I have a bash variable which is stored as

var="\"abc\""

So, when i locally print this variable it gives me proper output(with double quotes)

echo "$var"
"abc"

What, I want to do is print it using 'bash -c' option.. but when I do the same, it prints only value without double quotes.

bash -c "echo \"$var\""
abc
bash -c "echo $var"
abc

Can anyone help me on how to preserve the double quotes in my string, when I use it in 'bash -c'. And what does -c actually mean?

CodePudding user response：

In the statement

bash -c "echo $var"

the following happens:

(1) var is expanded (resulting into "abc" including the quotes).

(2) A bash child process is invoked, receiving as first parameter -c and as second paramter echo "abc"

(3) This child process runs the command, i.e. echo "abc", and according to the rules about quote removal, this is equivalent to echo abc, and you don't see any quotes in the output.

You may be tempted to do a

bash -c 'echo "$var"'

instead. In this case, the following happens:

(1) A bash child process is invoked, receiving as first parameter -c and as second paramter echo "$var". Note that $var is not expanded yet, because it is between single quotes.

(2) This child process runs the command, i.e. echo "$var". However, since we are in a child process, the variable var does not exist, and the command is equivalent to echo "", i.e. only a newline is printed.

You can however combine both solution by doing a

export var
bash -c 'echo "$var"'

In this case, var is made available to the child processes of your script, and you will see the quotes being printed.

CodePudding user response：

From info bash

-c If the -c option is present, then 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 -c option executes the commands within the quotes.

To preserve your double quotes while calling via bash -c, you would need to quote the variable seperately.

$ bash -c "echo '"$var"'"
"abc"