What does "$#" mean in bash?

Question

I have a script with this:

login {
    # checking parameters -> if not ok print error and exit script
    if [ $# -lt 2 ] || [ $1 == '' ] || [ $2 == '' ]; then
        echo "Please check the needed options (username and password)"
        echo ""
        echo "For further Information see Section 13"
        echo ""
        echo "Press any key to exit"
        read
        exit
    fi

  } # /login

But I really dont know what the $# means on the third line.

@devnull Makes sense since this function is called like this: ```login $USER $PASS``` Thanks! — dominique120, Apr 11 '14 at 18:52
Also http://stackoverflow.com/questions/11323847/what-do-and-stand-for-in-shell-script/ — devnull, Apr 11 '14 at 18:59
@devnull dupe, yes, but this question has the *far* superior title. Editing linked question now. — kojiro, Apr 11 '14 at 19:05
@kojiro Symbols and special characters cannot be searched anyways. (And marking a question as duplicate doesn't _restrict_ anyone from posting answers anyways.) — devnull, Apr 11 '14 at 19:07
@devnull I think it's very relevant for the human reading the summary, though. — kojiro, Apr 11 '14 at 19:08

score 10 · Accepted Answer · answered Apr 11 '14 at 18:54

The pound sign counts things.

If it's just $#, it's the number of positional parameters, like $1, $2, $3. (Not counting $0, mind you.)
If it's ${#var}, it's the number of characters in the expansion of the parameter. (String length)
If it's ${#var[@]}, it's the number of elements in the array. Since bash arrays are sparse, this can be different from the index of the last element plus one.

score 2 · Answer 2 · answered Apr 11 '14 at 18:51

2

It's the number of arguments passed.

You can read it here, search for "Detecting command line arguments"

answered Apr 11 '14 at 18:51

Andreas Wederbrand

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What does "$#" mean in bash?

2 Answers2

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