using function without argument

Question

Let's take a little example:

$ cat fu.sh
#!/usr/local/bin/bash

function lsl () {
   ls -1
}

function grps () {
  lsl | grep fu.sh
}

echo "This file is: `grps`"

Let's run it to see it works as we want:

$ ./fu.sh
This file is: fu.sh

So - we have function lsl which called from second function grps. Also - we called function grps alone at the end of script.

Here is a question - is there anything "unusual", "unsafe" or may be "irritant" and "not Feng Shui" - call function without any option or argument passed to it?

Your question is definitely not bash-specific, but no - there is nothing wrong with such a function, as long as the function does exactly what you intended for it to do. — admdrew
– admdrew, Commented Nov 19, 2013 at 19:11
there's nothing wrong with it. if you need a function and you don't need arguments, that's what you're going to use. — smcg
– smcg, Commented Nov 19, 2013 at 19:12
On a related note, avoid using both the function keyword and brackets (cf. here). — Matthieu
– Matthieu, Commented Nov 19, 2013 at 19:13
and the function is not very useful: if you rename your script to "foo.sh" and forget to edit the inside, it will be incorrect. You cuold just do: function grps () { echo "$(basename "$0")" ; } instead — Olivier Dulac
– Olivier Dulac, Commented Nov 19, 2013 at 20:27

Vishal R · Accepted Answer · 2013-11-19 19:13:02Z

1

No, there is noting wrong in the script. It will work properly. You need to go through basics of shell scripting.

answered Nov 19, 2013 at 19:13

Vishal R

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