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}
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2 Answers
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
You can see more information in Shell Programming: What's the difference between $(command) and command.
Tests
$ ./a
$
$ ./a a b c
1
2
3
3 Comments
Niels B.
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.
Thank you. So seq is actually an executable and not a language construct. It makes good sense.
fedorqui
Exactly! You explained it better than me :) See for example
$ which seq that returns something like /usr/bin/seq.Brace expansion occurs before variable expansion
http://www.gnu.org/software/bash/manual/bashref.html#Shell-Expansions
