How to access command line arguments of the caller inside a function?

bash command-line-arguments argv
DonGar picture DonGar · Apr 29, 2010 · Viewed 151.4k times · Source

I'm attempting to write a function in bash that will access the scripts command line arguments, but they are replaced with the positional arguments to the function. Is there any way for the function to access the command line arguments if they aren't passed in explicitly?

# Demo function
function stuff {
  echo $0 $*
}

# Echo's the name of the script, but no command line arguments
stuff

# Echo's everything I want, but trying to avoid
stuff $*

Answer

stigi picture stigi · Apr 29, 2010

If you want to have your arguments C style (array of arguments + number of arguments) you can use $@ and $#.

$# gives you the number of arguments.
$@ gives you all arguments. You can turn this into an array by args=("$@").

So for example:

args=("$@")
echo $# arguments passed
echo ${args[0]} ${args[1]} ${args[2]}

Note that here ${args[0]} actually is the 1st argument and not the name of your script.

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