Quoted printing of shell variables

Question

I would like to print the contents of a variable in such a way, that the printed output could be pasted directly into a shell to get the original content of the variable.

This is trivial if the content doesn't contain any special characters, esp no quotes. e.g.

$ x=foo
$ echo x=${x}
x=foo

In the above example i can take the output (x=foo and paste it into a new terminal to assign foo to x).

If the variable content contains spaces, things get a bit trickier, but it's still easy:

$ x="foo bar"
$ echo x=\"${x}\"
x="foo bar"

Now trouble starts, if the variable is allowed to contain any character, e.g.:

$ x=foo\"bar\'baz
$echo ${x}
foo"bar'baz
$ echo x=\"${x}\"
x="foo"bar'baz"
$ x="foo"bar'baz"
>

(and the terminal hangs, waiting for me to close the unbalanced ")

What I would have expected was an output like the following:

x=foo\"bar\'baz

How would I do that, preferably POSIX compliant (but if it cannot be helped, bash only)?

Answer 1

Use declare -p:

x=foo\"bar\'baz
declare -p x

declare -- x="foo\"bar'baz"

Even with an array:

arr=(a 'foo' 'foo bar' 123)
declare -p arr

declare -a arr='([0]="a" [1]="foo" [2]="foo bar" [3]="123")'

The -p option will display the attributes and values of each variable name that can be used to directly set the variable again.

---

As per comment, doing this in POSIX without help of declare -p would be:

set | grep '^x='

Answer 2

You could use the printf command with its %q flag it supports to print the associated argument shell-quoted

x=foo\"bar\'baz

printf 'x=%q' "$x"
x=foo\"bar\'baz

Again though %q is not a POSIX extension but added to most recent versions of bash shell. If this is what you where looking for look up this to see POSIX sh equivalent for Bash's printf %q