mind, at least) is to use seq:

for value in $(seq 1 5)
do
    echo $value
done

seq is a simple program included in basically every Unix that generates the sequence of numbers dictated by its two arguments.

answered Sep 19 '16 at 03:57

1

For `bash`: Using `seq` for this is overkill. It is wasting a process, for something that is built-in. Use `seq` only if one of them (1 or 5) is a variable. For `sh`: there is no simpler option than `seq`. – anishsane Sep 19 '16 at 07:19
@Sebastian thanks for the suggestion. I knew about the seq but I wasn't sure why the curly brackets didn't work. – Bi Bi Sep 19 '16 at 12:48
2

With bash, the better practice is `for ((i=1; i<5; i++)); do echo "$i"; done`. `seq` is not built into bash, and it is not part of the POSIX specification. There is no guarantee it will exist at all, and if it _does_ exist, there is no guarantee it will behave any particular way. – Charles Duffy Nov 11 '20 at 16:25

1 Answers1