Using $# in bash loops

Question

I am trying to understand why this loop does not print a number for each arguments supplied to the script.

#!/bin/bash

for i in {1..$#}; do
  echo $i
done

Instead, when supplied e.g. 3 arguments, it outputs

{1..3}

score 3 · Accepted Answer · edited May 23 '17 at 12:05

3

The expression {} does not accept variables.

To do so, you need to work with for example seq. The following will make it::

#!/bin/bash

for i in $(seq 1 $#); do
  echo $i
done

Note that $() is equivalent to ``. That is, it performs a command substitution. For example:

$ d=$(echo "hello")
$ echo $d
hello

$ ./a
$

$ ./a a b c
1
2
3

edited May 23 '17 at 12:05

Community

answered Sep 24 '13 at 10:57

fedorqui

Thank you. Can you explain what the dollarsign does in this context? I understand the dollarsign as a prefix to refer to variables, but I don't see any assignment in the parentheses. – Niels B. Sep 24 '13 at 11:00
Thank you. So seq is actually an executable and not a language construct. It makes good sense. – Niels B. Sep 24 '13 at 11:07
Exactly! You explained it better than me :) See for example `$ which seq` that returns something like `/usr/bin/seq`. – fedorqui Sep 24 '13 at 11:13

score 1 · Answer 2 · answered Sep 24 '13 at 12:23

1

Brace expansion occurs before variable expansion

answered Sep 24 '13 at 12:23

glenn jackman

2 Answers2