How do I escape a series of backslashes in a bash printf?

Question

The following script yielded an unexpected output:

printf "escaped slash: \\ \n"
printf "2 escaped slashes: \\\\ \n"
printf "3 escaped slashes: \\\\\\ \n"
printf "4 escaped slashes: \\\\\\\\ \n"

Run as a bash script under Ubuntu 14, I see:

escaped slash: \
2 escaped slashes: \ 
3 escaped slashes: \\ 
4 escaped slashes: \\

Err.. what?

Eugeniu Rosca · Accepted Answer · 2015-07-04T20:07:43.857

12

Assuming that printf FORMAT string is surrounded by double quotes, printf takes one additional level of expansion, compared to e.g. echo (both being shell builtin commands).

What you expect from printf can actually be achieved using single quotes:

printf '1 escaped slash:   \\ \n'
printf '2 escaped slashes: \\\\ \n'
printf '3 escaped slashes: \\\\\\ \n'
printf '4 escaped slashes: \\\\\\\\ \n'

outputs:

1 escaped slash:   \
2 escaped slashes: \\
3 escaped slashes: \\\
4 escaped slashes: \\\\

edited Jul 04 '15 at 20:07

answered Jul 04 '15 at 20:00

Eugeniu Rosca

5,177
16
45

Note this works because you prevent `bash` from removing any backslashes before `printf` sees the argument. – chepner Jul 05 '15 at 02:45
@chepner: I agree. It seems the order of evaluation is: 1) `bash` expands the `FORMAT` string *(only if surrounded by double quotes!)*. 2) `printf` combines the `ARGUMENTS` and the `FORMAT` into one string. 3) The final string is expanded again (i.e. \\ becomes \ ). 4) The result is finally printed. – Eugeniu Rosca Jul 05 '15 at 12:03
Close; `printf` only process the `FORMAT` before inserting the arguments; compare `printf '\n'` with `printf '%s' '\n'`. – chepner Jul 05 '15 at 13:01

Cyrus · Answer 2 · 2015-07-04T20:18:34.720

5

printf is a bash builtin. Look at help printf:

printf [-v var] format [arguments]
      Formats and prints ARGUMENTS under control of the FORMAT.

You should pass the format and the argument. So add the format "%s\n" before the argument:

printf "%s\n" "escaped slash: \\"
printf "%s\n" "2 escaped slashes: \\\\"
printf "%s\n" "3 escaped slashes: \\\\\\"
printf "%s\n" "4 escaped slashes: \\\\\\\\"

Output:

escaped slash: \ 
2 escaped slashes: \\ 
3 escaped slashes: \\\ 
4 escaped slashes: \\\\

edited Jul 04 '15 at 20:18

answered Jul 04 '15 at 19:58

Cyrus

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14
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@Yuri G.: Thank you for your additional description. – Cyrus Jul 04 '15 at 20:15
Note this works because you avoid `printf` applying another round of interpretation of backslashes. – chepner Jul 05 '15 at 02:45

score 1 · Answer 3 · answered Apr 25 '22 at 03:21

An additional note on what Cyrus said:

If you quote the printf's ARGUMENT with single quotes, you should save a lot of backslashes. For example,

printf "%s\n" 'escaped slash: \'
printf "%s\n" '2 escaped slashes: \\'
printf "%s\n" '3 escaped slashes: \\\'
printf "%s\n" '4 escaped slashes: \\\\'

outputs

1 escaped slash:   \
2 escaped slashes: \\
3 escaped slashes: \\\
4 escaped slashes: \\\\

How do I escape a series of backslashes in a bash printf?

3 Answers3