sanjay kumar via Bug reports for the GNU Bourne Again SHell
<bug-bash@gnu.org> writes:
Somehow all the line breaks in your report were lost. As I reconstruct
it, you wrote:
> cat test.sh
#!/bin/bash
echo "argv[0] = ${0}"
echo "argv[1] = ${1}"
echo "argv[2] = ${2}"
echo "count = ${#}"
===================
> bash -c ./test.sh abc def
argv[0] = ./test.sh
argv[1] =
argv[2] =
count = 0
===================
> bash ./test.sh abc def
argv[0] = ./test.sh
argv[1] = abc
argv[2] = def
count = 2
===================
> bash -version
GNU bash, version 5.0.17(1)-release (x86_64-pc-linux-gnu)
Copyright (C) 2019 Free Software Foundation, Inc.
License GPLv3+: GNU GPL version 3 or later
<http://gnu.org/licenses/gpl.html>
This is free software; you are free to change and redistribute it.There is
NO WARRANTY, to the extent permitted by law.
The problem is a little subtle. More exactly, the reason that the first
example didn't produce an obvious error is a little subtle. The
definition of "-c" is "commands are read from the first non-option
argument command_string. If there are arguments after the
command_string, the first argument is assigned to $0 and any remaining
arguments are assigned to the positional parameters."
So the first step in processing "bash -c ./test.sh abc def" is that Bash
assigns "abc" to $0 and "def" to $1. Then Bash starts processing
commands from the string "./test.sh". Assuming that ./test.sh has
execute permission, "./test.sh" is a valid command line, and Bash then
executes ./test.sh.
That command, however, has no arguments, so ./test.sh produces the
output that you see.
Dale
