Using $# in bash loops

问题

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}

回答1:

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

回答2:

Brace expansion occurs before variable expansion

http://www.gnu.org/software/bash/manual/bashref.html#Shell-Expansions

来源：https://stackoverflow.com/questions/18979681/using-in-bash-loops