How to use positional parameters with “bash -c” command?

Question

I\'d like to know the best way of using positional parameters when using the command bash -c.

The man pages indicates for the -c option tha

Answer 1

bash -c 'printf "%s %s %s\n" $0 $1 $2' param1 param2 param3

The above works but many would consider it a bad habit to get into. If you were to copy code from the -c string to a script, it would fail. Similarly, if you were to copy code from a script to a -c string, it would fail.

By contrast, with the following form, $1 means the same thing in the -c string that it would mean in a script or shell function:

bash -c 'printf "%s %s %s\n" $1 $2 $3' _ param1 param2 param3

Consistency of programming style reduces bugs.

The shell treats $0 differently

One customarily refers to all of a script's arguments with $@ or $*. Note that these variables do not include $0:

$ bash -c 'echo "$*"' param1 param2 param3
param2 param3
$ bash -c 'echo "$@"' param1 param2 param3
param2 param3

$0 is the program name

In regular scripts, $0 is the name of the script. Consequently, when using bash -c, some people prefer to use some meaningful name for the $0 parameter, such as:

bash -c 'printf "%s %s %s\n" $1 $2 $3' bash param1 param2 param3

Or:

bash -c 'printf "%s %s %s\n" $1 $2 $3' printer param1 param2 param3

This approach has a clear advantage if the -c string generates an error. For example, consider this script:

$ cat script.sh
#!/bin/bash
bash -c 'grepp misspelling "$1"' BadPgm file.txt

If we run the script, the following output is produced:

$ ./script.sh 
BadPgm: grepp: command not found

This identifies the source of the error as the command in the bash -c string.