bash -c in linux? What does it actually do?

[shivajikobardan]

Before you ask me to RTFM, Quoting from the manpages:

Quote

-c string If the -c option is present, then commands are read from string. If there are arguments after the string, they are assigned to the positional parameters, starting with $0.

But it's not very clear. I've read almost all stackoverflow and stackexchange questions about it and they've just complicated it for me.

I want to know a real life use case of it. And please upvote the correct answer so that I can know it.

bash -c 'echo $SHELL $HOME $USER'
env -i bash -c 'echo $SHELL $HOME $USER'

This is an example scenario where bash -c has been used in my tutorial that I'm following. I want to know its deep meaning and mainly the need of bash -c. I asked chatgpt but it was telling me  unintelligble stuffs.

 

[Kilrah]

It will ensure bash is the shell that executes this command instead of whatever shell the user is using. 

[shivajikobardan]

I think bash alone is what you mean. bash -c is more complicated than that. 

[manikyath]

1 minute ago, shivajikobardan said:

I think bash alone is what you mean. bash -c is more complicated than that. 

'bash' starts a bash shell.

'bash -c ***' starts a bash shell, executing command ***.

 

so:

7 minutes ago, shivajikobardan said:

'echo $SHELL $HOME $USER'

executes the command

 

and

7 minutes ago, shivajikobardan said:

bash -c 'echo $SHELL $HOME $USER'

executes the command in bash

[shivajikobardan]

Is it that simple? Then why's the whole of internet confused on this issue?

[manikyath]

45 minutes ago, shivajikobardan said:

Is it that simple? Then why's the whole of internet confused on this issue?

you appear to be the first one i've ever encountered that's confused about it.

[Eigenvektor]

5 hours ago, shivajikobardan said:

-c string — If the -c option is present, then commands are read from string.

This simply says, if you add a "-c" then the (first) string that follows the "-c" is interpreted as a command. So e.g.

bash -c "echo Hello"

will simply use the Bash shell to execute "echo Hello", causing Hello to be printed.

 

Quote

If there are arguments after the string, they are assigned to the positional parameters, starting with $0.

This says how additional arguments are treated, meaning if you add more than one string after the "-c". Additional arguments are used to replace $0, $1, $2, … if they appear in the command.

 

For example

bash -c 'echo $0 $1' "Hello" "Test"

will run the command "echo $0 $1", replacing "$0" with "Hello" and replacing "$1" with "Test". That's all there is to it.

 

Note however, how I used single quotes around the first string, rather than double quotes. This prevents my native shell from replacing $0 and $1, and instead passes them to its command (i.e. bash) verbatim. If I use double quotes, this is printed instead (using zsh here):

zsh

This means my native shell replaces $0 and $1, causing "echo $0 $1" to turn into "echo zsh" before the string is passed to bash (via -c), which then executes "echo zsh", causing zsh to be printed.