bash command to print string in unambiguous form

Question

I am wondering whether there exists a command in bash to print a string in a way exposes the special character it contains.

For example, suppose that a=$'\a\0\b\e'; does there exists a function to print \a\0\b\e literally from $a?

The closest I have got so far is by using the l command from sed:

echo "$a"  | sed -n 'l'

which returns \a\000\b\033$, but the notation is different from that inside $'', and it doesn't work if the string contains newlines.

As variant the cat -A. It shows non-printable characters, but in the another form. echo $'\t\b\e\a' | cat -A output: ^I^H^[^G — MiniMax
– MiniMax, Commented Oct 15, 2017 at 20:35
Also, od -c: echo $'\t\b\e\a' | od -c. Output: 0000000 \t \b 033 \a \n. — MiniMax
– MiniMax, Commented Oct 15, 2017 at 20:50
A slightly closer option is printf %s "$a" | hexdump -c, which gives me … \a \0 \b 033 …, i.e. just mangles the \e. — Sparhawk
– Sparhawk, Commented Oct 15, 2017 at 21:16
Can a string contain \0? I don't think so. @Sparhawk, what is your bash version? — glenn jackman
– glenn jackman, Commented Oct 16, 2017 at 0:48

Hauke Laging · Accepted Answer · 2017-10-15 21:31:10Z

5

var=$'a b \10 c'
printf %q "$var"
    $'a b \b c'

This works in bash. I do not know how compatible this is.

answered Oct 15, 2017 at 21:31

Hauke Laging

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