
/bin/bash - Proper Whitespace Handling - Whitespace Safety - End-of-Options Parameter Security
Supporting multiple command line parameters with spaces in wrapper scripts and End-of-Options Parameter (--
) for better security.
Safe ways to print[edit]
There is no safe usage of echo
, use printf '%s'
instead.
shellcheck bug reports:
Please note that printf
does not have a default format specifier, but treats the first positional parameter as the format. When the format is missing, the data is treated as if the format specifier is %b
. It is always recommended to be explicit on the format being used to avoid this mistake.
Normally, there is no need to interpret the escape sequences of a variable, therefore use the printf format specifier %s
when the data is not printed to the terminal:
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If you require escapes to be interpreted, interpret them on a per-need basis:
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Escapes that are already interpreted can be printed with %s
without making a difference:
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And this is why you should use stprint
when printing to the terminal, as it will sanitize unsafe characters while simply using printf '%s'
is not safe when escapes are already interpreted:
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Rule of thumb:
echo
: Never!printf
: Whenever the printed data is not used by a terminal.- Format
%b
: Only for trusted data - Format
%s
: With any data
- Format
stprint
: Whenever the printed data is used by a terminal.
Resources:
- https://github.com/anordal/shellharden/blob/master/how_to_do_things_safely_in_bash.md#echo--printf
- https://unix.stackexchange.com/questions/65803/why-is-printf-better-than-echo
- https://pubs.opengroup.org/onlinepubs/9799919799/utilities/echo.html
Bash Proper Whitespace Handling[edit]
- Quote variables
- Build parameters using arrays
- Enforce
nounset
#!/bin/bash ## https://yakking.branchable.com/posts/whitespace-safety/ set -eu app_user=user lib_dir="/tmp/test/lib/program with space/something spacy" main_app_dir="/tmp/test/home/user/folder with space/abc" mkdir -p "$lib_dir" mkdir -p "$main_app_dir" declare -a cmd cmd+=("cp") cmd+=("-r") cmd+=("${lib_dir}") cmd+=("${main_app_dir}/") "${cmd[@]}"
Why nounset
[edit]
Because it is better to be explicit if a variable should be empty or not:
rm -f "/$UNSET_VAR"
Will return:
rm: cannot remove '/': Is a directory
Setting UNSET_VAR=""
would not fix this issue, but that is another problem, checking if every used variable can be empty or not.
POSIX array[edit]
On a POSIX shell, there is one array, the $@
, which have different scopes by function or main script. You can build it with set --
:
Add items to an array:
set -- a b c
Add items to the beginning or end of the array:
set -- b set -- a "$@" c
Use of End-of-Options Parameter (--)[edit]
The end-of-options parameter "--
" is crucial because otherwise inputs might be mistaken for command options. This might even be a security risk. Here are examples using the `sponge` command:
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Result: OK. This works because "testfilename
" doesn't look like an option.
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Result: Fail. The command interprets "--testfilename
" as a series of options:
sponge: invalid option -- '-' sponge: invalid option -- 't' sponge: invalid option -- 'e' ...
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Result: OK. The `--
` signals that "--testfilename
" is a filename, not an option.
Conclusion:
- The "
--
" parameter marks the end of command options. - Use "
--
" at the end of a command to prevent misinterpretation. - This technique is applicable to many Unix/Linux commands, not just
sponge
.
nounset - Check if Variable Exists[edit]
#!/bin/bash set -x set -e set -o nounset ## Enable for testing. #unset HOME if [ -z "${HOME+x}" ]; then printf '%s\n' "Error: HOME is not set." >&2 fi printf '%s' "$HOME"
Safely Using Find with End-Of-Options[edit]
Example:
Note: Variable could be different. Could be for example --/usr
.
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Of if safe_echo_nonewline
is available from helper-scripts.
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misc[edit]
base_name="${file_name##*/}" file_extension="${base_name##*.}"
coding style[edit]
- use:
- shellcheck
safe-rm
scurl
str_replace
append-once
- use
${variable}
style - use shell options
set -o errexit set -o nounset set -o errtrace set -o pipefail
- do not use:
which
, usecommand -v
instead. This is becausewhich
is an external binary (minor reason) and produces stdout if a binary was found, which can be slightly confusing (major reason).
See Also[edit]

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