In a bash script, what would $'\0' evaluate to and why?

Question

In various bash scripts I have come across the following: $'\0'

An example with some context:

while read -r -d $'\0' line; do
    echo "${line}"
done <<< "${some_variable}"

What does $'\0' return as its value? Or, stated slightly differently, what does $'\0' evaluate to and why?

It is possible that this has been answered elsewhere. I did search prior to posting but the limited number of characters or meaningful words in dollar-quote-slash-zero-quote makes it very hard to get results from stackoverflow search or google. So, if there are other duplicate questions, please allow some grace and link them from this question.

Answer 1

In bash, $'\0' is precisely the same as '': an empty string. There is absolutely no point in using the special Bash syntax in this case.

Bash strings are always NUL-terminated, so if you manage to insert a NUL into the middle of a string, it will terminate the string. In this case, the C-escape \0 is converted to a NUL character, which then acts as a string terminator.

The -d option of the read builtin (which defines a line-end character the input) expects a single character in its argument. It does not check if that character is the NUL character, so it will be equally happy using the NUL terminator of '' or the explicit NUL in $'\0' (which is also a NUL terminator, so it is probably no different). The effect, in either case, will be to read NUL-terminated data, as produced (for example) by find's -print0 option.

In the specific case of read -d '' line <<< "$var', it is impossible for $var to have an internal NUL character (for the reasons described above), so line will be set to the entire value of $var with leading and trailing whitespace removed. (As @mklement notes, this will not be apparent in the suggested code snippet, because read will have a non-zero exit status, even though the variable will have been set; read only returns success if the delimiter is actually found, and NUL cannot be part of a here-string.)

Note that there is a big difference between

read -d '' line

and

read -d'' line

The first one is correct. In the second one, the argument word passed to read is just -d, which means that the option will be the next argument (in this case, line). read -d$'\0' line will have identical behaviour; in either case, the space is necessary. (So, again, no need for the C-escape syntax).

Answer 2

To complement rici's helpful answer:

<sup>Note that this answer is about bash. ksh and zsh also support $'...' strings, but their behavior differs:
* zsh does create and preserve NUL (null bytes) with $'\0'.
* ksh, by contrast, has the same limitations as bash, and additionally interprets the first NUL in a command substitution's output as the string terminator (cuts off at the first NUL, whereas bash strips such NULs).</sup>

$'\0' is an ANSI C-quoted string that technically creates a NUL (0x0 byte), but effectively results in the empty (null) string (same as ''), because any NUL is interpreted as the (C-style) string terminator by Bash in the context of arguments and here-docs/here-strings.

As such, it is somewhat misleading to use $'\0' because it suggests that you can create a NUL this way, when you actually cannot:

• You cannot create NULs as part of a command argument or here-doc / here-string, and you cannot store NULs in a variable:

  • echo $'a\0b' | cat -v # -> 'a' - string terminated after 'a'
  • cat -v <<<$'a\0b' # -> 'a' - ditto

• In the context of command substitutions, by contrast, NULs are stripped:

  • echo "$(printf 'a\0b')" | cat -v # -> 'ab' - NUL is stripped

• However, you can pass NUL bytes via files and pipes.

  • printf 'a\0b' | cat -v # -> 'a^@b' - NUL is preserved, via stdout and pipe
  • Note that it is printf that is generating the NUL via its single-quoted argument whose escape sequences printf then interprets and writes to stdout. By contrast, if you used printf $'a\0b', bash would again interpret the NUL as the string terminator up front and pass only 'a' to printf.

If we examine the sample code, whose intent is to read the entire input at once, across lines (I've therefore changed line to content):

while read -r -d $'\0' content; do  # same as: `while read -r -d '' ...`
    echo "${content}"
done <<< "${some_variable}"

This will never enter the while loop body, because stdin input is provided by a here-string, which, as explained, cannot contain NULs.
Note that read actually does look for NULs with -d $'\0', even though $'\0' is effectively ''. ^{In other words: read by convention interprets the empty (null) string to mean NUL as -d's option-argument, because NUL itself cannot be specified for technical reasons.}

In the absence of an actual NUL in the input, read's exit code indicates failure, so the loop is never entered.

However, even in the absence of the delimiter, the value is read, so to make this code work with a here-string or here-doc, it must be modified as follows:

while read -r -d $'\0' content || [[ -n $content ]]; do
    echo "${content}"
done <<< "${some_variable}"

However, as @rici notes in a comment, with a single (multi-line) input string, there is no need to use while at all:

read -r -d $'\0' content <<< "${some_variable}"

This reads the entire content of $some_variable, while trimming leading and trailing whitespace (which is what read does with $IFS at its default value, $' \t\n').
@rici also points out that if such trimming weren't desired, a simple content=$some_variable would do.

Contrast this with input that actually contains NULs, in which case while is needed to process each NUL-separated token (but without the || [[ -n $<var> ]] clause); find -print0 outputs filenames separated by a NUL each):

while IFS= read -r -d $'\0' file; do
    echo "${file}"
done < <(find . -print0)

^{Note the use of IFS= read ... to suppress trimming of leading and trailing whitespace, which is undesired in this case, because input filenames must be preserved as-is.}