I know I can do this in Bash:

wc -l <<< "${string_variable}"

Basically, everything I found involved <<< Bash operator.

But in POSIX shell, <<< is undefined, and I have been unable to find an alternative approach for hours. I am quite sure there is a simple solution to this, but unfortunately, I didn't find it so far.


The simple answer is that wc -l <<< "${string_variable}" is a ksh/bash/zsh shortcut for printf "%s\n" "${string_variable}" | wc -l.

There are actually differences in the way <<< and a pipe work: <<< creates a temporary file that is passed as input to the command, whereas | creates a pipe. In bash and pdksh/mksh (but not in ksh93 or zsh), the command on right-hand side of the pipe runs in a subshell. But these differences don't matter in this particular case.

Note that in terms of counting lines, this assumes that the variable is not empty and does not end with a newline. Not ending with a newline is the case when the variable is the result of a command substitution, so you'll get the right result in most cases, but you'll get 1 for the empty string.

There are two differences between var=$(somecommand); wc -l <<<"$var" and somecommand | wc -l: using a command substitution and a temporary variable strips away blank lines at the end, forgets whether the last line of output ended in a newline or not (it always does if the command outputs a valid nonempty text file), and overcounts by one if the output is empty. If you want to both preserve the result and count lines, you can do it by appending some known text and stripping it off at the end:

output=$(somecommand; echo .)
line_count=$(($(printf "%s\n" "$output" | wc -l) - 1))
printf "The exact output is:\n%s" "${output%.}"
  • 1
    @Inian Keeping wc -l is exactly equivalent to the original: <<<$foo adds a newline to the value of $foo (even if $foo was empty). I explain in my answer why this may not have been what was wanted, but it's what was asked. – Gilles Nov 20 '18 at 7:20

Not conforming to shell built-ins, using external utilities like grep and awk with POSIX compliant options,


Doing with grep to match start of lines

printf '%s' "${string_variable}" | grep -c '^'

And with awk

printf '%s' "${string_variable}" | awk 'BEGIN { count=0 } NF { count++ } END { print count }'

Note that some of the GNU tools, especially, GNU grep does not respect POSIXLY_CORRECT=1 option to run the POSIX version of the tool. In grep the only behavior affected by setting the variable will be the difference in processing of the order of the command line flags. From the documentation (GNU grep manual), it seems that


If set, grep behaves as POSIX requires; otherwise, grep behaves more like other GNU programs. POSIX requires that options that follow file names must be treated as file names; by default, such options are permuted to the front of the operand list and are treated as options.

See How to use POSIXLY_CORRECT in grep?

  • 2
    Surely wc -l is still viable here? – Michael Homer Nov 20 '18 at 7:05
  • @MichaelHomer: From what I've observed, wc -l needs a proper newline delimited stream (having a trailing '\n` at the end to count properly). One cannot use a simple FIFO to use with printf, e.g. printf '%s' "${string_variable}" | wc -l might not work as expected but <<< would because of the trailing \n appended by the herestring – Inian Nov 20 '18 at 7:14
  • 1
    That was what printf '%s\n' was doing, before you took it out... – Michael Homer Nov 20 '18 at 7:18

The here-string <<< is pretty much a one-line version of the here-document <<. The former isn't a standard feature, but the latter is. You can use << too in this case. These should be equivalent:

wc -l <<< "$somevar"

wc -l << EOF

Though do note that both add an extra newline at the end of $somevar, e.g. this prints 6, even though the variable only has five lines :

wc -l <<< "$s"

With printf, you could decide if you want the additional newline or not:

printf "%s\n" "$s" | wc -l         # 6
printf "%s"   "$s" | wc -l         # 5

But then, do note that wc only counts complete lines (or the number of newline characters in the string). grep -c ^ should also count the final line fragment.

printf "%s" "$s" | wc -l           # 0 !

printf "%s" "$s" | grep -c ^       # 1

(Of course you could also count the lines entirely in the shell by using the ${var%...} expansion to remove them one at a time in a loop...)

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