I have a FreeBSD 9.3 RELEASE server and I am trying to write a shell script using sh. I am trying to get the length of a variable that is passed to a function in a shell script, like so:


myfunc() {
  echo $varlength

test="aaaaa aaaaa"
myfunc $test

The above script should return 11, which is the length of the test variable, but ${#1} does not seem to work. I have also tried many other ways to accomplish that but i can't figure it out.

varlength=$(expr ${#1})  does not work
varlength=$(${#1})  does not work
varlength=$(expr \( "X$1" : ".*" \) - 1)  does not work
varlength=$({#1})  does not work

and many many other attempts I made failed.

  • Can you try declaring the script as bash in the she-bang? Like this: #!/usr/bin/env bash – ryekayo Sep 30 '14 at 20:37
  • 1
    Instead of trying to do it with a shell built-in, you could always echo it and pipe it to the "wc -c" program to count the characters. Slightly more expensive of course. – BobDoolittle Sep 30 '14 at 20:37
  • 1
    ${#1} is defined in posix. I don't know what shell you're using, but it should work. What is happening when you run it? Error, no output, what? The result I expect is 5. – Patrick Sep 30 '14 at 20:43
  • The script is rather long and declaring it as bash would mean that I have to spend some time making sure it works, since it is written with sh in mind. Piping it to wc would require that I export the variable to a file and passing that file as an argument to wc which is indeed more expensive. ${#1} is indeed defined in POSIX. The result I get is also 5, but as you can see it is wrong. The actual length of the variable is 11. After 11 hours scripting, I might be missing something that is obvious to someone else, that is why I asked here. Thank you for your replies. – Zerotronic Zerotronicus Sep 30 '14 at 21:17
  • No, piping to wc does not require writing the variable to a file; just do varlength=$(echo "$1" | wc -c). (You'll want to subtract 1 from the result, because it will count the newline. Or, do echo -n "$1" or echo -e "$1\c" -- but not if $1 might contain backslashes.) – G-Man Oct 1 '14 at 21:51

because $test contains whitespace, when you say

myfunc $test # without quotes

your function receives > 1 argument. myfunc receives 2 arguments here, aaaaa and aaaaa.

You want this:

myfunc "$test" # with quotes

Rule of thumb: always quote your "$variables" unless you know exactly when and why not to.

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